\section{Spectral Theory}
\label{sec:spectral-theory-c*-algebras}

\subsection{Spectral Theory in Banach Algebras}

Throughout this section, let $A$ be a unital Banach algebra. We actually do not need to require $A$ to be a $C^*$-algebra for the foundations of spectral theory, but the existence of a unit is essential. We will adopt the shorthand of writing $\lambda$ for $\lambda 1$ for all $\lambda \in \C$.

\begin{definition}
Given $a \in A$, we define the \emph{resolvent set} of $a$ as
\begin{equation}
\rho(a) = \qty{\lambda \in \C: \lambda - a \in A^\times}
\end{equation}
An element of $\rho(a)$ is called a \emph{regular value} of $a$. If $\rho(a) \neq \varnothing$, the map $r_a:\rho(a) \rightarrow A$ defined as
\begin{equation}
r_a(\lambda) = (\lambda - a)^{-1}
\end{equation}
is called the \emph{resolvent} of $a$.

Likewise, we define the \emph{spectrum} of $a$ as
\begin{equation}
\sigma(a) = \qty{\lambda \in \C: \lambda - a \notin A^\times} = \C \setminus \rho(a).
\end{equation}
An element of $\sigma(a)$ is called a \emph{spectral value} of $a$. If $\sigma(a) \neq \varnothing$, we define the \emph{spectral radius} of $a$ as
\begin{equation}
r(a) = \sup_{\lambda \in \sigma(a)} \abs{\lambda}.
\end{equation}
\end{definition}

A priori we do not know that $\rho(a)$ or $\sigma(a)$ is nonempty. The following exposition will establish that in fact both of them are nonempty, so the resolvent $r_a(\lambda)$ and the spectral radius $r(a)$ are always defined.

\begin{theorem}\label{thm:Neumann_series}
Let $a \in A$. If $\lambda \in \C$ such that $\norm{a} < \abs{\lambda}$, then the \emph{Neumann series} $\sum_{n=0}^\infty (a/\lambda)^n$ converges, $\lambda \in \rho(a)$, and
\begin{equation}
r_a(\lambda) = \frac{1}{\lambda}\sum_{n=0}^\infty \qty(\frac{a}{\lambda})^n.
\end{equation}
Furthermore,
\begin{equation}\label{eq:resolvent_norm_bound}
\norm{r_a(\lambda)} \leq \frac{1}{\abs{\lambda} - \norm{a}}.
\end{equation}
\end{theorem}

\begin{proof}
We begin by showing that the sequence of partial sums is a Cauchy sequence. Given $M,N \in \N$ with $M < N$, we have
\begin{equation}
\norm{\sum_{n=0}^N \qty(\frac{a}{\lambda})^n - \sum_{n = 0}^M \qty(\frac{a}{\lambda})^n} = \norm{\sum_{n=M+1}^N \qty(\frac{a}{\lambda})^n} \leq \sum_{n=M+1}^N \norm{\qty(\frac{a}{\lambda})^n} \leq \sum_{n=M+1}^N \qty(\frac{\norm{a}}{\abs{\lambda}})^n,
\end{equation}
where we have used submultiplicativity in the last step. Since $\norm{a}/\abs{\lambda} < 1$, the rightmost expression can be made arbitrarily small by taking $M$ and $N$ to be large. Thus, the sequence of partial sums is Cauchy, hence convergent, since $A$ is complete.

Now, for any $N \in \N$, we have
\begin{equation}
(\lambda - a)\qty[\frac{1}{\lambda}\sum_{n=0}^N \qty(\frac{a}{\lambda})^n] = \qty[\frac{1}{\lambda}\sum_{n=0}^N \qty(\frac{a}{\lambda})^n](\lambda - a) = 1 - \qty(\frac{a}{\lambda})^{N+1}
\end{equation}
Submultiplicativity and $\norm{a/\lambda} < 1$ imply $(a/\lambda)^n \rightarrow 0$. Thus, taking the limit as $N \rightarrow \infty$ of the above line yields
\begin{equation}
(\lambda - a)\qty[\frac{1}{\lambda}\sum_{n=0}^\infty \qty(\frac{a}{\lambda})^n] = \qty[\frac{1}{\lambda}\sum_{n=0}^\infty \qty(\frac{a}{\lambda})^n](\lambda - a) = 1,
\end{equation}
as desired.

Finally, we note that for $N \in \N$, using the formula for a geometric series yields
\begin{equation}
\norm{\frac{1}{\lambda}\sum_{n=0}^N \qty(\frac{a}{\lambda})^n}  \leq \frac{1}{\abs{\lambda}}\sum_{n=0}^N \qty(\frac{\norm{a}}{\abs{\lambda}})^n \leq  \frac{1}{\abs{\lambda}}\frac{1}{1 - \norm{a}/\abs{\lambda}} = \frac{1}{\abs{\lambda} - \norm{a}}.
\end{equation}
Taking the limit as $N \rightarrow \infty$ yields \eqref{eq:resolvent_norm_bound}.
\end{proof}

The following corollary rephrases some of the key points of the above theorem.

\begin{corollary}
Given $a \in A$, the resolvent set $\rho(a)$ is nonempty and
\begin{equation}
r(a) \leq \norm{a}.
\end{equation}
\end{corollary}

\begin{corollary}
Given $a \in A$, the resolvent $r_a:\rho(a) \rightarrow A$ is continuous.
\end{corollary}

\begin{proof}
Since $\rho(a) \neq \varnothing$, the resolvent is defined. Continuity then follows from continuity of addition, scalar multiplication, and inversion.
\end{proof}

\begin{corollary}\label{cor:resolvent_is_open}
Let $a \in A$ and $\lambda_0 \in \rho(a)$. If $\abs{\lambda - \lambda_0} < \norm{r_a(\lambda_0)}^{-1}$, then $\lambda \in \rho(a)$ and
\begin{equation}
r_a(\lambda) = \sum_{n=0}^\infty (\lambda_0 - \lambda)^n r_a(\lambda_0)^{n+1}.
\end{equation}
In particular, it follows that
\begin{equation}
B_{\norm{r_a(\lambda_0)}^{-1}}(\lambda_0) \subset \rho(a)
\end{equation}
for all $\lambda_0 \in \rho(a)$, so $\rho(a)$ is open in $\C$.
\end{corollary}

\begin{proof}
Let $\abs{\lambda - \lambda_0} < \norm{r_a(\lambda_0)}^{-1}$. Then $\norm{(\lambda_0 - \lambda)r_a(\lambda_0)} < 1$, so $1 - (\lambda_0 - \lambda)r_a(\lambda_0)$ is invertible by Theorem \ref{thm:Neumann_series}. Since $\lambda_0 - a$ is invertible, the product
\begin{equation}
(\lambda_0 - a)\qty[1 - (\lambda_0 - \lambda)r_a(\lambda_0)] = \lambda_0 - a - (\lambda_0 - \lambda) = \lambda - a
\end{equation}
is also invertible, so $\lambda \in \rho(a)$. Furthermore, using the Neumann series for the inverse of $1 - (\lambda_0 - \lambda)r_a(\lambda_0)$, we obtain
\begin{equation}
\begin{aligned}
r_a(\lambda) &= \qty[1 - (\lambda_0 - \lambda)r_a(\lambda_0)]^{-1} r_a(\lambda_0)\\
&= \qty[\sum_{n=0}^\infty (\lambda_0 - \lambda)^n r_a(\lambda_0)^n]r_a(\lambda_0)\\
&= \sum_{n=0}^\infty (\lambda_0 - \lambda)^n r_a(\lambda_0)^{n+1},
\end{aligned}
\end{equation}
as desired.
\end{proof}

In a similar vein, we have the following Corollary.

\begin{corollary}
The set $A^\times$ is open in $A$.
\end{corollary}

\begin{proof}
Let $a \in A^\times$ and let $b \in A$ such that $\norm{b - a} < \norm{a^{-1}}^{-1}$. This implies that
\begin{equation}
\norm{1 - ba^{-1}} \leq \norm{a - b}\norm{a^{-1}} < 1,
\end{equation}
so $1 - (1 - ba^{-1}) = ba^{-1}$ is invertible, which implies $b$ is invertible.
\end{proof}

Having studied a few properties of the resolvent set, we now turn to the spectrum. In particular, we want to show that the spectrum is nonempty. We will be aided by a few algebraic properties of the resolvent. We denote the commutator of two elements $a,b \in A$ by $[a,b] = ab - ba$.

\begin{lemma}
Let $a \in A$. For all $\lambda, \mu \in \rho(a)$, we have:
	\begin{enumerate}
		\item[\tn{(i)}] $[a, r_a(\lambda)] = 0$,
		\item[\tn{(ii)}] $r_a(\mu) - r_a(\lambda) = (\lambda - \mu)r_a(\mu)r_a(\lambda)$
		\item[\tn{(iii)}] $[r_a(\lambda), r_a(\mu)] = 0$.
	\end{enumerate}
\end{lemma}

\begin{proof}
(i). It is clear that $[\lambda - a, a] = 0$. So, since $r_a(\lambda) = (\lambda - a)^{-1}$, we have
\begin{equation}
0 = r_a(\lambda)[\lambda - a, a]r_a(\lambda) = r_a(\lambda)a -ar_a(\lambda) = [r_a(\lambda), a].
\end{equation}

(ii). We compute
\begin{equation}
\begin{aligned}
\qty[r_a(\mu) - r_a(\lambda)](\lambda - a)(\mu - a) &= r_a(\mu)(\lambda - a)(\mu - a) - (\mu - a)\\
&= (\lambda - a) - (\mu - a)\\
&= \lambda - \mu.
\end{aligned}
\end{equation}
We used the fact that $[r_a(\mu), \lambda - a] = 0$ in the second step. Multiplying by $r_a(\mu)r_a(\lambda)$ on the left now yields the desired result.

(iii). If $\lambda = \mu$, the result is trivial. If $\lambda \neq \mu$, we may divide both sides of (ii) by $\lambda - \mu$ to obtain
\begin{equation}
r_a(\mu) r_a(\lambda) = \frac{r_a(\mu) - r_a(\lambda)}{\lambda - \mu}.
\end{equation}
The right hand side is invariant under exchange of $\mu$ and $\lambda$, so the result follows.
\end{proof}

\begin{lemma}\label{lem:resolvent_derivative}
Let $a \in A$. If $f$ is in the continuous dual $A^*$, then $f \circ r_a :\rho(a) \rightarrow \C$ is holomorphic %and
%\begin{equation}\label{eq:resolvent_derivative}
%(f \circ r_a)^{(n)}(\lambda) = (-1)^n n! f\qty(r_a(\lambda)^{n+1}).
%\end{equation}
\end{lemma}

\begin{proof}
Let $\lambda \in \rho(a)$. For any $\mu \in \rho(a)$, $\mu \neq \lambda$, we have
\begin{equation}
\frac{f(r_a(\mu)) - f(r_a(\lambda))}{\mu - \lambda} = f\qty(\frac{r_a(\mu) - r_a(\lambda)}{\mu - \lambda}) = -f\qty(r_a(\mu)r_a(\lambda)).
\end{equation}
Since $f$ and $r_a$ are continuous, the limit of the above as $\mu \rightarrow \lambda$ exists and is
\begin{equation}
(f \circ r_a)'(\lambda)  = -f\qty(r_a(\lambda)^2).
\end{equation}
This proves that $f \circ r_a$ is holomorphic. % and proves \eqref{eq:resolvent_derivative} for $n = 1$.
%
%Suppose \eqref{eq:resolvent_derivative} is true for some $n \in \N$. Then for distinct numbers $\mu, \lambda \in \C$, we have
%\begin{equation}
%\begin{aligned}
%\frac{(f \circ r_a)^{(n)}(\mu) - (f \circ r_a)^{(n)}(\lambda)}{\mu - \lambda} &= (-1)^n n! f\qty(\frac{r_a(\mu)^{n+1} - r_a(\lambda)^{n+1}}{\mu - \lambda})\\
%&= (-1)^n n! f\qty[\qty(\frac{r_a(\mu) - r_a(\lambda)}{\mu - \lambda})\sum_{k=0}^n r_a(\mu)^{n-k}r_a(\lambda)^k]\\
%&= (-1)^{n+1} n! f\qty(r_a(\mu)r_a(\lambda)\sum_{k=0}^n r_a(\mu)^{n-k}r_a(\lambda)^k)
%\end{aligned}
%\end{equation}
%where the expansion in the second step relies on the fact that $[r_a(\mu),r_a(\lambda)] = 0$. Taking the limit as $\mu \rightarrow \lambda$ yields
%\begin{equation}
%(f \circ r_a)^{(n+1)}(\lambda) = (-1)^{n+1} (n+1)! f\qty(r_a(\lambda)^{n+2}).
%\end{equation}
%This concludes the inductive step and completes the proof.
\end{proof}

\begin{corollary}
Given $a \in A$, the spectrum $\sigma(a)$ is nonempty and compact.
\end{corollary}

\begin{proof}
We know $\sigma(a)$ is closed and bounded since $\rho(a)$ is open and $r(a) \leq \norm{a}$, so it just remains to show that $\sigma(a)$ is nonempty. If $\sigma(a) = \varnothing$, then $\rho(a) = \C$, so $f \circ r_a$ is entire for all $f \in A^*$. Since $f$ is bounded, we have
\begin{equation}
\abs{(f \circ r_a)(\lambda)} \leq \norm{f} \norm{r_a(\lambda)}.
\end{equation}
Furthermore, since $\lambda \mapsto \norm{r_a(\lambda)}$ is continuous, it is bounded by a constant for $\abs{\lambda} \leq 1 + \norm{a}$ by the extreme value theorem. For $\abs{\lambda} > 1 + \norm{a}$, we have a bound from the Neumann series:
\begin{equation}
\norm{r_a(\lambda)} \leq \frac{1}{\abs{\lambda} - \norm{a}} \leq 1.
\end{equation}
Thus, $f \circ r_a$ is bounded and entire, so by Liouville's theorem it is constant. But if $\mu \neq \lambda$, then $r_a(\mu) \neq r_a(\lambda)$, for otherwise we would have
\begin{equation}
0 = r_a(\mu) - r_a(\lambda) = (\lambda - \mu)r_a(\mu)r_a(\lambda),
\end{equation}
which would imply that $r_a(\mu) = r_a(\lambda) = 0$, but $0$ is not invertible. By the Hahn-Banach theorem, there must be some $f \in A^*$ such that $f \circ r_a$ is not constant, which is a contradiction. Therefore $\rho(a) \neq \C$.
\end{proof}

In fact, using Lemma \ref{lem:resolvent_derivative}, we can say exactly what the spectral radius $\rho(a)$ is.

\begin{theorem}\label{thm:spectral_radius_formula}
Let $a \in A$. The spectral radius is given by
\begin{equation}
r(a) = \lim_{n \rightarrow \infty} \norm{a^n}^{1/n},
\end{equation}
where the limit on the right hand side is guaranteed to exist.
\end{theorem}

\begin{proof}
The result is trivial if $a = 0$, so assume $a \neq 0$. We will show that
\begin{equation}\label{eq:spectral_radius_lim_inequality}
\limsup \norm{a^n}^{1/n} \leq r(a) \leq \liminf \norm{a^n}^{1/n},
\end{equation}
which immediately yields the result.

