Functional Analysis Lecture 5

[[lecture-data]]

← Lecture 4 | Lecture 6 → Lecture Notes: Rodriguez, page 22

1. Norm and Banach Spaces

Las time, we talked about the Hahn-Banach theorem.

Recall from last time:

Zorn's Lemma

If every chain in a nonempty, partially ordered set $E$ has an upper bound, then $E$ contains a maximal element.

[!def] Hamel basis

A Hamel basis $H\subset V$ is a linearly independent set such that every element of $V$ is a finite linear combination of elements in $H$.

Example

1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\}$$ is a [[Hamel basis]] for $\mathbb{R}^2$.

It is not immediately obvious that every vector space has a Hamel basis, so we start by proving this via Zorn’s lemma.

Theorem

If $V$ is a vector space, then $V$ has a Hamel basis.

Proof

Let $E$ be the set of linearly independent subsets of $V$. Define a partial order on $E$, $⪯$ via inclusion: $e,e^{\prime }\in V,e⪯e^{\prime }⟺e\subseteq e^{\prime }$ Towards applying Zorn, let $C$ be a chain in $E$. Define $c={\bigcup }_{e\in C}e$ This is a linearly independent subset:

• Let $v_1,\dots ,v_n\in c$. Thus there exist $e_1,\dots ,e_n\in C$ where for all $j$ we have $v_j\in e_j$.
• Since $C$ is a chain, there exists some $J$ such that for all $j=1,\dots ,n$ we have $e_j⪯e_J$ (ie $e_j\subseteq e_J$). Thus $v_1,\dots ,v_n\in e_J$.

Now, since $e_J$ is a linearly independent subset, this implies that $v_1,\dots ,v_n$ are themselves linearly independent. Thus $c\in E$

Thus for all $e\in C$, we have $e⪯c$. ie, there is an upper bound of $C$.

By Zorn’s lemma, $E$ has a maximal element $H$. I claim that $H$ spans $V$.

Suppose BWOC that $H$ does not span $V$. Then there exists some $v\in V$ such that $v$ cannot be written as a finite linear combination of elements in $H$. Then $H\cup \{v\}$ is linearly independent. But then $H\subseteq H\cup \{v\}$ and $H\prec H\cup \{v\}$ so $H$ is not maximal $⟹⟸$

Thus $H$ spans $V$.

Thus every vector space has a Hamel basis \tag*{$\blacksquare$}

see every vector space has a hamel basis

Hahn-Banach theorem

Let $V$ be a normed vector space and $M\subset V$ a subspace. If $u:M\to \mathbb{C}$ is a linear map such that for all $t\in M$ $|u(t)|\le C||t||$ (ie, we have a bounded linear functional), then there exists a continuous extension $U:V\to \mathbb{C}$ such that $U\in \mathcal{B}(V,\mathbb{C})$ and $U{|}_M=u,||U(t)||\le C||t||\forall t\in V$ (with the same $t$ as above)

Lemma

If $V$ is a normed vector space and $M\subset V$ is a subspace and $u:M\to \mathbb{C}$ is linear such that $|u(t)|\le C||t||$ for all $t\in M$, and

• $x\notin M$ Then there exists a function $u^{\prime }:M^{\prime }\to \mathbb{C}$ which is linear
• (Where $M^{\prime }=M+\mathbb{C}x=\{t+ax:t\in M,a\in \mathbb{C}\}$ ) Such that $u^{\prime }{|}_M=u$ and for all $t^{\prime }\in M^{\prime }$ we have $|u^{\prime }(t^{\prime })|\le C||t^{\prime }||$

See we can always extend functions on subspaces

Strategy for using this lemma to prove Hahn-Banach:

1. Place a partial order on all continuous extensions of $u$
2. Apply Zorn’s lemma to this set, giving us a maximal element $U$
3. Use the lemma to show that this maximal extension is defined on all of our desired space $V$
   • the proof of this will be similar to our proof that we indeed had a basis in the proof that every vector space has a hamel basis

Proof (of Hahn-Banach)

Let $E$ be the set of all continuous extensions of $u$: $E=\{(v,N):N\text{ is a subspace of }V,M\subset N,v\text{ cont. ext. of }u\text{ to }N\}$ ie, each element is a bounded linear functional on $N$ with the same bound $C$ as the original functional $u$.

Now, define the partial order $⪯$ on $E$ as follows: $(v_1,N_2)⪯(v_2,N_2)\text{ if }{N}_1\subset N_2\text{ and }{v}_2{|}_{N_1}=v_1$ Let $C$ be a chain in $E$ indexed by the set $I$: $C=\{(v_i,N_i):i\in I\}$ Then for all $i_1,i_2\in I$ either $(v_{i_1},N_{i_1})⪯(v_{i_2},N_{i_2})$ or vice versa. Now, let $N={\bigcup }_{i\in I}{N}_i$ be the union of all such subspaces $N_i$. We can show that $N$ is indeed a subspace.

Let $x_1,x_2\in N$ and $a_1,a_2\in \mathbb{C}$. Then there exist $i_1,i_2\in I$ such that $x_1\in N_{i_1}$ and $x_2\in N_{i_2}$. Since $C$ is a chain, WLOG, $N_{i_1}\subset N_{i_2}$.

Then $x_1,x_2\in N_{i_2}⟹a_1{x}_1+a_2{x}_2\in N_{i_2}\subset N⟹N$ is a subspace of $V$.

Now we can define $v:N\to \mathbb{C}$ by

• if $t\in N_i$ then $v(t)=v_i(t)$

Is this well-defined? ie, if $t\in N_{i_1}\cap N_{i_2}$ does this mean $v_{i_1}(t)=v_{i_2}(t)$?

