MATH 206 Notes: Functional Analysis

Fall 2024
By Aathreya Kadambi
Expanding on Lectures by Professor Algebraic Topology

I’m taking algebraic topology by Professor Zworski!

Story 1: Linear Algebra

Relevant Lectures: Lecture 1

We started out by defining cosets and quotient spaces, and then the first isomorphism theorem for vector spaces. Hahaha this is a homework problem I did for COMPSCI 189 yesterday!

We study the rest of mathematics because linear algebra is too hard. - Taylor, according to Zworski.

Why is this interesting? Suppose we start with $P : H_1 \rightarrow H_2$ and we can find $R_{-} : H_- \rightarrow H_2$ and $R_+ : H_1 \rightarrow H_+$ such that $\begin{pmatrix} P & R_{-} \\ R_+ & 0 \end{pmatrix} : H \oplus H_- \rightarrow H\oplus H_+$ is invertible and $\begin{pmatrix} P & R_- \\ R_+ & 0 \end{pmatrix}^{-1} = \begin{pmatrix} E & E_+ \\ E_- & E_{-+}\end{pmatrix} : H_2 \oplus H_+ \rightarrow H_1 \oplus H_-$ then $P$ is invertible if and only if the stuff from the theorem. Then we have reduced the invertibility of $P$ to something else. It is used in linear algebra for reducing huge systems in linear algebra to much smaller finite ones.

This problem of something is called a Grushih problem. “The key here is stability under perturbations and so on”. Although I have no clue what perturbations he is talking about.

Story 2: Dimension

Relevant Lectures: 1

Dimension of $V$ being $n$ is the same as saying there exists a corresponding basis of $n$ vectors for $V$. This is for finite idmensional vector spaes, and if this is not possible, we say that the dimension is infinite. Hormander actually has another different amusing definition of dimension for vector spaces.

Equivalently, $\dim$ is a functor from the category of vector spaces to $\N$ such that

• $\dim K^n = n$,
• $T : V_1 \rightarrow V_2$ is surjective is equvialent to $\dim V_1 \ge \dim V_2$,
• $T : V_2 \rightarrow V_1$ is injective is equivalent to $\dim V_1 \le \dim V_2$. Zworski kept laughing while writing this.

We say: $\text{Rank }T = \dim \text{Im }T = \text{codim } \ker T$ where $\text{codim } W = \dim V/W$ with $W \subseteq V$.

Proof.

Since $\dim W \le \dim V$ and $\text{codim } W \le \dim V$, we can assume that $\dim W$ and $\text{codim } W$ are finite, because otherwise we just have infinities summing together.

$T_1 : K^n \rightarrow V$ and $T_2 : K^m \rightarrow V/W$ bijection. There exists $\tilde{T}_2 : K^m \rightarrow V$ such that $\tilde{T}_2 \circ \pi = T_2$, $\pi : V \rightarrow V/W$. The tilde mappings are called “lifts” of the non-tilde mappings. $T_2 e_j = x_j + W, \tilde{T}_2(\sum a_j e_j) := \sum a_j x_j$ $e_j$ a basis of $K^m$.

$T: T_1 \oplus \tilde{T}_2 : K^{n+m} \ni (a,b) \mapsto T_1 a + \tilde{T}_2 b \in V$

If $\dim V_j < \infty$, then we define the index: $\text{ind}(T) L= \dim \ker T - \text{codim } \text{Im }T = \dim V_1 - \dim V_2$

Story 3: Index of a Transformation

Relevant Lectures: Lecture 1, Lecture 2

Suppose $\dim \ker T < \infty$ or $\dim (V_2/\text{Im }T) < \infty$. Then $\text{ind}(T) = \dim \ker T - \text{codim } \text{Im }T$

$0 \rightarrow V_0 \xrightarrow{T_0} V_1 \xrightarrow{T_1} \dots \xrightarrow{T_{N-1}} V_N \rightarrow 0$ is called a complex if $T_{k+1}T_k = 0$ or $\ker T_{k+1} \supseteq \text{Im }T_k$.

An example of a complex from MATH 53:

A complex is called exact if $\text{Im }T_k = \ker T_{k+1}$.

This result will allow us to show Theorem 2 by constructing an appropriate exact complex.