Suppose $\lambda \in \sigma(a)$. If $\lambda^n \in \rho(a^n)$ for some $n \in \N$, then
\begin{equation}
\lambda^n - a^n = (\lambda - a) \qty(\lambda^{n-1} + \lambda^{n-2}a + \cdots + \lambda a^{n-2} + a^{n-1})
\end{equation}
is invertible. Let $b$ be the rightmost term in parentheses and note that $b$ commutes with $\lambda - a$. But then
\begin{equation}
(\lambda - a)b(\lambda^n - a^n)^{-1} = 1 = (\lambda^n - a^n)^{-1}b (\lambda - a),
\end{equation}
so $\lambda - a$ has a left inverse and a right inverse. Hence $\lambda - a$ has a two-sided inverse, so $\lambda \in \rho(a)$, which is a contradiction. Therefore $\lambda^n \in \sigma(a^n)$. Since $\rho(a^n) \leq \norm{a^n}$, we see that
\begin{equation}
\abs{\lambda} = \qty(\abs{\lambda}^n)^{1/n} \leq \norm{a^n}^{1/n}.
\end{equation}
This is true for all $\lambda \in \rho(a)$ and $n \in \N$, so
\begin{equation}
\rho(a) \leq \inf_{n \in \N} \norm{a^n}^{1/n} \leq \liminf \norm{a^n}^{1/n}.
\end{equation}

To prove the other half of \eqref{eq:spectral_radius_lim_inequality}, suppose $\rho(a) > 0$, let $f \in A^*$, and consider the function $g:B_{\rho(a)^{-1}}(0) \rightarrow \C$ defined as
\begin{equation}
g(\lambda) = \begin{cases} \hfil f(r_a(\lambda^{-1})) &: \lambda \neq 0\\ \hfil 0 &: \lambda = 0\end{cases}
\end{equation}
If $\rho(a) = 0$, we may define $g$ in this way on all of $\C$. This function is holomorphic on the deleted disk $B_{\rho(a)^{-1}}(0) \setminus \qty{0}$ by Lemma \ref{lem:resolvent_derivative} and the fact that $\lambda \mapsto \lambda^{-1}$ is holomorphic on $\C \setminus \qty{0}$. Furthermore, for $\abs{\lambda} < \norm{a}^{-1}/2$, we have
\begin{equation}
\abs{g(\lambda)} \leq \norm{f} \norm{r_a(\lambda^{-1})} \leq \frac{\norm{f}}{\abs{\lambda}^{-1} - \norm{a}} = \frac{\norm{f}\abs{\lambda}}{1 - \norm{a}\abs{\lambda}} \leq 2 \norm{f}\abs{\lambda},
\end{equation}
so $g$ is continuous at zero. It is then a consequence of Morera's theorem that $g$ is holomorphic on the whole disk $B_{\rho(a)^{-1}}(0)$.

Furthermore, for $0 < \abs{\lambda} < \norm{a}^{-1}$, we can use the Neumann series to obtain a power series expansion:
\begin{equation}
g(\lambda) = f\qty(\lambda \sum_{n=0}^\infty \qty(\lambda a)^n ) = \lambda \sum_{n=0}^\infty f(a^n) \lambda^n.
\end{equation}
Since $g$ is holomorphic on $B_{\rho(a)^{-1}}(0)$ or on $\C$ if $\rho(a) = 0$, the above gives its power series expansion on its entire domain by the unique representability of $g$ by a power series. The radius of convergence of this power series is therefore at least $\rho(a)^{-1}$, so the series converges absolutely for every $\lambda$ in the domain of $g$. Hence, for any $f \in A^*$ and $\lambda \in \C$ with $\abs{\lambda} < \rho(a)^{-1}$, the sequence $\abs{\lambda^n f(a^n)}$ is bounded as $n$ varies across the natural numbers.

Recall that the map $\Psi: A \rightarrow A^{**}$ defined as $\Psi(a)(f) = f(a)$ is a linear isometry. Then the boundedness of $\abs{\lambda^n f(a^n)}$ for all $f \in A^*$ indicates that the family $\Psi(\lambda^n a^n)$ is pointwise bounded. By the uniform boundedness principle, the family $\Psi(\lambda^n a^n)$ is uniformly bounded, i.e.\ for each $\lambda$ there exists $M_\lambda > 0$ such that
\begin{equation}
\abs{\lambda^n f(a^n)} < M_\lambda,
\end{equation}
for all $f \in A^*$ with $\norm{f} \leq 1$. By the Hahn-Banach theorem, there exists $f \in A^*$ with $\norm{f} \leq 1$ such that $\abs{f(a^n)} = \norm{a^n}$. Thus, we have $\abs{\lambda}^n \norm{a^n} < M_\lambda$ for all $n \in \N$, or
\begin{equation}
\norm{a^n}^{1/n} < M_\lambda^{1/n} \abs{\lambda}^{-1},
\end{equation}
assuming $\abs{\lambda} > 0$. Taking the limit supremum of both sides yields
\begin{equation}
\limsup \norm{a^n}^{1/n} \leq \limsup M^{1/n}_\lambda \abs{\lambda}^{-1} = \lim_{n \rightarrow \infty} M_\lambda^{1/n} \abs{\lambda}^{-1} = \abs{\lambda}^{-1}.
\end{equation}
If $\rho(a) = 0$, this is valid for all $\abs{\lambda} > 0$, which implies $\limsup \norm{a^n}^{1/n} = 0 = \rho(a)$. If $\rho(a) > 0$, this is valid for all $\abs{\lambda}^{-1} > \rho(a)$, which implies that
\begin{equation}
\limsup \norm{a^n}^{1/n} \leq \rho(a),
\end{equation}
as desired.
\end{proof}

The proof that $r(a) \leq \liminf \norm{a^n}^{1/n}$ in Theorem \ref{thm:spectral_radius_formula} contains some interesting observations worth highlighting.

\begin{lemma}\label{lem:invert_product}
If $a_1,\ldots, a_n, b \in A$ such that
\begin{equation}
b = a_1 a_2 \cdots a_n,
\end{equation}
and $[a_i, a_j] = 0$ for all $i,j$, then $b$ is invertible if and only if each $a_i$ is invertible.
\end{lemma}

\begin{proof}
It is obvious that $b$ is invertible if each $a_i$ is invertible. If $b$ is invertible, then for any $i\leq n$, we have
\begin{equation}
\qty(b^{-1}\prod_{j \neq i} a_j)a_i = b^{-1}b = 1 =bb^{-1} =  a_i \qty(\prod_{j \neq i} a_j)b^{-1}.
\end{equation}
Thus, $a_i$ has a left inverse and a right inverse, which must be equal.
\end{proof}

\begin{theorem}\label{thm:polynomial_spectrum}
Let $p = \sum_{i=0}^n \alpha_i z^i$ be a complex polynomial and let $a \in A$. Then
\begin{equation}
\sigma(p(a)) = p(\sigma(a)).
\end{equation}
\end{theorem}

\begin{proof}
Fix $\lambda \in \C$ and factorize $\lambda - p(a)$:
\begin{equation}
\lambda - p(z) = \beta_0 \prod_{i=1}^n (\beta_i - z)
\end{equation}
for some $\beta_0,\ldots, \beta_n \in \C$. Then
\begin{equation}
\lambda - p(a) = \beta_0 \prod_{i=1}^n (\beta_i - a).
\end{equation}
By Lemma \ref{lem:invert_product}, $\lambda \in \sigma(p(a))$ if and only if $\beta_i \in \sigma(a)$ for some $i \geq 1$. But there exists $i \geq 1$ such that $\beta_i \in \sigma(a)$ if and only if $\lambda \in p(\sigma(a))$, so we're done.
\end{proof}

Theorem \ref{thm:polynomial_spectrum} gives one example of how algebraic manipulations in $A$ affect the spectra of the elements being manipulated. Let us give more results in this vein.

\begin{theorem}
If $a, b \in A$, then
\begin{equation}\label{eq:spectrum_of_product}
\sigma(ab) \cup \qty{0} = \sigma(ba) \cup \qty{0}.
\end{equation}
If $a \in A^\times$, then
\begin{equation}
\sigma(a^{-1}) = \sigma(a)^{-1}.
\end{equation}
\end{theorem}

\begin{proof}
Suppose $\lambda \in \rho(ab)$. Then using $(\lambda - ba)b = b(\lambda - ab)$ we compute
\begin{equation}
\begin{aligned}
(\lambda - ba)\qty[1 + b(\lambda - ab)^{-1}a] = \qty(\lambda - ba) + ba = \lambda.
\end{aligned}
\end{equation}
and likewise
\begin{equation}
\qty[1 + b(\lambda - ab)^{-1}a] (\lambda - ba) = (\lambda - ba)  + ba = \lambda.
\end{equation}
Therefore $\lambda \in \rho(ba)$ if $\lambda \neq 0$. Of course, the same result holds with $a$ and $b$ switched. In other words,
\begin{equation}
\rho(ab) \setminus \qty{0} = \rho(ba) \setminus \qty{0}
\end{equation}
Taking complements yields \eqref{eq:spectrum_of_product}.

If $a \in A^\times$, then clearly $0 \notin \sigma(a)$ and $0 \notin \sigma(a^{-1})$. If $\lambda \neq 0$, then
\begin{equation}
\lambda^{-1} - a = \lambda^{-1} a(a^{-1} - \lambda),
\end{equation}
which implies $\lambda^{-1} \in \sigma(a)$ if and only if $\lambda \in \sigma(a^{-1})$ by Lemma \ref{lem:invert_product}. Since $\lambda^{-1} \in \sigma(a)$ if and only if $\lambda \in \sigma(a)^{-1}$, this is the desired result.
\end{proof}

\subsection[Spectral Theory in C$^*$-Algebras]{Spectral Theory in C$^*$-Algebras}

We continue where we left off in the previous section by showing how the spectrum behaves with respect to the star operation. We now let $A$ be a unital $C^*$-algebra for the rest of this section.

\begin{proposition}\label{prop:star_spectrum}
Let $a \in A$. Then
\begin{equation}
\sigma(a^*) = \sigma(a)^*.
\end{equation}
\end{proposition}

\begin{proof}
We note that $\lambda - a^*$ is invertible if and only if $\lambda - a^*$ is invertible by Proposition \ref{prop:inverse_star_commute}. Hence, $\lambda \in \sigma(a^*)$ if and only if $\lambda^* \in \sigma(a)$ if and only if $\lambda \in \sigma(a)^*$.
\end{proof}

We now investigate the spectra of several special classes of elements of $A$.