• Suppose $t\in N_{i_1}\cap N_{i_2}$ and $(v_{i_1},N_{i_1})⪯(v_{i_2},N_{i_2})$. Since $v_{i_1}{|}_{N_{i_1}}=v_{i_1}$ then $v_{i_1}(t)=v_{i_2}(t)$. Thus $(v_{i_2},N_{i_2})$ is an extension of $(v_{i_1},N_{i_1})$

Similarly, we can show that $v$ is linear and also the extension of any $v_i$.

So then for all $i\in I$, we have $(v_i,N_i)⪯(v,N)$. ie, $(v,N)$ is an upper bound for $C$.

By Zorn, $E$ has a maximal element $(U,N)$ and claim $N=V$. Suppose BWOC this is not the case. Then let $x\notin N$. Then since we can always extend functions on subspaces, there exists a continuous extension $v$ of $U$ for $N+\mathbb{C}x$ (ie, $(v,N+\mathbb{C}x)\in E$). But then we have $(U,N)⪯(v,N+\mathbb{C}x)$ so $(U,N)$ is not maximal $⟹⟸$ \tag*{$\blacksquare$}

Proof (of Lemma)

First, note that if $t^{\prime }\in M^{\prime }(=M+\mathbb{C}x)$ then there exists unique $t\in M$ and $a\in \mathbb{C}$ such that $t^{\prime }=t+ax$.

(uniqueness) $t+ax=\tilde{t}+\tilde{a}x⟹(a-\tilde{a})x=\tilde{t}-t\in M$. And if $a-\tilde{a}\ne 0$ then $x\in M$ so we must have $a=\tilde{a}$. But then we have $\tilde{t}=t$, so $t,a$ must be unique.

If

Thus, upon choosing $\lambda \in \mathbb{C}$, the map $u^{\prime }(t+ax)=u(t)+a\lambda t\in M,a\in \mathbb{C}$ is well-defined on $M^{\prime }$ and $u^{\prime }:M^{\prime }\to \mathbb{C}$ is linear.

WLOG, suppose $C=1$. We want to choose $\lambda \in \mathbb{C}$ such that $\forall t\in M,a\in \mathbb{C}|u^{\prime }(t)+a\lambda |\le ||t+a\lambda ||$ Once $\lambda$ is fixed, $u^{\prime }$ is our continuous extension. When $a=0$, this inequality holds regardless of $\lambda$ (because it holds on $M$). Thus we only need to choose $\lambda$ for $a\ne 0$. When $a\ne 0$, we have

\lvert u'(t) + a\lambda \rvert \leq \lvert \lvert t + ax \rvert \rvert&\iff \left\lvert \left\lvert u\left( \frac{t}{-a} \right) -\lambda \right\rvert \right\rvert \leq \left\lvert \left\lvert \frac{t}{-a} -x \right\rvert \right\rvert \quad\forall t \in M \\ &\iff \lvert u(t) - \lambda \rvert \leq \lvert \lvert t - x \rvert \rvert \quad\forall t \in M \end{align}$$ Since $\frac{t}{-a} \in M$. We first show there exists some $\alpha \in \mathbb{R}$ such that $$\lvert w(t) - \alpha \rvert \leq \lvert \lvert t-x \rvert \rvert \quad \forall t \in M$$ where $w(t) = \frac{u(t) - \overline{u(t)}}{2} = \mathrm{Re}(u(t))$. Now, note that $$\begin{align} \lvert w(t) \rvert & = \lvert \mathrm{Re}(u(t)) \rvert \\ & \leq \lvert u(t) \rvert \\ & \leq \lvert \lvert t \rvert \rvert \\ \implies w(t_{1}) - w(t_{2}) & =w(t_{1}-t_{2}) \quad \forall t_{1},t_{2} \in M\\ & \leq \lvert w(t_{1}-t_{2}) \rvert \\ & \leq \lvert \lvert t_{1}-t_{2} \rvert \rvert \\ & \leq \lvert \lvert t_{1}-x \rvert \rvert + \lvert \lvert t_{2}-x \rvert \rvert\\ \implies w(t_{1}) - \lvert \lvert t_{1}-x \rvert \rvert & \leq w(t_{2}) + \lvert \lvert t_{2}-x \rvert \rvert \quad\forall t_{1},t_{2}\in M \\ \implies \sup_{t \in M} w(t) -\lvert \lvert t-x \rvert \rvert & \leq w(t_{2}) + \lvert \lvert t_{2} - x \rvert \rvert \quad \forall t_{2} \in M\\ \implies \sup_{t \in M} w(t) -\lvert \lvert t-x \rvert \rvert & \leq\inf_{t \in M} w(t) + \lvert \lvert t-x \rvert \rvert \end{align}$$ So we can choose an $\alpha$ between these two quantities such that for all $t \in M$ we have $$\begin{align} w(t) - \lvert \lvert t-x \rvert \rvert &\leq \alpha &\leq w(t) + \lvert \lvert t-x \rvert \rvert \\ \implies - \lvert \lvert t-x \rvert \rvert &\leq \alpha - w(t) &\leq \lvert \lvert t-x \rvert \rvert \\ \implies \lvert w(t) - \alpha \rvert &\leq \lvert \lvert t-x \rvert \rvert \end{align}$$ Using the same process for the imaginary part of $u(t)$ and repeating the argument with $ix$, we can find a $\lambda$ such that the desired bound holds. > [!NOTE] > To see this, we can verify that since the bound holds on both the real and imaginary components of $x$, it holds for all complex multiples of $x$ This defines our function $$u'(t+ax) = u(t) + a\lambda$$ on all of $M +\mathbb{C}x$. Thus we are done. $$\tag*{$\blacksquare$}$$

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