Proof.

Let $R_{k+1} = \ker T_{k+1} = \text{Im }T_k$. Then $\dim V_k = \dim R_k + \dim R_{k+1}$ Now, $\sum \dim V_{2k} = \sum \dim R_{2k} + \dim R_{2k+1} == \sum \dim V_{2k+1}$

Story 4: Fedholm Operations

Recall that an operator $T : V_1 \rightarrow V_2$ is Fredholm when $\dim \ker T < \infty$ and $\dim \text{coker } T < \infty$.

$\text{ind}(T) = \dim \ker T - \dim \text{coker } T$

\dim \ker T_n = \left\{ \begin{array}\{0\} & n \le 0 \\ n & n > 0 \end{array} \right.

Last time, we said $V_1 \supseteq V$ and $x \not\in V_1$ implies there is a hyperplane $V_2$ such that $V_2$ contains $V_1$ and $x$ isn’t in $V_2$.

Proof Sketch.

Take $x \in W_1 \backslash \{0\}$. There is a hyperplane $H_1$ such that $x \not \in H_1$, so that $H_1 + W_1 = V$, and so

$\dim (W_1 \cap H_1) + \text{codim }H_1 = \text{codim }(W_1 + H_1) + \dim W_1$ so $\dim(W_1 \cap H_1) = \dim W_1 - 1$. So now repeating this process, we can find: $\dim(W_1 \cap H_1 \cap H_2 \cap ... \cap H_k) = \dim W_1 - k$ for $0 \le k \le \dim W_1 = d$, so that $W_2 = H_1 \cap ... \cap H_d$. Now we get: $\dim (W_1 \cap W_2) + \text{codim }W_2 = \text{codim}(W_1 + W_2) + \dim W_1$ and now $\dim (W_1 \cap W_2) = 0$. Also, $\text{codim}(H_1 \cap H_2) \le \text{codim}(H_1 \cap H_2) + \text{codim}(H_1 + H_2) = \text{codim}(H_1) + \text{codim}(H_2)$ so inductively, $\text{codim }W_2 \le \sum_{j=1}^d \text{codim }H_j = d$ and also $\text{codim }W_2 \ge \dim W_1 = d$ so $\text{codim }W_2 = d$.

For the first direction, we use Theorem 1 to get a $W_2$ satisfying $\ker T \cap W_2 = 0$, $W_2 + \ker T = V$, $\dim \ker T = n_+$. Lte $x_1,\dots,x_{n_+}$ be a basis of $\ker T$. There exists $L_j : V_1 \rightarrow K$ such that $L_j(x_i) = \delta_{ij}$, $y \in V_1$, $y = \sum a_j x_j + \overline{y}$, $\overline{y} \in W_2$, $L_j(y) = a_j$.

$V_+ := K^{n_+}$, $R_+ : V_1 \rightarrow K^{n_+}$. $y \mapsto (L_1(y),...,L_{n_+}(y))$.

Choose a basis of $V_2/\text{Im }T$.

Similarly, we do $V_= := K^{n_-}$, and $(b_1,...,b_{n_-}) \mapsto \sum_{j=1}^{n_-} b_j y_j$.

We now wish to show that $R_+$ is surjective. So

$\begin{pmatrix} T & R_{-} \\ R_{+} & 0\end{pmatrix} : V_1 \oplus V_- \rightarrow V_2 \oplus V_+$ is surjective. So what we have shown (I’ve omitted most of the proof about $Tu + \sum_{j=1}^n b_j y_j = 0$ and stuff) is that if the operator is Fredholm, then we can set up this thing so that it is invertible. We also get that $E_{-+} : K^{n_+} \rightarrow K_{n_-}$ is clearly Fredholm since $n_+$ and $n_-$ are finite.