\begin{definition}
An element $a \in A$ is
\begin{enumerate}
	\item[(i)] \emph{normal} if $[a,a^*] = 0$,
	\item[(ii)] an \emph{isometry} if $a^*a = 1$, and
	\item[(iii)] \emph{unitary} if $a^*a = aa^* = 1$, i.e.\ $a \in A^\times$ and $a^{-1} = a^*$.
\end{enumerate}
Note that both unitary and self-adjoint elements are normal. Furthermore, if $a$ is an isometry or a unitary, then $\norm{a} = 1$ by the $C^*$-property.
\end{definition}

\begin{corollary}
If $a \in A$ is normal, then $\rho(a) = \norm{a}$.
\end{corollary}

\begin{proof}
We claim that
\begin{equation}\label{eq:normal_inductive_step}
\norm{a^{2^n}}^2 = \norm{a}^{2^{n+1}}.
\end{equation}
for all normal $a \in A$. For $n = 0$ this is trivial. Suppose it is true for some $n = k-1$ where $k \in \N$. Using normality of $a$, the $C^*$-property, and the fact that $a^*a$ is self-adjoint, we compute
\begin{equation}
\begin{aligned}
\norm{a^{2^k}}^2  &= \norm{\qty(a^{2^k})^*a^{2^k}} = \norm{\qty(a^*)^{2^k}a^{2^k}} = \norm{(a^*a)^{2^k}} \\
&= \norm{(a^*a)^{2^{k-1}}}^2  = \norm{a^*a}^{2^k} = \norm{a}^{2^{k+1}}.
\end{aligned}
\end{equation}
This proves \eqref{eq:normal_inductive_step}. Now, using the formula for the spectral radius, we obtain
\begin{equation}
r(a) = \lim_{n \rightarrow \infty} \norm{a^{2^n}}^{1/2^n} = \lim_{n \rightarrow \infty} \norm{a}^{2^{n}/2^n} = \norm{a},
\end{equation}
as desired.
\end{proof}

\begin{corollary}
If $a \in A$ is isometric, then $r(a) = 1$.
\end{corollary}

\begin{proof}
We note that
\begin{equation}
\norm{a^n}^2 = \norm{\qty(a^n)^* \qty(a^n)} = \norm{(a^*)^n \qty(a^n)} = \norm{1} = 1.
\end{equation}
Thus,
\begin{equation}
r(a) = \lim_{n \rightarrow \infty} \norm{a^n}^{1/n} = \lim_{n \rightarrow \infty} 1 = 1,
\end{equation}
as desired.
\end{proof}

\begin{corollary}
If $a \in A$ is unitary, then $\sigma(a) \subset S^1$.
\end{corollary}

\begin{proof}
We have
\begin{equation}
\sigma(a)^{-1} = \sigma(a^{-1}) = \sigma(a^*) = \sigma(a)^*.
\end{equation}
Thus, if $\lambda \in \sigma(a)$, then $\lambda^{-1} = \mu^*$ for some $\mu \in \sigma(a)$. Since $r(a) \leq 1$, we know $1 \leq \abs{\lambda}^{-1} = \abs{\mu} \leq 1$, so $\abs{\lambda} = 1$.
\end{proof}

\begin{theorem}
If $a \in A$ is self-adjoint, then
\begin{equation}
\sigma(a) \subset [-\norm{a}, \norm{a}].
\end{equation}
In particular, $\sigma(a^2) = \sigma(a)^2 \subset [0, \norm{a}^2]$.
\end{theorem}

\begin{proof}
Let $\lambda \in \R$ such that $\lambda^{-1} = \abs{i\lambda^{-1}} > \norm{a}$. Then $\lambda^{-1} \in \rho(a)$, so $1 + i\lambda a = -i\lambda(i\lambda^{-1}  - a)$ is invertible. Note that $(1 + i\lambda a)^* = 1 - i \lambda a$ is invertible as well. Define
\begin{equation}
u = (1 - i\lambda a)(1 + i \lambda a)^{-1}.
\end{equation}
Observe that $u^* = (1 - i\lambda a)^{-1}(1 + i \lambda a)$. Using the fact that $1 - i \lambda a$ commutes with $1 + i \lambda a$, we see that
\begin{equation}
u^*u = (1- i\lambda a)^{-1} (1 + i \lambda a)(1 - i \lambda a) (1 + i \lambda a)^{-1} = 1.
\end{equation}
Since $u$ is the product of two invertible elements, we know $u$ is invertible, and the above shows that $u^{-1} = u^*$, so $u$ is unitary.

Given $\mu \in \C$, $\mu \neq i\lambda^{-1}$, observe that
\begin{equation}
\abs{\frac{1 - \cplxi \, \lambda \mu}{1 + \cplxi \, \lambda \mu}} = \sqrt{\frac{(1 + \lambda \, \Im \mu)^2 + (\lambda \, \Re \mu)^2}{(1 - \lambda \, \Im \mu)^2 + (\lambda \, \Re \mu)^2}},
\end{equation}
which equals $1$ if and only if $\Im \mu = 0$, i.e.\ $\mu \in \R$. Since $\sigma(u) \subset S^1$, we see that if $\mu \in \C \setminus \R$, then $(1-i \lambda \mu)(1 + i \lambda \mu)^{-1} \in \rho(u)$. In particular, if $\mu \neq i \lambda^{-1}$, then
\begin{equation}
\begin{aligned}
(1 - i \lambda \mu)(1 + i \lambda \mu)^{-1} - u &= (1 + i \lambda \mu)^{-1}\qty[(1 - i \lambda \mu)(1 + i \lambda a) - (1 + i \lambda \mu)(1 - i\lambda a)](1 + i \lambda a)^{-1}\\
&= 2i\lambda(1 + i \lambda \mu)^{-1} (a - \mu)(1 + i \lambda a)^{-1}.
\end{aligned}
\end{equation}
If $\mu \in \C \setminus \R$, then the left hand side is invertible, so $\mu - a$ is invertible, so $\mu \in \rho(a)$. Furthermore, $i\lambda^{-1} \notin \sigma(a)$ since $\abs{i\lambda^{-1}} > \norm{a}$. Thus, $\sigma(a) \subset \R$.
\end{proof}

\begin{corollary}\label{cor:unique_star_norm}
If $a \in A$, then
\begin{equation}
\norm{a} = \sqrt{r(a^*a)}.
\end{equation}
Hence, the norm is completely determined by the algebraic operations, i.e.\ the $C^*$-norm is unique.
\end{corollary}

\begin{proof}
Note that $a^*a$ is normal, so
\begin{equation}
\norm{a}^2 = \norm{a^*a} = r(a^*a).
\end{equation}
The result follows by taking a square root.
\end{proof}

\begin{theorem}[Spectral Permanence]
Let $B$ be a unital $C^*$-subalgebra of $A$. For any $a \in B$, the spectrum of $a$ in $B$ is the same as the spectrum of $a$ in $A$.
\end{theorem}

\begin{proof}
Let us temporarily use the notation $\sigma_A(a)$ and $\sigma_B(a)$ to distinguish the spectrum of $a$ in $A$ and $B$ respectively. We shall use the notation $\rho_A(a)$, $\rho_B(a)$, $r_A(a)$, and $r_B(a)$ similarly. If $\lambda \in \sigma_A(a)$, then $\lambda - a$ is not invertible in $A$, so $\lambda - a$ is certainly not invertible in $B$. Thus, $\sigma_A(a) \subset \sigma_B(a)$.

The reverse inclusion $\sigma_B(a) \subset \sigma_A(a)$ is trivial if $a = 0$, so suppose $a \neq 0$. It is easy to check that $\sigma_B(a) \subset \sigma_A(a)$ follows from the inclusion $B \cap A^\times \subset B^\times$, and this is what we will show. First suppose $a \in B \cap A^\times$ and $a$ is self-adjoint. Let $\lambda_0 = 2i \norm{a}$ and note that $\lambda_0 \in \rho_B(a)$ since $\abs{\lambda_0} > \norm{a}$. We know from Corollary \ref{cor:resolvent_is_open} that $B_{\norm{r_a(\lambda_0)}^{-1}}(\lambda_0) \subset \rho_B(a)$; thus we want to show that $0 \in B_{\norm{r_a(\lambda_0)}^{-1}}(\lambda_0)$ to show that $a \in B^\times$.

Observe that self-adjointness of $a$ implies that $r_a(\lambda_0)^* = r_a(\lambda_0^*)$, from which it follows that $r_a(\lambda_0)$ is normal since $[r_a(\lambda), r_a(\mu)] = 0$ for all $\lambda, \mu \in \rho_B(a)$. Normality then yields
\begin{equation}
\norm{r_a(\lambda_0)} = r_A\qty(r_a(\lambda_0)).
\end{equation}
But we also know that
\begin{equation}
\sigma_A(r_a(\lambda_0)) = \sigma_A(\lambda_0 - a)^{-1} = \qty[\lambda_0 - \sigma_A(a)]^{-1}
\end{equation}
Hence,
\begin{equation}
r_A(r_a(\lambda_0)) = \tn{dist}(\lambda_0, \sigma_A(a))^{-1}.
\end{equation}
But since $\lambda_0$ is purely imaginary and $\sigma_A(a) \subset \R$, we know that $\tn{dist}(\lambda_0, \sigma_A(a)) \geq \abs{\lambda_0}$! In fact,  we know $0 \notin \sigma_A(a)$ since $a \in A^\times$, so this is a strict inequality and taking inverses yields
\begin{equation}
\norm{r_a(\lambda_0)} = \tn{dist}(\lambda_0, \sigma_A(a))^{-1} < \abs{\lambda_0}^{-1}.
\end{equation}
This implies $0 \in B_{\norm{r_a(\lambda_0)}^{-1}}(\lambda_0)$, proving the theorem for self-adjoint $a$.

Finally, consider $a \in B \cap A^\times$, not necessarily self-adjoint. However, we see that $a^*a \in B \cap A^\times$ and $a^*a$ is self-adjoint, so $a^*a \in B^\times$. Defining
\begin{equation}
b = (a^* a)^{-1}a^*,
\end{equation}
we see that $ba = 1$, so $b = a^{-1}$ since $a$ was assumed to be invertible in $A$. Since $b \in B$ manifestly, we conclude that $a \in B^\times$, as desired.
\end{proof}

Let us now consider how the spectrum of an element behaves under $*$-homomorphisms.

\begin{proposition}\label{prop:star_hom_inverse}
Let $A$ and $B$ be unital $C^*$-algebras and let $\pi:A \rightarrow B$ be a unital $*$-homomorphism. Then $\pi(A^\times)\subset B^\times$ and
\begin{equation}
\pi(a^{-1}) = \pi(a)^{-1}
\end{equation}
for all $a \in A^\times$. If $\pi$ is bijective, then $\pi(A^\times) = B^\times$.
\end{proposition}

\begin{proof}
Let $a \in A^\times$ and observe
\begin{equation}
\pi(a^{-1})\pi(a) = \pi(a^{-1}a) = \pi(1) = 1 =\pi(1) = \pi(aa^{-1}) = \pi(a)\pi(a^{-1}).
\end{equation}
This proves that $\pi(a) \in B^\times$ and $\pi(a)^{-1} = \pi(a^{-1})$. If $\pi$ is bijective, then $\pi^{-1}(B^\times) \subset A^\times$ by the same argument, so $\pi \pi^{-1}(B^\times) = B^\times \subset \pi(A^\times)$.
\end{proof}

\begin{proposition}\label{prop:star_hom_spectrum}
Let $A$ and $B$ be unital $C^*$-algebras and let $\pi:A \rightarrow B$ be a unital $*$-homomorphism. Then
\begin{equation}
\sigma(\pi(a)) \subset \sigma(a)
\end{equation}
and
\begin{equation}
\pi(r_a(\lambda)) = r_{\pi(a)}(\lambda)
\end{equation}
for all $\lambda \in \rho(a)$. If $\pi$ is bijective, then $\sigma(\pi(a)) = \sigma(a)$.
\end{proposition}

\begin{proof}
Let $\lambda \in \rho(a)$. Then by Proposition \ref{prop:star_hom_inverse}, we know $\pi(\lambda - a) = \lambda - \pi(a) \in B^\times$ and
\begin{equation}
r_{\pi(a)}(\lambda) = (\lambda - \pi(a))^{-1} = \pi(\lambda - a)^{-1} = \pi\qty(r_a(\lambda)).
\end{equation}
We see that $\rho(a) \subset \rho(\pi(a))$, so $\sigma(\pi(a)) \subset \sigma(a)$ by taking complements. If $\pi$ is bijective, then $\sigma(a) = \sigma(\pi^{-1}(\pi(a))) \subset \sigma(\pi(a))$ as well.
\end{proof}

\begin{proposition}\label{prop:star_hom_cont}
Let $A$ and $B$ be unital $C^*$-algebras and let $\pi:A \rightarrow B$ be a unital $*$-homomorphism. Then
\begin{equation}
\norm{\pi(a)} \leq \norm{a}
\end{equation}
for all $a \in A$. In particular, $\pi$ is continuous and $\norm{\pi} = 1$.
\end{proposition}

\begin{proof}
By Proposition \ref{prop:star_hom_spectrum}, we know that $r(\pi(a)) \leq r(a)$ for all $a \in A$. Then by Corollary \ref{cor:unique_star_norm},
\begin{equation}
\norm{\pi(a)} = \sqrt{r(\pi(a)^*\pi(a))} = \sqrt{r\qty(\pi(a^*a))} \leq \sqrt{r(a^*a)} = \norm{a}.
\end{equation}
This shows that $\pi$ is continuous, and $\norm{\pi} = 1$ since $\norm{\pi(1)} = \norm{1} = 1$.
\end{proof}

With these simple propositions in hand, we can prove a powerful theorem, known as the \emph{continuous functional calculus} for self-adjoint elements.

\begin{theorem}
Let $A$ be a unital $C^*$-algebra and let $a \in A$ be self-adjoint. There exists a unique unital $*$-homomorphism $C(\sigma(a)) \rightarrow A$, $f \mapsto f(a)$ such that $p(a) = \sum_{i=0}^n \alpha_{i} a^i$ for all complex polynomials $p(z) = \sum_{i=0}^n \alpha_i z^i $. Furthermore, for all $f \in C(\sigma(a))$, we have
	\begin{enumerate}
		\item[\tn{(i)}] $\norm{f(a)} = \norm{f}$,
		\item[\tn{(ii)}] $f(a)$ is in the $C^*$-algebra generated by $1$ and $a$. In particular, $[f(a),a] = 0$,
		\item[\tn{(iii)}] $\pi(f(a)) = f(\pi(a))$ for any unital $*$-homomorphism $\pi:A \rightarrow B$,
		\item[\tn{(iv)}] $\sigma(f(a)) = f(\sigma(a))$.
	\end{enumerate}
Finally, if $g \in C(\sigma(f(a)))$, then
	\begin{enumerate}
		\item[\tn{(v)}] $(g \circ f)(a) = g(f(a))$.
	\end{enumerate}
\end{theorem}

\noindent Note that $f(\pi(a))$ is well-defined in (iii) since $\sigma(\pi(a)) \subset \sigma(a)$, and $(g \circ f)(a)$ is well-defined by (iv).

\begin{proof}
It clear that the map $p \mapsto p(a)$ defined on polynomials $p(z) = \sum_{i,j=0}^n \alpha_i z^i $ is linear. Furthermore, since $p(a)$ is self-adjoint for any polynomial $p$, we have
\begin{equation}
\begin{aligned}
\norm{p(a)} &= r(p(a)) \\
&= \sup\qty{\abs{\lambda}: \lambda \in \sigma(p(a))} \\
&= \sup\qty{\abs{\lambda}: \lambda \in p(\sigma(a))} \\
&= \sup\qty{\abs{p(\lambda)}: \lambda \in \sigma(a)}\\
&= \norm{p},
\end{aligned}
\end{equation}
so the map $p \mapsto p(a)$ is continuous. Since $\sigma(a)$ is a compact subset of $\R$, the Weierstrass approximation theorem (and the Tietze extension theorem) imply that the set of polynomials is dense in $C(\sigma(a))$. Therefore the map $p \mapsto p(a)$ extends uniquely to a linear map $f \mapsto f(a)$ on $C(\sigma(a))$. It follows by standard continuity arguments that $\norm{f(a)} = \norm{f}$ for all $f \in C(\sigma(a))$ since $\norm{p(a)} = \norm{p}$ for all polynomials.

It is clear that for polynomials $p$ and $q$, we have $(pq)(a) = p(a)q(a)$ and $(p^*)(a) = (p(a))^*$, the latter relying on self-adjointness of $a$. That $(fg)(a) = f(a)g(a)$ and $(f^*)(a) = (f(a))^*$ for arbitrary $f, g \in C(\sigma(a))$ again follows by standard continuity arguments using the fact that the polynomials are dense in $C(\sigma(a))$. Thus, $f \mapsto f(a)$ is a unital $*$-homomorphism.

Once again, (ii) and (iii) clearly hold for polynomials. Thus, (ii) holds for $f$ by a standard argument using density of polynomials and completeness of the $C^*$-algebra generated by $1$ and $a$. Likewise, (iii) holds by density of polynomials and by continuity of $*$-homomorphisms as shown in Proposition \ref{prop:star_hom_cont}.

To prove (iv), let $(p_n)$ be a sequence of polynomials such that $p_n \rightarrow f$. Given $\lambda \in \sigma(a)$, we know
\begin{equation}
p_n(\lambda) \in p_n(\sigma(a)) = \sigma(p_n(a)),
\end{equation}
so $p_n(\lambda) - p_n(a)$ is not invertible. Since the complement of $A^\times$ is closed, taking the limit as $n\rightarrow \infty$ yields $f(\lambda) - f(a) \notin A^\times$, so $f(\lambda) \in \sigma(f(a))$. Hence $f(\sigma(a)) \subset \sigma(f(a))$. On the other hand, if $\lambda \notin f(\sigma(a))$, then $\lambda - f$ is invertible in $C(\sigma(a))$ with inverse $g \in C(\sigma(a))$. Then
\begin{equation}
(\lambda - f(a))g(a) = g(a)\qty(\lambda - f(a)) = 1,
\end{equation}
so $\lambda - f(a)$ is invertible, i.e.\ $\lambda \notin \sigma(f(a))$. This proves $\sigma(f(a)) \subset f(\sigma(a))$, as desired.

To prove (v), we note that $g \mapsto g \circ f$ is a unital $*$-homomorphism $C(\sigma(f(a))) \rightarrow C(\sigma(a))$, so $g \mapsto g \circ f \mapsto (g \circ f)(a)$ is a unital $*$-homomorphism $C(\sigma(f(a))) \rightarrow A$. Furthermore, if $p$ is a polynomial, then $(p \circ f)(a) = p(f(a))$ since the map $C(\sigma(a)) \rightarrow A$ respects addition and multiplication. Therefore $(g \circ f)(a) = g(f(a))$ by uniqueness of the $*$-homomorphism $C(\sigma(f(a))) \rightarrow A$.
\end{proof}

\begin{corollary}
Let $A$ and $B$ be unital $C^*$-algebras. If $\pi:A \rightarrow B$ is an injective unital $*$-homomorphism, then $\pi$ is an isometry:
\begin{equation}
\norm{\pi(a)} = \norm{a}
\end{equation}
for all $a \in A$.
\end{corollary}

\begin{proof}
Since $\norm{\pi(a)} \leq \norm{a}$ by Proposition \ref{prop:star_hom_cont}, we need only show the reverse inequality. First we show $\norm{a} \leq \norm{\pi(a)}$ for all self-adjoint elements $a \in A$. Suppose $a$ is self-adjoint and $\norm{\pi(a)} < \norm{a}$. Recall that $\rho(a) = \norm{a}$ and $\sigma(a) \subset \R$, so $\norm{a} \in \sigma(a)$ or $-\norm{a} \in \sigma(a)$, and likewise for $\pi(a)$. Choose $f :[-\norm{a}, \norm{a}] \rightarrow \R$ such that $f$ vanishes on $[-\norm{\pi(a)}, \norm{\pi(a)}]$ and $f(\norm{a}) = f(-\norm{a}) = 1$. Then $f(\pi(a)) = \pi(f(a)) = 0$ but $\norm{f(a)} = \norm{f} > 1$, contradicting injectivity of $\pi$. Therefore $\norm{a} = \norm{\pi(a)}$ for self-adjoint $a$.

For arbitrary $a \in A$, we have
\begin{equation}
\norm{\pi(a)}^2 = \norm{\pi(a)^*\pi(a)} = \norm{\pi(a^*a)} = \norm{a^*a} = \norm{a}^2,
\end{equation}
which concludes the proof.
\end{proof}

\subsection{Positive Elements}

We shall continue to let $A$ be a unital $C^*$-algebra.

\begin{definition}
An element $a \in A$ is \emph{positive} if $a$ is self-adjoint and $\sigma(a) \subset [0,\infty)$. We let $A_+$ denote the set of all positive elements of $A$.
\end{definition}

\begin{proposition}\label{prop:positive_norm_condition}
Let $a \in A$ be self-adjoint and let $\lambda \in \R$ such that $\lambda \geq \norm{a}$. Then $a$ is positive if and only if $\norm{\lambda - a} \leq \lambda$.
\end{proposition}

\begin{proof}
If $a \in A_+$, then
\begin{equation}
\norm{\lambda - a} = r(\lambda - a) = \sup \qty{\abs{\mu}: \mu \in \sigma(\lambda - a) = \lambda - \sigma(a)} \leq \lambda
\end{equation}
since $\sigma(a) \subset [0,\norm{a}] \subset [0,\lambda]$.

Suppose $\norm{\lambda - a} \leq \lambda$. If $\mu \in \sigma(a)$, then
\begin{equation}
\abs{\lambda - \mu} \leq  r(\lambda - a) = \norm{\lambda - a} < \lambda,
\end{equation}
which implies that $ \mu \geq 0$, hence $a \in A_+$.
\end{proof}

\begin{proposition}
The set of positive elements $A_+$ is closed.
\end{proposition}

\begin{proof}
Let $(a_n)_{n \in \N}$ be a sequence in $A_+$ converging to $a \in A$. Since the star operation is continuous, we have $a_n^* \rightarrow a^*$, but since $a_n^* = a_n$ for all $n \in \N$, we see that $a$ is self-adjoint. Since  the norm is also continuous, we see that $\norm{a_n} \rightarrow \norm{a}$. In particular, there exists $M > 0$ such that $\norm{a_n} < M$ for all $n \in \N$. Then $\norm{a} \leq M$ and $\norm{M - a_n} \leq M$ for all $n \in \N$ by Proposition \ref{prop:positive_norm_condition}, so
\begin{equation}
\norm{M - a} = \lim_{n \rightarrow \infty} \norm{M - a_n} \leq M.
\end{equation}
Since $\norm{a} \leq M$ and $\norm{M - a} \leq M$, Proposition \ref{prop:positive_norm_condition} implies that $a$ is positive.
\end{proof}

\begin{proposition}
The sum of two positive elements is positive.
\end{proposition}

\begin{proof}
If $a, b \in A_+$,  then $a+b$ is self-adjoint and Proposition \ref{prop:positive_norm_condition} implies
\begin{equation}
\norm{\norm{a} + \norm{b} - (a + b)} \leq \norm{\norm{a} - a} + \norm{\norm{b} - b} \leq \norm{a} + \norm{b}.
\end{equation}
A second application of Proposition \ref{prop:positive_norm_condition} yields $a + b \in A_+$.
\end{proof}

\begin{proposition}\label{prop:positive_conditions}
Let $a \in A$ be self-adjoint. The following are equivalent.
	\begin{enumerate}
		\item[\tn{(i)}] The element $a$ is positive.
		\item[\tn{(ii)}] There exists a unique positive $b \in A$ such that $a = b^2$.
		\item[\tn{(iii)}] There exists a self-adjoint $b \in A$ such that $a = b^2$.
		\item[\tn{(iv)}] There exists $c \in A$ such that $a = c^*c$.
	\end{enumerate}
\end{proposition}

\begin{proof}
The implications (ii) $\Rightarrow$ (iii) and (iii) $\Rightarrow$ (iv) are trivial.

(i) $\Rightarrow$ (ii). By the continuous functional calculus, we can take $\sqrt{a}$ and by the composition property and the fact that $\qty(\sqrt{x})^2 = x$ on $\sigma(a)$, we have that $a = \qty(\sqrt{a})^2$. The fact that the square root is real-valued and that the continuous functional calculus is a $*$-homomorphism imply that $\sqrt{a}$ is self-adjoint, and $\sqrt{\sigma(a)} = \sigma(\sqrt{a})$ implies that $\sqrt{a}$ is positive. If $a = b^2$ for any other positive $b \in A$, then the fact that $\sqrt{x^2} = x$ for $x \in \sigma(a)$ and the composition property imply that $\sqrt{a} = \sqrt{b^2} = b$.

(iv) $\Rightarrow$ (i). First we prove a lemma. Given $d \in A$, we claim that $\sigma(-d^*d) \subset [0,\infty)$ implies $d = 0$. Write $d = d_1 + id_2$, where $d_1$ and $d_2$ are self-adjoint. Then
\begin{equation}
d^*d + dd^* =  2d_1^2 + 2d_2^2.
\end{equation}
If $\sigma(-d^*d) \subset [0,\infty)$, then $\sigma(-dd^*) \subset \sigma(-d^*d) \cup \qty{0} \subset [0,\infty)$. Thus $d^*d = 2d_1^2 + 2d_2^2 - dd^*$ is positive, since it is the sum of positive elements. But then $\sigma(d^*d) = \qty{0}$, which implies that $d^*d = 0$ and therefore $d = 0$ by the $C^*$-property of the norm.

Continuing, we suppose $a = c^*c$ for some $c \in A$. by the continuous functional calculus, the elements $a_+ = (\abs{a} + a)/2$ and $a_- = (\abs{a} - a)/2$ are positive and $a = a_+ - a_-$. Furthermore, observe that
\begin{equation}
a_+ a_- = \frac{1}{4}\qty(\abs{a}^2 - a^2) = 0.
\end{equation}
Defining $d = ca_-$, we compute
\begin{equation}
-d^*d = -a_- c^* c a_- = -a_-(a_+ - a_-)a_- = (a_-)^3,
\end{equation}
which implies that $-d^*d$ is positive. By the previous paragraph, we know $d = 0$. Thus,
\begin{equation}
0 = c^* d = c^*c a_- = aa_- = - (a_-)^2
\end{equation}
which implies that $a_- = 0$, for example by self-adjointness of $a_-$ and the $C^*$-property of the norm. Thus, $a = a_+$ is positive.
\end{proof}

\begin{definition}
We define a partial ordering on $A_+$ by setting $a \leq b$ for $a, b \in A_+$ if and only if $b - a \in A_+$. Reflexivity and antisymmetry are easy to check and transitivity follows since $(c - b) + (b - a) = c - a \in A_+$ given $c - b, b - a \in A_+$. In fact, $A_+$ is a directed set since given $a, b \in A_+$, we have $a+b \in A_+$ and $a, b \leq a + b$.
\end{definition}

\begin{proposition}\label{prop:inner_aut_preserves_order}
Let $a, b \in A_+$ and let $c \in A$. If $a \leq b$, then $c^*ac \leq c^* bc$.
\end{proposition}

\begin{proof}
Since $a \in A_+$, there exists $d \in A$ such that $a = d^*d$. Then $c^*ac = c^*d^*dc = (dc)^*dc$, so $c^*ac \in A_+$. Likewise, $c^*bc \in A_+$ since $b \in B_+$. Likewise, $c^*(b - a)c \in A_+$ since $b -a \in A_+$, so $c^*ac \leq c^*bc$.
\end{proof}

\begin{proposition}\label{prop:positive_respects_norm}
If $a \in A_+$ and $\lambda \geq 0$, then $a \leq \lambda$ if and only if $\norm{a} \leq \lambda$.
\end{proposition}

\begin{proof}
We have the following equivalences:
\begin{equation}
a \leq \lambda \quad \Longleftrightarrow \quad \sigma(\lambda - a) = \lambda - \sigma(a) \subset [0, \infty) \quad \Longleftrightarrow \quad r(a) = \norm{a} \leq \lambda,
\end{equation}
as desired.
\end{proof}