To show the other direction, first we observed that $R_+$ is surjective and $R_-$ is injective by actually multiplying out $\begin{pmatrix}T & R_- \\ R_+ & 0 \end{pmatrix}\begin{pmatrix} E & E_+ \\ E_- & E_{-+} \end{pmatrix}$ and $\begin{pmatrix} E & E_+ \\ E_- & E_{-+} \end{pmatrix}\begin{pmatrix}T & R_- \\ R_+ & 0 \end{pmatrix}.$

Then, we took $\begin{pmatrix} u \\ u_-\end{pmatrix} \in V_1 \oplus V_-$ and $\begin{pmatrix} v \\ v_+\end{pmatrix} \in V_2 \oplus V_+$, and saw that $Tu = v$, $u_- = 0$, $Ev + E_+ v_+ = u$ and $E_- v + E_{-+} v_+ = 0$.

Since $v$ is in the image of $T$, $E_-v$ is in the image of $E_{-+}$. This allows us to make a well defined map from $E_- : \text{Im }T \rightarrow \text{Im }E_{-+}$:

$E_-^\# : V_2/\text{Im }T \rightarrow V_- / \text{Im }E_{-+}$ Then by playing around with the above results, we get injectivity. It is also surjective, so the dimensions of $V_-/\text{Im }E_{-+}$ and $V_2/\text{Im }T$ are isomorphic.

Now we can define a map $E_+ : \ker E_{-+} \rightarrow \ker T$. We get $E_+$ is injective on all of $V_+$ and thus on $\ker E_{-+}$ from the fact that $R_+ E_+$ is identity. Surjectivity comes from the same two equations in the iff statement.

The main helpful idea was that $Tu = v, u_- = 0$ was equivalent to $Ev + E_+ v_+ = u$ and $E_- v + E_{-+} v_+ = 0$.

We can purturb our operator by a finite rank operator and stuff.

Proof similar to above.

We can write $V_1 \simeq \ker T \oplus W_1$ and $V_2 \simeq W_2 \oplus \text{Im }T$, so that the matrix of $T$ is

$T = \begin{pmatrix} 0 & 0 \\ 0 &\tilde{T} \end{pmatrix}$ $S = \begin{pmatrix} \tilde{S} & 0 \\ 0 & 0 \end{pmatrix}$ with $\tilde{T} : W_1 \rightarrow \text{Im }T$ and $\tilde{S} : \ker T \rightarrow W_2$ invertible.

Convex Sets

Now we take $K = \R$.

A convex set $A \subseteq V$, $K = \R$ is convex if for all $x, y\in A$, $\R \supseteq \{t \in \R : x + ty \in A\}$ is an interval in $\R$.

The reason we define it weirdly like this as opposed to the canonical definition is that we want to say that a convex set is linearly open if that interval is open.

Hahn-Banach Theorem

This one was taught by Zhongkai.

Proof.

First looked at $V = \R^2$ and made observation that for any $\in V$, half line $\{t x : t\ ge 0\}$ if it intersects $A$, then $\{tx : t\le 0\} \cap A = \varnothing$ (or else $0 \in A$ by convexity).

He did a ton of stuff and then had to use Zorn’s Lemma.

Now we will stop this topic for a while and go back to Hahn Banach with a version in topology or something.

Metrics and Topologies

Defined topological space and topological structure. Defined Metric space as well, and how metric space induces a topology.

Defined closed and also what it means for a set to be complete.

A set $A\subseteq E$ is called nowhere dense if its closure $\overline{A}$ doesn’t have interior. A set is called first category if it is a countable union of nowhere dense sets. A set is called second category if it is NOT first category.

Baire Category Theorem

$(E,d)$ is a complete metric space, $\{U_n\}_{n=1}^\infty$ countable family of open dense sets. Then $\cap_{n=1}^\infty U_n$ is dense in $E$.

Definition. $A \subseteq E$ is called generic if it contains the countable intersection of open dense sets.

When discussing something, he did a proof by picture where he drew a zig zaggy function with slope on any interval being either $2n$ or $-2n$.

Tychonov Theorem

Locally convex, balanced, absorbing.

• $M \subseteq V$ is called balanced if for all $x \in M$, $|a| \le 1$, $ax \in M$.
• $M\subseteq V$ is called absorbing if for all $x \in V$, $|a|$ sufficiently small, $ax \in M$.