If we replace $\lambda$ in the above proposition by an arbitrary element, then we only have an implication in one direction.

\begin{proposition}
If $a, b \in A_+$ and $a \leq b$, then $\norm{a} \leq \norm{b}$.
\end{proposition}

\begin{proof}
We know $b \leq \norm{b}$ by Proposition \ref{prop:positive_respects_norm}, so $a \leq \norm{b}$ by transitivity. But this implies $\norm{a} \leq \norm{b}$ by another application of Proposition \ref{prop:positive_respects_norm}.
\end{proof}

\begin{proposition}
If $a, b \in A_+ \cap A^\times$ and $a \leq b$, then $b^{-1} \leq a^{-1}$.
\end{proposition}

\begin{proof}
Note that $a^{-1}, b^{-1} \in A_+$ by the continuous functional calculus. Since $\sqrt{a^{-1}}$ is self-adjoint, we have
\begin{equation}
1 = \sqrt{a^{-1}}a \sqrt{a^{-1}} \leq \sqrt{a^{-1}}b\sqrt{a^{-1}}.
\end{equation}
Thus, $\sigma(\sqrt{a^{-1}}b \sqrt{a^{-1}}) \subset [1, \infty)$ and by the continuous functional calculus,
\begin{equation}
1 \geq \qty(\sqrt{a^{-1}}b\sqrt{a^{-1}})^{-1} = \sqrt{a^{-1}}^{-1} b^{-1} \sqrt{a^{-1}}^{-1}.
\end{equation}
Multiplying by $\sqrt{a^{-1}}$ to the left and right as in the first step now yields $b^{-1} \leq a^{-1}$.
\end{proof}


\begin{itemize}
	\item $(E,\skal{.}{.})$ complex Hilbert space.
	\item $A \in \mathcal{B(E,E)}$.
	\item Consider the equation
\[
	x = A(x)+ y, \qquad y \in E.
\]
\[
	(I-A)(x)= y.
\]
\item Consider this problem for $\lambda \in \mathbb{C}$. \\
\item Set \[
	\rho(A) := \set[\lambda \in \mathbb{C}]{(A- \lambda I)^{-1} \in \mathcal{B}(E,E)}
\]
\item $\rho(A)$ is called the resolvent set for $A$.
\item Set \[
	\sigma(A) = \mathbb{C} \setminus \rho(A).
\]
\item $\sigma(A)$ is called the spectrum of $A$.
\item Clearly, a necessary condition for $(A-\lambda I)^{-1} \in \mathcal{B}(E,E)$ is that
\[
	A - \lambda I:E \to E
\]
is a bijection.
\item Linearity for $(A-\lambda I)^{-1}$ follows from the linearity of $A-\lambda I$.

\end{itemize}
\begin{theorem}[Banachs's inverse mapping theorem]
	$(E,\norm{.})$ Banach space, \\ $A \in \mathcal{B}(E,E)$. $A-\lambda I: E \to E$ bijection. Then
	\[
		\Rightarrow (A- \lambda I)^{-1} \in \mathcal{B}(E,E)
	\]
\end{theorem}
\begin{beweis}
	based on the Open mapping theorem. Proof is omitted. Assume $\lambda \in \sigma(A)$. Then $A-\lambda I:E \to E$ is not a bijection.

\begin{itemize}
	\item If $A-\lambda I: E \to E$ is not injective then there exists $0 \neq x \in  E $ such that
	\[
		(A- \lambda I)(x) = 0,
	\]
	i.e. $\lambda$ is an eigenvalue of $A$.
	Set
	\[
		\sigma_p(A) = \set[\lambda \in \mathbb{C}]{\lambda \text{eigenvalue of $A$}}.
	\]
	\item If $A-\lambda I$ is injective, densely defined but not bounded then $\lambda \in \sigma(A)$. The set of such $\lambda$'s is called the continuous spectrum of $A$, denoted $\sigma_c(A)$
	\item If $A-\lambda I$ is not surjective then the set of such $\lambda$'s is called the residual spectrum, denoted $\sigma _r(A)$.
\end{itemize}
\end{beweis}
 \begin{lemma*}
 	$(E,\norm{.})$ Banach space, $A \in \mathcal{B}(E,E)$ with $\norm{A} < 1$. Then
	\[
		(I-A)^{-1} \in \mathcal{B}(E,E)
	\]
	and
	\[
		(I-A)^{-1} = I + \sum_{n=1}^{\infty} A^n.
	\]
	This series is called a Neumannseries.
 \end{lemma*}
\begin{beweis}
	Observe
	\[
		\norm{A^n} = \norm{A \cdot A \cdots A} \leq \norm{A}^n, \qquad n = 1,2,\dots
	\]
	and
	\[
		\sum_{n=1}^{\infty}\norm{A^n}< \infty.
	\]
	Since $E$ is a Banach space we have
	\[
		\sum_{n=1}^{\infty}A^n
	\]
	converges in $\mathcal{B(E,E)}$. Since $E$ Banach space implies $\mathcal{B}(E,E)$ is a Banach space. \\
	Note
	\[
		(I-A)\left( I+ \sum_{n=1}^{N} A^n \right) = I-A^{N+1} \to I, \qquad \text{in }\mathcal{B}(E,E).
	\]
	\[
		\left(I + \sum_{n=1}^{N}A^n \right)(I-A) = I-A^{N+1} \to I, \qquad \text{in }\mathcal{B}(E,E).
	\]
	We get
	\[
		\left( I + \sum_{n=1}^{\infty} \right)(I-A) = I = (I-A)(I + \sum_{n=1}^{\infty}A^n).
	\]
	We have $(I-A)^{-1} $ exists and is equal to $I+ \sum_{n=1}^{A^n}$.
\end{beweis}
 \begin{lemma*}
 	$(E,\norm{.})$ Banach space and $A \in \mathcal{B}(E,E)$. Then
	\begin{enumerate}
		\item $\sigma(A) \neq \emptyset$.
		\item $\sigma(A)$ closed set in $\mathbb{C}$.
		\item $\sigma(A) \subseteq \overline{B(0,\norm{A})}$
	\end{enumerate}
 \end{lemma*}
\begin{beweis}
	\begin{enumerate}
		\item omitted.
		\item Enough to prove that $\rho(A)$ is an open set in $\mathbb{C}$. \\
		Fix $\lambda_0 \in \rho(A)$. So $(A- \lambda_0 I)^{-1} \in \mathcal{B}(E,E)$. \\ Note:
		\begin{align*}
			A-\lambda I &= A - \lambda_0 I - (\lambda - \lambda_0)I \\
			&= \underset{\substack{\text{invertible} \\ \text{ since }\lambda_0 \in \rho(A)}}{\underbrace{(A-\lambda_0 I)}} \underset{\substack{ \text{ invertible if} \\ \norm{(\lambda-\lambda_0)(A- \lambda_0 I)^{-1}}<1 \\ \text{by previous lemma, i.e.} \\ \abs{\lambda - \lambda_0} < \frac{1}{\norm{(A- \lambda_0 I)^{-1}}}}}{\underbrace{\left( I - (\lambda - \lambda_0)(A- \lambda_0 I)^{-1} \right)}}.
		\end{align*}
		Clearly, $A-\lambda I$ is invertible if
		\[
			\abs{\lambda - \lambda_0} < \frac{1}{\norm{(A- \lambda_0 I)^{-1}}}.
		\]
		\item It is enough to show that $\lambda \in \rho(A)$ if \[
			\abs{\lambda} > \norm{A}.
		\]
		Note
		\[
			A- \lambda I = - \lambda (I - \frac{1}{\lambda}A).
		\]
		Here
		\[
			\norm{- \frac{1}{\lambda}A} = \frac{1}{\abs{\lambda}} \norm{A} < 1.
		\]
		$I - \frac{1}{\lambda}A$ is invertible by previous lemma. So $\rho(A)$.
	\end{enumerate}
\end{beweis}
Now assume $(E,\skal{.}{.})$ is a complex Hilbert space with infinite dimension. $A \in \mathcal{K}(E,E)$ (We don't assume $A$ is self-adjoint). Then
\begin{enumerate}
	\item $\lambda \in \sigma(A) \setminus \set{0}$ $\qquad $ $\Rightarrow$ is an eigenvalue of $A$.
	\item $\lambda \in \sigma(A) \setminus \set{0}$ $\qquad $ $\Rightarrow $ $\dim \set[x \in E]{A(x) = \lambda x} < \infty$.
	\item $O$ is the only cluster point for $\sigma(A)$
	\item $0 \in \sigma(A)$ since if $0 \not \in \sigma(A)$ then $A^{-1} \in \mathcal{B}(E,E)$ and \[
		\underset{\in \mathcal{K}(E,E)}{\underbrace{\underset{\in \mathcal{K}(E,E)}{\underbrace{A}} \underset{\in \mathcal{B}(E,E)}{\underbrace{A^{-1}}}}} = I.
	\]
	But $I \not \in \mathcal{K}(E,E)$ since $E$ $\infty$-dimensional. Just take an ON-sequence $(x_n)_{n=1}^{\infty}$ in $E$. Then \[
		x_n \rightharpoonup 0, \qquad \text{ in }E
	\]
	but $\norm{x_n}=1$ for all $n$ and if $I \in \mathcal{K}(E,E)$ then
	\[
		x_n = I(x_n) \to I(0) = 0, \qquad  \text{ in }E
	\]
	which implies that $\norm{x_n}\to 0$ for $n \to \infty$. Moreover (by Hilbert-Schmidt theorem) $(E, \skal{.}{.})$ complex Hilbert space, seperable and $\infty$-dim. $A \in \mathcal{K}(E,E)$ and self-adjoint it follows
	\[
		\Rightarrow \qquad (u_n)_{n=1}^{\infty} \text{ ON-basis for $E$ where}
	\]
	\[
		A(u_n) = \lambda_n u_n, \qquad n=1,2,\dots.
	\]
	($\lambda_n$ eigenvalue of $A$ with normalised eigenvector $u_n$) with
	\[
		\lim_{n \to \infty}\lambda_n = 0.
	\]
	For $x \in E$
	\[
		x = \sum_{n=1}^{\infty} \skal{x}{u_n}u_n
	\]
	and
	\[
		A(x) = \sum_{n=1}^{\infty}\lambda \skal{x}{\lambda_n}u_n
	\]
\end{enumerate}
\minisec{Fredholm Alternativ:}
$E,A$ as above. Then
\begin{enumerate}
	\item $x = A(x) + y$ is seperable for all $y \in E$. \\
	iff
	\item $x = A(x)$ has no non-trivial solution $x \in E$.
\end{enumerate}
Exactly one of the statements hold:
\begin{enumerate}
	\item (1) from above
	\item (2) has a non-trivial solution $x \in E$.
\end{enumerate}
In general (1) is seperable for $y \in E$ iff
\[
	y \in \set[x \in E]{A(x)=x}^{\perp}.
\]
If so: If $x$ is a solution to (1) then also $x + \tilde x$ is a solution to (1) where
\[
	\tilde x \in \set[x \in E]{A(x)=x}
\]
\begin{beweis}
	Look at (1). Let $(u_n)_{n=1}^{\infty}$ be the ON-basis from the previous theorem.
	\[
		x = \sum_{n=1}^{\infty}\skal{x}{u_n}u_n, \qquad y = \sum_{n=1}^{\infty}\skal{y}{u_n}u_n.
	\]
	\[
		A(x) = \sum_{n=1}^{\infty} \lambda_n \skal{x}{u_n}u_n.
	\]
	(1) taked the form
	\[
		\sum_{n=1}^{\infty} \left( \skal{x}{u_n} - \lambda_n \skal{x}{u_n} - \skal{y}{u_n} \right) u_n = 0.
	\]
	This implies
	\[
		(I- \lambda)\skal{x}{u_n}- \skal{y}{u_n} = 0, \qquad n=1,2,\dots.
	\]
	If $\lambda_n \neq 1$ then
	\[
		\skal{x}{u_n} = \frac{\skal{y}{u_n}}{1 - \lambda_n}.
	\]
	If $\lambda_n = 1$ then $-y$ must be orthogonal to every $u_n$ corresponding by the eigenvalue $1$.
	\[
		\sum_{n=1}^{\infty} \frac{\skal{y}{u_n}}{1- \lambda_n}u_n \in E
	\]
	since
	\[
		(\frac{\skal{y}{u_n}}{1- \lambda_n})_{n=1}^{\infty} \in l^2
	\]
	since
	\[
		\sup_{\substack{n \\ \lambda_n \neq 1}} \abs{\frac{1}{1- \lambda_n}} < \infty
	\]
	since
	\[
		\lim_{n \to \infty}\lambda_n = 0
	\]
	and
	\[
		(\skal{y}{u_n})_{n=1}^{\infty} \in l^2.
	\]
\end{beweis}
\newpage
\section{Boundary Value Problems for ODE's}
\label{sec:boundary_value_problems_for_ode_s}
Consider
\[
	(*)\qquad \begin{cases}
		Lu &=f \in C([0,1]) \\
		R_ju &= 0 \qquad j=1,2,\dots,n
	\end{cases}
\]
(homogenuous boundary conditions),
where
\[
	Lu := u^{(n)} + C_{n-1}(x)u^{(n-1)} + \dots + c_1(x)u' + c_0(x)u, \qquad u \in C^n([0,1])
\]
with
\[
	c_0(x),c_1(x), \dots, c_{n-1}(x) \in C([0,1])-
\]
\[
	R_j = \sum_{k=0}^{n-1} \left( \alpha_{jk} u^{(k)}(0) + \beta_{jk}u^{(k)}(1) \right), \qquad j=1,2,\dots,n
\]
with
\[
	\alpha_{jk}, \beta_{jk} \in \mathbb{C}, \qquad j=1,\dots,n, \qquad k=0,\dots,n-1
\]
Reformulate (*).
\[
	u(x) = \int_{0}^{1} \underset{\substack{\text{Green's function} \\ \text{for $L$ and $R_j$} \\ j=1,\dots,n}}{\underbrace{g(x,y)}}f(y) \,\mathrm{d}y \qquad \in C^n([0,1])
\]
and satisfies the boundary conditions $R_j = 0$ for $j=1,2,\dots,n$. \\
Consider the problem
\[
(**) \qquad 	\begin{cases}
		Lu &= f(x,u), \qquad x \in [0,1) \\
		R_ju &= 0, \qquad j=1,2,\dots,n.

	\end{cases}
\]
The reformulation above gives
\[
	u(x) = \int_{0}^{1}g(x,y)f(y,u(y)) \,\mathrm{d}y, \qquad x \in [0,1].
\]
To find a solution set
\[
	T(u)(x) = \int_{0}^{1}g(x,y)f(y,u(y)) \,\mathrm{d}y, \qquad x \in [0,1].
\]
\[
	T: C([0,1]) \to C([0,1])
\]
A fixed point to $T$ gives a solution to $(**)$. Note that if $u \in C([0,1])$ then
\[
	T(u) \in C^n([0,1])
\]
and satisfies $R_j=0$ for $j=1,2,\dots$. \\
Given $L$ and $R_j$ for $j=1,2,\dots,n$ find the corresponding Green's function.

\begin{beispiel}
	\[
		\begin{cases}
			Lu &= u''-u, \qquad \text{ on }[0,1]\\
			R_1u &=u(0)=0 \\
			R_2u &=u(1)=0
		\end{cases}
	\]
\end{beispiel}
	\begin{theorem}
		$Lu=f \in C([0,1])$, where
		\[
			Lu: = u^{(n)}+ c_{n-1}(x)u^{(n-1)} + \dots + c_1(x)u' + c_0(x)u
		\]
		and $\xi = (\xi_1, \dots, \xi_n) \in \mathbb{C}^n$. Then for $x_0 \in [0,1]$
		\[
			\Rightarrow \qquad \exists\,!\,\,u \in C^n([0,1]) \text{ with } Lu=f.
		\]
		and
		\[
			(u,u',\dots,u^{(n-1)}) \big|_{x_0}^{} = \xi.
		\]
	\end{theorem}
	\begin{beweis}
		Reformulate the problem as a system of first order differential equations.
	\[
		\begin{cases}
			Lu &=f \\
			(u,u',\dots,u^{(n-1)})  \big|_{x_0}^{} &= \xi
		\end{cases}
	\]
	corresponds to
	\[
		\begin{cases}
			\tilde u' = \tilde f \\
			\tilde u(x_0) = \xi
		\end{cases}
	\]
	and is equivalent to
	\[
		\tilde u(x) = \xi + \int_{x_0}^{x} \tilde f(s) \,\mathrm{d}s.
	\]
	$\tilde f$ contains $\tilde u$ implicitly. The statement of the proof follows from an application of Banach's fixed point theorem. (See course homepage and proof of picard's existence theorem.)
	\end{beweis}
Set
\[
 	\mathcal{N}(L) = \set[u \in C^n((0,1))]{Lu=0}
\]
\textbf{Claim:} \text{    }$\dim \mathcal{N}(L)=n$ \\
Set \[
	C_R^n([0,1]) = \set[u \in C^n((0,1))]{R_ju = 0, j = 1,2,\dots,n}
\]
and $L_0 = L  \big|_{C^n_R([0,1])}^{}$. Let $u_1,\dots,u_m \in \mathcal{N}(L)$

\begin{theorem}
	The following statements are equivalent. Let $u_1,\dots,u_n \in \mathcal{N}(L)$
	\begin{enumerate}
		\item $W(x) \neq 0$ for all $x \in [0,1]$.
		\item $W(x)\neq  0$ for some $x \in [0,1]$.
		\item $u_1,u_2, \dots,u_n$ is a basis for $\mathcal{N}(L)$.
	\end{enumerate}
	where
	\[
		W(x) = \det \left( \begin{pmatrix}
			u_1(x) & \dots & u_n(x) \\
			u'_1(x) & \dots & u'_n(x) \\
			\vdots & & \vdots \\
			u_1^{(n-1)} & \dots & u_n^{(n-1)(x)}
		\end{pmatrix} \right), \qquad x \in [0,1].
	\]
\end{theorem}
\begin{theorem}
	With the notation from above the following statements are equivalent.
	\begin{enumerate}
		\item $L_0 : C^n_R( [0,1]) \to C([0,1])$ is a bijection.
		\item $\det(R_ju_k)_{1 \leq j,k \leq n} \neq 0$.
	\end{enumerate}
\end{theorem}

\begin{beispiel}[continue]
	From the example above we get
	\[
		u_1(x) = e^x, \qquad u_2(x) = e^{-x}.
	\]
	\[
		u(x) = A e^x + B e^{-x}
	\]
	and
	\begin{align*}
		R_1u_1 &= u_1(0)=e^0 =1 \\
		R_1u_2 &= u_2(0)=e^0 =1 \\
		R_2u_1 &= u_1(1)=e \\
		R_2u_2 &= u_2(1)=\frac{1}{e}
	\end{align*}
	and
	\[
		\det(R_ju_k) = \det( \begin{pmatrix}
			1 & 1 \\ e & \frac{1}{e}
		\end{pmatrix}) = \frac{1}{e} - e \neq 0.
	\]
\end{beispiel}

\begin{theorem}
	Assume $u_1, \dots,u_n$ basis for $\mathcal{N}(L)$ and $\det(R_ju_k) \neq 0$. Set $G=L_0^{-1}$. \[
		\Rightarrow \qquad \exists\,!\, \text{continuous }g \in C([0,1] \times [0,1])
	\]
	such that
	\[
		G(f) = \int_{0}^{1}g(x,y)f(y) \,\mathrm{d}y
	\] is a solution of
	\[
		\begin{cases}
			Lu &= f \\
			R_ju &=0, \qquad j=1,\dots,n
		\end{cases}.
	\]
	Here
	\[
		g(x,y)= \underset{\equiv e(x,y)}{\underbrace{\left( \sum_{k=1}^{n}a_k(y)u_k(x) \right)}} \theta(x-y)+ \sum_{k=1}^{n}b_k(y)u_k(x).
	\]
	where
	\begin{align*}
		e^{(k)}_x (y,y) &= 0, \qquad k = 0,1,\dots,n-2 \\
		e^{(n-1)}_x(y,y) &=1
	\end{align*}
	Note
	\[
		Lu = 1 u^{(n)} + c_{n-1}u^{(n-1)} + \dots + c_0 u.
	\]
	and
	\begin{align*}
		R_j(g(.,y)) = 0, \qquad 0 < y < 1, \qquad j=1,2,\dots,n
	\end{align*}
\end{theorem}
Note
\begin{align*}
	\int_{0}^{1}g(x,y)f(y) \,\mathrm{d}y &= \int_{0}^{1}e(x,y) \theta(x-y)f(y) \,\mathrm{d}y + \int_{0}^{1} \sum_{k=1}^{n} b_k(y)u_k(x)f(y) \,\mathrm{d}y \\
	&= \underset{L[...]=f}{\underbrace{\underset{=L [...] = f}{\underbrace{\int_{0}^{x}\sum_{k=1}^{\infty}a_k(y)u_k(x) f(y) \,\mathrm{d}y}} + \underset{L[...]= 0}{\underbrace{\sum_{k=1}^{N} \int_{0}^{1}b_k(y)f(y) \,\mathrm{d}y u_k(x)}}}}
\end{align*}
Calculate $g(x,y)$ for $n=2$: \\
Set \begin{align*}
	e(x,y) = a_1(y)u_1(x)+ a_2(y)u_2(x)
\end{align*}
\[
	\begin{cases}
		e(y,y) &= a_1(y)e^y + a_2(y)e^{-y} = 0 \\
		e'_x(y,y) &= a_1(y)e^y - a_2(y) e^{-y}=1 \\
	\end{cases}
\]
So we get
\begin{align*}
	a_1(y) &= \frac{1}{2} e^{-y} \\
	a_2(y) &= - \frac{1}{2} e^{-y}
\end{align*}
and
\begin{align*}
	e(x,y) &= \frac{1}{2} e^{-y} e^x - \frac{1}{2} e^{y}e^{-x}  \\
	&= \frac{1}{2} (e^{x-y} - e^{y-x}), \qquad (x,y) \in [0,1] \times [0,1]
\end{align*}
Set
\[
	g(x,y) = e(x,y)\theta(x-y) + b_1(y)u_1(x)+ b_2(y)u_2(x)
\]
For $0<y<1$
\begin{align*}
	R_1g(.,y)=0, &\text{ i.e. }g(0,y)=0, \qquad \text{for }y \in (0,1), \\
	 &\text{ i.e. }b_1(y)u_1(0)+ b_2u_2(0) = 0 \text{ for }y \in (0,1), \\
	 &\text{ So $b_1(y) + b_2(y) = 0$}.
\end{align*}
\begin{align*}
	R_2g(.,y)=0, &\text{ i.e. }g(1,y)=0, \qquad \text{for }y \in (0,1), \\
	 &\text{ i.e. }e(1,y) + b_1(y)u_1(1)+ b_2(y)u_2(1) = 0 \text{ for }y \in (0,1), \\
	 &\text{ So }\frac{1}{2}\left( e^{1-y}- e^{y-1} \right) + b_1(y)e + b_2(y)e^{-1}=0 \text{ for }y \in (0,1).
\end{align*}
So we have in total
\[
	\begin{cases}
		b_1(y) + b_2(y) &= 0 \\
		\frac{1}{2}\left( e^{1-y}- e^{y-1} \right) + b_1(y)e + b_2(y)e^{-1} &=0
	\end{cases}.
\]
We obtain
\[
	\begin{cases}
		b_1(y) &= -b_2(y) \\
		b_2(y)\left(  e^{-1} - e \right) &= \frac{1}{2}\left( e^{y-1}- e^{1-y} \right)
	\end{cases}.
\]
So
\[
	b_2(y) = \frac{\frac{1}{2}(e^{y-1}-e^{1-y})}{\left( e^{-1}-e \right)} = \frac{1}{2} \frac{e^{1-y}-e^{y}}{e^2-1}
\]
and
\[
	b_1(y) = \frac{1}{2} \frac{e^y-e^{2-y}}{e^2 -1}.
\]
We obtain
\[
	g(x,y) = \frac{1}{2}(e^{x-y}- e^{y-x})\theta(x-t) + \frac{1}{2} \frac{e^{x+y-e^{x+2-y}}}{e^2-1} + \frac{1}{2} \frac{e^{2-y-x}-e^{y-x}}{e^2-1}.
\]

Question: $g(x,y)=g(y,x)$ for all $x,y \in [0,1]$? \\
In general, we say that $L_0 = L  \big|_{C^n_R([0,1])}^{}$ is \underline{symmetrie} if
\[
	\skal{L_0(u)}{v}_{L^2} = \skal{u}{L_0(v)}_{L^2}, \qquad \forall\, u,v \in C_R^n([0,1])
\]

\begin{beispiel}[continue]
	As above we have
	\[
		L(u) = u'' - u
	\] with boundary conditions
	\[
		u(0)= u(1)=0
	\]
	Set $u,v \in C_R^2([0,1])$
	\begin{align*}
		\skal{L_0(u)}{v}_{L^2} &= \int_{0}^{1} L_0(u)\bar{v} \,\mathrm{d}x  \\
		&= \int_{0}^{1}u'' \bar{v}- u \bar{v} \,\mathrm{d}x \\
		&= - \int_{0}^{1} u' \bar{v} + u \bar{v} \,\mathrm{d}x
		+ \underset{=u'(1)\underset{=0}{\underbrace{\bar{v}(1)}}-u'(0) \underset{=0}{\underbrace{\bar{v}(0)}}}{\underbrace{u'\bar{v}  \Big|_{0}^{1}}} \\
		&= - \int_{0}^{1}\left( u' \bar{v}' + u \bar{v} \right) \,\mathrm{d}x \\
		&= \int_{0}^{1} u( \bar{v}'' - \bar{v}) \,\mathrm{d}x \\
		&= \int_{0}^{1} u \overline{L_0v} \,\mathrm{d}x \\
		&= \skal{u}{L_0v}_{L^2}
	\end{align*}
\end{beispiel}


%!TEX root = ../../report.tex

\subsubsection{Fractals} % (fold)
\label{ssub:fractals}

A fractal is defined in \cite{Ebert2002} as ``a geometrically complex object, the complexity of which arises through the repetition of a given form over a range of scales''.
This concept is observed in some forms that exist in nature. From trees, mountains, coastlines to the network of neurons on a human cortex can be seen as examples of fractals. Natural shapes tend to be irregular and fragmented and exhibit a complexity incomparable to regular geometry \cite{mandelbrot1984fractal}.
In \cite{Ebert2002} is proposed to think of fractals as a new form of symmetry, \emph{Dilation Symmetry}, which is when an object is invariant over a change of scale. This invariance migth be only qualitatively and not exact. For instance, a river network exhibit dilation symmetry if \textit{zooming in} in some part looks the same as the whole image. As this example, many others show dilated symmetry. As clouds, tree branches and some vegetables as shown in Figure~\ref{fig:NFractals}. These examples are fractals.

\begin{figure}
        \centering
        \begin{subfigure}[b]{0.4\textwidth}
                \includegraphics[width=\textwidth]{img/Theory/Fractals/Leaf.png}
                \label{fig:Fleaf}
        \end{subfigure}%
        ~ %add desired spacing between images, e. g. ~, \quad, \qquad, \hfill etc.
          %(or a blank line to force the subfigure onto a new line)
        \begin{subfigure}[b]{0.4\textwidth}
                \includegraphics[width=\textwidth]{img/Theory/Fractals/Fractal_Broccoli.jpg}
                \label{fig:Fbrocoli}
        \end{subfigure}
        \caption{Fractals in Nature}\label{fig:NFractals}
\end{figure}

This idea was applied in maths with the evolution of a new area in this science called fractal mathematics. The objective of this field is to describe this very complex shapes. With really simple rules as repeating a substitution pattern.

\begin{figure}[htbp]
	\centering
	\includegraphics[width=0.9\textwidth]{img/Theory/Fractals/Fractal1_1000.png}
	\caption{Geometric Fractals}
	\label{fig:GFractals}
\end{figure}

In the Figure~\ref{fig:GFractals} there are four examples of Geometric Fractals, with the first five iterations of each one. All of them are built by the substitution of a part of the image by another one.