Locally convex: $N$ is open, balanced, and absorbing. $P_n(x) := \inf\{t > 0 : \frac{x}{t} \in N\}$

Seminorms $N_{i,j} = \{x \in V : p_i(x) < \epsilon_j\}$, $\epsilon_j \rightarrow 0$. Hausdorff implies $\cap_{i,j}N_{ij} = \{0\}$, which is how we can guarantee that $d(x,y) = 0 \Rightarrow p_i(x-y) = 0\Rightarrow x - y = 0$. Some classical consutrction $d(x,y) = \sum_{n=1}^\infty 26{-n} \frac{p_n(x-y)}(1+p_n(x-y))$ Triangle inequality is a bit tricky but we can show it from the regular triangle inequality on $p_n$.

TBS is a supset of locally convex TBS which is a supset of seminorm space which is a supset of Frechet space which is a supset of Banach space is a supset of Hilbert space.

Metrizable Locally Convex Topological Veector Spaces

As a consequence of a bunch of theorems, a topology is defined using a countable family of seminorms $p_j : V \rightarrow [,\infty)$, $p_j(x+y) \le p_j(x) + p_j(y)$, $p_j(a_x) = |a|p_j(x)$, i.e., $x_n \rightarrow x$ is equivalent to $p_j(x_n - x) \rightarrow 0$ as $n \rightarrow \infty$ for all $j$.

$V, \{p_j\}$ is called a Frechet space if and only if it is complete.

Seminormal space (i.e. top space defined by one seminorm).

The $\|f\|_p$ norm is a seminorm based on Minkowski’s theorem or something.

$W \subseteq V$, we would like to define a norm on $V/W$. $\tilde{p}(x+W) := \inf_{y \in x + W} p(y)$ $\tilde{p}$ is continuous if and only if $W$ is closed.

Issue in theorem statement, basically we have a seminorm, but we want to show that when we divide it by points that go to zero, we get a norm.

Now notice that $C(\R^n)$ is not a normed space. Why? Suppose there exists a norm $p$ such that $\forall j \exists c_j$ such that $p_j(x) \le c_j p(x)$. Just take $f(j) = jc_j$ (connected by linear pieces in between), and then the condition would be $j c_j \le c_j p(f)$ which is impossible if it has to happen over all $j$.

We have a whole bunch of different Hahn-Banach theorems. These are all based on the geometric version. There is one version though here:

In Rudin’s Real and Complex Analysis, in an appendix, there is a proof that you only need it to be holomorphic on the interior of $K$. $\|u\| = \sup_{x \in K} |u(x)|$.

There exists a nontrivial fact which is that every continuous linear functional $f$ from … $f(u) = \int u(z) \; d\mu(z)$ for a finite complex supported something. (I think he’s saying we can find $\mu$ so that this works for any functional $f$). The proof is also in the book by Rudin that he mentioned apparently.

Suppose $v$ is compactly supported and smooth in $\mathbb{C}$.

$\overline{\partial} = \frac{1}{2}(\partial_x + i\partial_y)$ $u \in C^1(U)$ $u \in \mathcal{O}(U) \Leftrightarrow \overline{\partial} u = 0$

$\int_{\mathbb{C}} \overline{\partial}v(z) \; dm(z) = 0$ $-\frac{1}{\pi}\int_{\mathbb{C}} \overline{\partial} v(z) (\zeta - z)^{-1} \; dm(z) = v(\zeta)$ Basically proved using Green’s formula to reduce it to Cauchy’s integral formula.

From Hahn-Banach, we need that $\int \zeta^n \; d\mu(\zeta) = 0$ finite for all measure supported in $K$ for all $n$ implies $\int u(\zeta) d\mu(\zeta) = 0$. This left hand side of the implication is the condition that it vanishes on the sapce of polynomials.

Eventually, $z \rightarrow \int (\zeta - z)^{-1} d\mu(\zeta)$ is holomorphic on $K^c$.

Next time we’ll solve PDEs using Hahn-Banach theorem :eyes:.

Recollection of Theorems

• If $K = \R$, apply gen? theory so $H \supseteq F$, $H \cap A = \varnothing$, $\text{codim }H = 1$, all we need is $\overline{H} = H$, since $\overline{H} \neq V$ as $A$ is open and nonempty.
• If $K = \mathbb{C}$, take it as a real vector space to get some $H$ from the first part, then consider $H_1 = H \cap (iH)$. Then it is not too hard to show that it works.

As a corollary,