The example of the second row is known as the Koch snowflake. In this example, at each iteration, all the line segments are replaced by four segments with 1/3 of the size of the original one with the two in the middle being placed in a angle forming a equilateral triangle with the original line that is removed.

It's clear that the detail that is presented in each iteration increases as the scale changes. To try to mesure this evolution there is the idea of fractal dimension in which the detail in a pattern changes in comparison with the scale in which it is measured ($Fractal\_dimension$).

As stated before, the world is visualy very complex, so when synthetizing worlds, ``\emph{complexity} equals \emph{work}''\cite{Ebert2002}. This work can be done by the programer/artist or by a computer. Some cases are not easy to model procedurally. This is not the case of the fractals, as being defined as a simple mathematical function, it's relatively easy to implement a procedure to model fractals.

This techique is used to model many natural forms that present fractal properties. Mountains, as an instance, have been modeled with the use of fractals. Other natural forms that present fractal properties are trees, river systems, ligthning or vascular systems in living beings.

% subsubsection fractals (end)


\section{Uniform Convergence}
\begin{definition}
    A complex sequence $x_n$ is said to converge to a complex number $x$ if $\forall\epsilon>0,\exists N\in\mathbb N$ such that $\forall n>N$, $|x-x_n|<\epsilon$.
\end{definition}
\begin{definition}
    Let $S$ be a set and let $f_n:S\to\mathbb C$ be a sequence of functions.
    Let $f:S\to\mathbb C$ be a function.
    We say $f_n\to f$ pointwise if for any $x$, $f_n(x)\to f(x)$.
    In other words, $\forall x\in S, \forall\epsilon>0, \exists N\in\mathbb N, \forall n>N$, $|f(x)-f_n(x)|<\epsilon$.
\end{definition}
\begin{example}
    Let $S$ be the closed interval $[0,1]$ and $f_n(x)=x^n$, then $f_n\to f$ pointwise where
    $$f(x)=
    \begin{cases}
        1\textit{, if $x=1$}\\
        0\textit{, otherwise}
    \end{cases}$$
\end{example}
Note that in this example, despite the fact that all of $f_n$ are continuous, even smooth, the resulting limit $f$ needs not be continuous.\\
Here is another example:
\begin{example}
    Let $S=\mathbb R_{\ge 0}$ and let $f_n(x)=x^2e^{-nx}$, then $f_n\to 0$ pointwise, since
    $$0\le |f_n(x)|=\frac{x^2}{e^{nx}}= \frac{x^2}{1+nx+\frac{n^2x^2}{2}+\frac{n^3x^3}{6}\cdots}\le\frac{x^2}{nx}=\frac{x}{n}\to0$$
    as $n\to\infty$.
\end{example}
There is another form of convergence, called uniform convergence, which is defined as follows:
\begin{definition}
    Let $S$ be a set and let $f_n:S\to\mathbb C$ be a sequence of functions.
    Let $f:S\to\mathbb C$ be a function.
    We say $f_n\to f$ uniformly if $\forall\epsilon>0, \exists N\in\mathbb N, \forall n>N, \forall x\in S, |f(x)-f_n(x)|<\epsilon$.
\end{definition}
Note that the only difference between pointwise and uniform convergence is that the large integer $N$ does not depend on $x$ if the convergence is uniform.
Note also that uniform convergence implies pointwise convergence, but not the other way around.
Although it does not seem to be such a great difference in definition, in practice, it makes all the difference in the world.
\begin{proposition}
    The sequence in the first example, i.e. $f_n:[0,1]\to\mathbb R$ with $f_n(x)=x^n$, does not converge uniformly.
\end{proposition}
\begin{proof}
    We can take for example $\epsilon=1/2$. Then for any $n\in\mathbb N$, we can take $x=\sqrt[n]{2/3}$, so that we have
    $$|f_n(x)-f(x)|=|f_n(x)|=\frac{2}{3}>\frac{1}{2}$$
    So the claimed $N$ does not exist.
    Therefore the sequence $f_n$ does not converge uniformly.
\end{proof}
\begin{proposition}
    The sequence in the second example, i.e. $f_n=\mathbb R_{\ge 0}\to\mathbb R$ where $f_n(x)=x^2e^{-nx}$, converge absolutely.
\end{proposition}
\begin{proof}
    Note that
    $$0\le f_n(x)=\frac{x^2}{e^{nx}}=\frac{x^2}{1+nx+\frac{n^2x^2}{2}+\cdots}\le\frac{x^2}{n^2x^2/2}=\frac{2}{n^2}$$
    Therefore for any $\epsilon>0$, we can take $N=\lceil\sqrt{2/\epsilon}\rceil$, so for any $x\ge 0, n>N$, we ahve
    $$|f(x)-f_n(x)|=|f_n(x)|=f_n(x)\le\frac{2}{n^2}<\frac{2}{N^2}\le\frac{2}{(\sqrt{2/\epsilon})^2}=\epsilon$$
    So $f_n\to 0$ uniformly.
\end{proof}
In fact, although continuous functions may not converge pointwise to a continuous function, they do converge uniformly to one.
\begin{theorem}
    Let $S\subset\mathbb C$ be open.
    Suppose that $f_n:S\to\mathbb C$ is a sequence of continuous functions.
    If $f_n\to f$ uniformly, then $f$ is continuous as well.
\end{theorem}
\begin{proof}[Informal sketch]
    Idea: Transfer the nice property of $f_n$ to $f$.
    Choose large enough $N$ such that $f_n-f$ is arbitratily small for all $n>N$.
    We can always choose $x'$ close to $x$ where $f_n(x)$ close to $f(x)$.
    Then just use triangle inequality.
    "3-$\epsilon$ proof"
\end{proof}
\begin{proof}
    $\forall\epsilon>0$, we can choose large enough $N$ such that $\sup|f_n-f|<\epsilon/3$
    We can choose $\delta>0$ such that $|x-x'|<\delta\implies |f_n(x)-f_n(x')|<\epsilon/3$.
    $$|f(x)-f(x')|\le |f(x)-f_n(x)|+|f(x')-f_n(x')|+|f_n(x)-f_n(x')|<3\frac{\epsilon}{3}=\epsilon$$
    As desired.
\end{proof}
\begin{remark}
    1. We can use this theorem to show that $x^n$ as in the previous example does not converge uniformly.\\
    2. It is not true that differentiability is preserved under unform convergence.
\end{remark}
\begin{theorem}
    Let $f_n:[a,b]\to\mathbb R$ be all Riemann integrable.
    Then if it converges uniformly, its limit is also Riemann integrable.
    Furthermore,
    $$\int_a^b\lim_{n\to\infty} f_n(x)\,\mathrm dx=\lim_{n\to\infty}\int_a^b f_n(x)\,\mathrm dx$$
\end{theorem}
Recall that a function is Riemann integerable if and only if the upper and lower sums of $f$ on the interval can be arbitratily close.
\begin{proof}
    Firstly $f$ is bounded. Since $f_n$ are bounded, we can just choose large enough $n$ such that $|f_n-f|<1$ and $|f_n|<M$, then
    $|f|\le|f-f_n|+|f_n|<\epsilon+M<M+1$ so $f$ is bounded.\\
    For $\epsilon>0$ choose $N$ such that $\sup|f_n-f|<\epsilon/(3(b-a))$ for any $n>N$.
    Since $f_n$ is integrable, there is some disection $D$ of the interval $[a,b]$ such that $U_D(f_n)-L_D(f_n)<\epsilon/3$.
    We have
    $$|L_D(f)-L_D(f_n)|=\sum_{(x_i)\in D}\left|\inf_{x\in[x_i, x_{i+1}]}f(x)-\inf_{x\in[x_i, x_{i+1}]}f_n(x)\right|(x_{i+1}-x_i)<\epsilon/3$$
    Similarly $|U_D(f)-U_D(f_n)|<\epsilon/3$.
    So
    \begin{align*}
        |U_D(f)-L_D(f)|&\le|U_D(f)-U_D(f_n)|\\
        &+|L_D(f_n)-L_D(f)|+|U_D(f_n)-L_D(f_n)|\\
        &<3\epsilon/3=\epsilon
    \end{align*}
    This shows that $f$ is integrable.
    Finally, we have
    $$|\int_a^bf(x)-f_n(x)\,\mathrm dx|\le \int_a^b\sup|f(x)-f_n(x)|\,\mathrm dx<\epsilon/3<\epsilon$$
    which completes the proof.
\end{proof}
\begin{remark}
    1. For uniform convergence, we can swap the integral and the limit.\\
    2. If $f_n\to f$ uniformly and that all $f_n$ is bounded, then $f$ is bounded.
\end{remark}
\begin{corollary}
    For uniform convergence, we can swap infinite sums and integral.
    That is, if $f_n:[a,b]\to\mathbb R$ is a sequence of integrable functions whose partial sum converges absolutely to some function $f$, then $f$ is integrable and
    $$\int_a^bf(x)\,\mathrm dx=\sum_{n=1}^\infty\int_a^bf_n(x)\,\mathrm dx$$
\end{corollary}
\begin{proof}
    Let
    $$F_n(x)=\sum_{k=1}^nf_k(x)$$
    so $F_n$ are integrable and $F_n\to f$ uniformly.
    Then we can just apply the preceding theorem.
\end{proof}
\begin{theorem}
    Let $f_n:[a,b]\to\mathbb R$ be continuously differentable on $[a,b]$.
    Assume that the sequence of partial sums of $f^\prime_n$ at every point converges uniformly.
    And there is an $c\in[a,b]$ such that
    $$\sum_{n=1}^\infty f_n(c)$$
    converges, then the sequence of partial sums of $f_n$ converges uniformly.
    Furthermore, the limit $f$ is continuously differentiable and
    $$f^\prime(x)=\sum_{n=1}^\infty f^\prime_n(x)$$
\end{theorem}
\begin{proof}[Sketch of proof]
    Let
    $$F_n(x)=\sum_{k=1}^nf_k(x), g(x)=\sum_{n=1}^\infty f^\prime_n(x)$$
    So we want to find a particular solution to the differential equation $f^\prime=g$, and show that $F_n$ converges unformly to it.
    So basically we want to do
    $$f(x)=\int_c^x g(t)\,\mathrm dt+\sum_{n=1}^\infty f_n(c)=\lim_{n\to\infty}F_n(x)=\sum_{n=1}^\infty f_n(x)$$
    rigorously and it would be done.
\end{proof}
\begin{proof}
    Let
    $$g(x)=\sum_{n=1}^\infty f^\prime_n(x)$$
    $g$ is continuous and hence Riemann integrable on $[a,b]$.
    Define $f:[a,b]\to\mathbb R$ by
    $$f(x)=\int_c^xg(t)\,\mathrm dt+\lambda$$
    where
    $$\lambda=\sum_{n=1}^\infty f_n(c)$$
    By FTC, $f$ is differentiable and $f^\prime(x)=g(x)$.
    Since $g$ is continuous, $f\in\mathcal C^1([a,b])$.
    It remains to show that the series sum of $f_n(x)$ converges uniformly to $f(x)$.
    Let $F_n(x)$ be the partial sum of the series, then by estimating its difference with $f$ and the fact that the partial sum of derivatives of $f_n$ converges uniformly (use FTC again), we can show that $F_n\to f_n$ uniformly.
    [Write details later]
\end{proof}
\begin{definition}
    Let $f_n$ be a sequence of scalar function on a set $S$.
    We say $f_n$ is uniformly Cauchy on $S$ if $\forall\epsilon>0,\exists N\in\mathbb N,\forall x\in X,\forall n,m>N$,
    $$|f_n(x)-f_m(x)|<\epsilon$$
\end{definition}
\begin{theorem}[General Principle of Uniform Convergence]\label{GP_UnifConv}
    A sequence of uniformly Cauchy scalar functions $f_n$ on $S$ converges uniformly.
\end{theorem}
\begin{proof}
    Firstly, we shall find a pointwise limit $f$ of $f_n$.
    The existence of $f$ is immediate since $f_n(x)$ is always Cauchy (hence converges) with $x$ fixed.\\
    Then we shall show that this convergence is uniform.
    Choose any $\epsilon>0$, $\exists N\in\mathbb N,\forall x\in X,\forall n,m>N,|f_n(x)-f_m(x)|<\epsilon/2$.
    Now we fix $x\in S,n>N$, since $f_n\to f$ pointwise, we can choose $m>N$ with $|f_m(x)-f(x)|<\epsilon/2$, then
    $$|f(x)-f_n(x)|\le |f(x)-f_m(x)|+|f_m(x)-f_n(x)|<2\epsilon/2<\epsilon$$
    So $f_n\to f$ uniformly.
\end{proof}
So what we did is to fix the $x$ and the $n$, then let that $m$ tend to infinity, then we can use the pointwise convergence to give the result.
This is how we get pass the dependence of $N$ on $x$ in the pointwise convergence result.
\begin{corollary}
    Let $f_n$ be a sequence of scalar functions on $S$, let
    $$\sum_{n=1}^\infty M_n$$
    be convergent with $M_n\ge 0$.\\
    If $\sup |f_n|\le M_n$ for any $n$, then
    $$\sum_{n=1}^\infty f_n$$
    converges uniformly.
\end{corollary}
\begin{proof}
    Let $F_n$ be the partial sum of $f_n$, $F_n$ is uniformly Cauchy due to the convergence of the series of $M_n$.
    Essentially, $\forall\epsilon>0,\exists N\in\mathbb N,\forall n,m>N$,
    $$\sum_{k=n+1}^mM_k<\epsilon$$
    so
    $$\sum_{k=n+1}^m|f_n(x)|<\epsilon$$
    Therefore it is unifomly Cauchy, so it converges uniformly.
\end{proof}
Now we consider the power series
$$\sum_{n=0}^\infty a_n(z-a)^n$$
$(a_n)_0^\infty$ be a sequence of complex number.
Let $R$ be the radius of convergence.
Now on the disk $|z-a|<R$, we consider
$$f(z)=\sum_{n=0}^\infty a_n(z-a)^n$$
The question is: is the convergence uniform?
\begin{example}
    Consider
    $$f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}$$
    where $R=1$.
    It does not converge uniformly.
    Indeed, the $N^{th}$ partial sum is bounded by $N+1$ but $1/(1-z)$ is unbounded.
\end{example}
\begin{theorem}
    For any $r$ with $0<r<R$, the series converges uniformly on $D(a,r)$.
\end{theorem}
\begin{proof}
    For $w\in\mathbb C$ such that $r<|w-a|<R$, there is an $M$ such that $|a_n(w-a)^n|<M$ for some $M>0$ and any $n$.
    We have $|z-a|/|w-a|<1$ for any $z\in D_(a,r)$, hence by taking $M_n=M(r/|w-a|)^n$ shows the result.
\end{proof}
The derivative of a power series (we can prove that it is complex differentiable, and we can do it term-by-term) has the same radius of convergence.
\begin{remark}
    If we fix $w\in D(a,R)$, we can choose $r$ such that $|w-a|<r<R$.
    Fix any $\delta>0$ such that $|w-a|+\delta<r$, then $D(w,\delta)\subset D(a,r)$, so
    $$\sum_{n=0}^\infty a_n(z-a)^n$$
    converges uniformly on $D(w,\delta)$.
    We say it is locally uniformly on $D(a,R)$.
\end{remark}


\section{Direct Limits}\label{limits}
Goals are UCT (universal coefficient theorem), which is about $ H_\ast(X;M)$ for varying $M$, the K\"{u}nneth theorem (which is about $ H_\ast(X\times Y)$), cohomology, and Poincar\'{e} duality.
\subsection{Two more things about $\Tor$}
If $R$ is a PID, then there's not so much to say about $\Tor$, because any submodule of a free module is free. So that means that any $R$-module has a free resolution $0\to F_1=\ker(f)\to F_0\xrightarrow{f}\to N\to 0$. It follows that $\Tor^R_n(M,N)=0$ for $n>1$. If you have a field, then tensoring is exact, so $\Tor^k_0(M,N)=0$ for $n>0$. That's why it's easy to work with fields. (There's also Pr\"ufer rings, where every module is flat.) By the way, this means that if you have a sexseq $0\to A\to B\to C\to 0$, then over a PID $R$, there's a six-term exact sequence $0\to\Tor^R_1(M,A)\to \Tor^R_1(M,B)\to \Tor^R_1(M,C)\to M\otimes_R A\to M\otimes_R B\to M\otimes_R C\to 0$.
\begin{example}
I want to give an example when you do have higher $\Tor$. Let $k$ be a field. Let $R=k[e]/(e^2)$. This is sometimes called the ``dual numbers'', or the exterior algebra over $k$. We're going to consider $R$-modules. Let's construct a projective resolution of $k$. Hmm. What is an $R$-module $M$? It's just a $k$-vector space $M$ with an operator $d$ that has action given by multiplication by $e$, and this satisfies $d^2=0$. And this is a chain complex! I guess it's not quite a chain complex because for us chain complexes are graded. So I guess it's an ungraded chain complex. I can consider $ H(M;d):=\ker d/\img d$. Here's an example of an $R$-module. There's the augmentation $R\to k$ sending $e\mapsto 0$. This makes $k$ an $R$-module, where $d=0$. Let's construct a free resolution of $k$.

Here we go. We're going to write $k=\bullet(=1)$ and $R:=(1=)\bullet\xrightarrow{d}\bullet(=e)$. Well, we have $R\to k$ given by $(\bullet\xrightarrow{d}\bullet)\to \bullet$. The kernel is not free, so we get $(\bullet\xrightarrow{d}\bullet)\to (\bullet\xrightarrow{d}\bullet)\to \bullet$. And this continues, so we get a projective resolution $\cdots\xrightarrow{e} R\xrightarrow{e} R\xrightarrow{e} R\to k$. What is $\Tor$? Well, $\Tor^R_\ast(M,k)$ is the homology of the following chain complex $\cdots\xrightarrow{d} M\xrightarrow{d} M\to 0$. What is that homology? Clearly $\Tor^R_0(M,k)=M\otimes_R k=M/dM=M/eM$. This is often called the module of indecomposables. And, $\Tor^R_n(M,k)= H(M;d)$.
\end{example}
Last comment about $\Tor$ is that there's a symmetry there. Of course, $M\otimes_R N\cong N\otimes_R M$. This uses the fact that $R$ is commutative. This leads right on to saying that $\Tor^R_n(M,N)\cong \Tor^R_n(N,M)$. We've been computing $\Tor$ by taking a resolution of the second variable. But I could equally have taken a resolution of the first variable. This follows from the fundamental theorem of homological algebra.
\subsection{Direct limits}
A long time ago, I said what a poset was. Let me tell you a joke about posets. I was at a conference, and Quillen was giving a talk. He was a student of Raoul Bott. Quillen was giving his talk, and he used the term ``poset''. This word was invented by Garrett Birkhoff(?). Then Bott objected and said ``What is this crazy word?'', and Quillen responded ``What are you talking about? Your colleague invented it!''. Anyway, it's not a joke. I guess it's just a piece of MIT and Harvard rivalry.

Anyway, a poset is a small category $\cI$ such that $\#\cI(i,j)\leq 1$ and isomorphism implies identity. I want to talk about a \emph{directed set}.
\begin{definition}
A poset $(\cI,\leq)$ is \emph{directed} if, for every $i,j$, there exists a $k$ such that $i\leq k$ and $j\leq k$.
\end{definition}
\begin{example}
For example, the natural numbers $\Z_{\geq 0}$ with equality. Another example: if $X$ is a space and $I$ is the set of open subsets of $X$. It's directed by saying that $U\leq V$ if $U\subseteq V$. This is because $U,U^\prime$ need not be comparable, but $U,U^\prime\subseteq U\cup U^\prime$. Another example is $\Z_{>0}$ where $i\leq j$ if $i|j$. This is because $i,j|(ij)$.
\end{example}
\begin{definition}
Let $\cI$ be a directed set. An $\cI$-directed diagram in $\cc$ is a functor $\cI\to\cc$. This means that for every $i\in \cI$, there is $X_i\in\cc$, and for every $i\leq j$, there's a map $X_i\xrightarrow{f_{ij}} X_j$ (and similarly for composition).
\end{definition}
\begin{example}\label{linear}
If $\cI=(\Z_{\geq 0},\leq)$, then you get $X_0\xrightarrow{f_{01}}X_1\xrightarrow{f_{12}}X_2\to\cdots$. This is the most important.
\end{example}
\begin{example}
Suppose $\cI=(\Z_{>0},|)$, i.e., the third example above. You can consider $\cI\to\mathbf{Ab}$, say assigning to each $i$ the integers $\Z$, and $f_{ij}:\Z\xrightarrow{j/i}\Z$. You get the picture.
\end{example}
These directed systems are a little complicated. But there's a simple one, namely the constant one.
\begin{example}
Let $\cI$ be any directed set. You have a constant functor $c_A:\cI\to\cc$ at some $A\in\cc$.
\end{example}
Of course, $\cI$-directed systems in $\cc$ are functors $\cI\to\cc$. They have natural transformations, and those are the morphisms in the category of $\cI$-directed systems. That just means that if I have two directed systems $X,Y:\cI\to\cc$, then a map from one to the other is a commuting diagram:
\begin{equation*}
\xymatrix{X_i\ar[r]\ar[d]^{g_i} & X_j\ar[d]^{g_j}\\
Y_i\ar[r] & Y_j}
\end{equation*}
for all $i\leq j$.

It'd be great if every directed system was constant, but this isn't true. This leads to direct limits.
\begin{definition}
A direct limit is an object $L$ and a map $X\to c_L$, which is initial among maps to constant systems. This means that I have some other map $X\to c_A$, then there's a unique induced map $c_L\to c_A$ that is induced from a map $L\to A$. We write $\varinjlim_{i\in \cI}X_i=\colim_{i\in I}X_i$. (My own note: it's also sometimes called an an inductive limit.)
\end{definition}
This is a universal property. So two different direct limits are canonically isomorphic.
\begin{example}
Consider $\cI=(\Z_{\geq 0},\leq)$, then you get $X_0\xrightarrow{f_{01}}X_1\xrightarrow{f_{12}}X_2\to\cdots$ in $\mathbf{Top}$. What is the direct limit? It's going to be $\bigcup_i X_i$. But what's the topology? Give it the finest topology so that all of the maps to the union are open. This just means that a subset is open in $\bigcup_i X_i$ if the preimage is open.
\end{example}
\begin{example}
Recall our example where we let $\cI=(\Z_{>0},|)$, i.e., the third example above. You can consider $\cI\to\mathbf{Ab}$, say assigning to each $i$ the integers $\Z$, and $f_{ij}:\Z\xrightarrow{j/i}\Z$. The colimit is $\QQ$, where you send $X_n\to\QQ$ via $1\mapsto \frac{1}{n}$.
\end{example}
\begin{lemma}
Let $X:\cI\to\mathbf{Ab}$ (or $\mathbf{Mod}_R$). A map $f:X\to c_L$ is the direct limit (we write $f_i:X_i\to L$) if and only if:
\begin{enumerate}
\item For every $x\in L$, there exists an $i$ and an $x_i\in X_i$ such that $f_i(x_i)=x$.
\item Let $x_i\in X_i$ be such that $f_i(x_i)=0$ in $L$. Then there exists some $j\geq i$ such that $f_{ij}(x_i)=0$ in $X_j$.
\end{enumerate}
\end{lemma}
\begin{proof}
Straightforward.
\end{proof}
It generalizes the observation that $\QQ$ is the colimit of the diagram we drew above for $\cI=(\Z_{>0},|)$.
\begin{corollary}
The direct limit $\varinjlim_I:\Fun(\cI,\mathbf{Ab})\to\mathbf{Ab}$ is exact. In other words, $X_\bullet\to Y_\bullet\xrightarrow{p} Z_\bullet$ is an exact sequence of $\cI$-directed systems (at every degree, we get an exact sequence of abelian groups), then $\varinjlim_IX_\bullet\xrightarrow{i} \varinjlim_IY_\bullet\xrightarrow{p} \varinjlim_IZ_\bullet$.
\end{corollary}
\begin{proof}
First of all, $X_\bullet\to Z_\bullet$ is zero. Thus it factors through the constant zero object, so that $\varinjlim_IX_\bullet\to \varinjlim_I Z_\bullet$ is zero. Let $y\in \varinjlim_IY_\bullet$, and suppose $y$ maps to $0$ in $\varinjlim_IZ_\bullet$. By the first condition, there exists $i$ such that $y=f_i(y_i)$ for some $y_i\in Y_i$. Then $p(y)=f_ip(y_i)$ because $p$ is a map of systems. This is zero. This means that there is $j\geq i$ such that $f_{ij}p(y_i)=0$. We have an element in $Y_j$ that maps to zero, so there is some $x_j$ that is the preimage of the element in $Y_j$. So we're done.
\end{proof}
This makes the world a nice place to live.