Functional Analysis Lecture 5

[[lecture-data]]

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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 partially ordered set $E$ has an [[upper bound]], then $E$ contains a [[maximal element]].

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

${[\begin{matrix} 1 \\ 0 \end{matrix}], [\begin{matrix} 0 \\ 1 \end{matrix}]} $ $ i s a [[H a m e l b a s i s]] f o r $ 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^{'} \in V, e ⪯ e^{'} ⟺ e \subseteq e^{'}

Towards applying Zorn, let $C$ be a [[chain]] in $E$ . Define

c = ⋃_{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 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 $H ≺ H \cup {v}$ so $H$ is not maximal $⟹ ⟸$

Thus $H$ spans $V$ .

Thus every [[vector space]] has a [[Hamel basis]]

\begin{matrix} ◼ \end{matrix}

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 $u : M \to C$ is a linear map such that for all $t \in M$

| u (t) | \leq C | | t | |

(ie, we have a bounded linear functional), then there exists a continuous extension $U : V \to C$ such that $U \in B (V, C)$ and

U |_{M} = u, | | U (t) | | \leq C | | t | | \forall t \in V

(with the same $t$ as above)

Lemma

$x \notin M$
Then there exists a function $u^{'} : M^{'} \to C$ which is linear
(Where $M^{'} = M + C x = {t + a x : t \in M, a \in C}$ )
Such that $u^{'} |_{M} = u$ and for all $t^{'} \in M^{'}$ we have $| u^{'} (t^{'}) | \leq C | | t^{'} | |$

See [[we can always extend functions on subspaces]]

Strategy for using this lemma to prove Hahn-Banach:

Place a continuous extensions of $u$
Apply [[Zorn's lemma]] to this set, giving us a [[maximal element]] $U$
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 is a subspace of V, M \subset N, v cont. ext. of u 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}) if N_{1} \subset N_{2} 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 = ⋃_{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 $a_{1}, a_{2} \in C$ . 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 $v : N \to C$ 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 continuous extension $v$ of $U$ for $N + C x$ (ie, $(v, N + C x) \in E$ ). But then we have $(U, N) ⪯ (v, N + C x)$ so $(U, N)$ is not maximal $⟹ ⟸$

\begin{matrix} ◼ \end{matrix}

Proof (of Lemma)

First, note that if $t^{'} \in M^{'} (= M + C x)$ then there exists unique $t \in M$ and $a \in C$ such that $t^{'} = t + a x$ .

(uniqueness)

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

Thus, upon choosing $λ \in C$ , the map

u^{'} (t + a x) = u (t) + a λ t \in M, a \in C

is well-defined on $M^{'}$ and $u^{'} : M^{'} \to C$ is linear.

WLOG, suppose $C = 1$ . We want to choose $λ \in C$ such that

\forall t \in M, a \in C | u^{'} (t) + a λ | \leq | | t + a λ | |

Once $λ$ is fixed, $u^{'}$ is our continuous extension. When $a = 0$ , this inequality holds regardless of $λ$ (because it holds on $M$ ). Thus we only need to choose $λ$ for $a \neq 0$ . When $a \neq 0$ , we have

\begin{aligned} | u^{'} (t) + a λ | \leq | | t + a x | | & ⟺ | | u (\frac{t}{- a}) - λ | | \leq | | \frac{t}{- a} - x | | \forall t \in M \\ ⟺ | u (t) - λ | \leq | | t - x | | \forall t \in M \end{aligned}

Since $\frac{t}{- a} \in M$ . We first show there exists some $α \in R$ such that

| w (t) - α | \leq | | t - x | | \forall t \in M

where $w (t) = \frac{u (t) - \overset{―}{u (t)}}{2} = Re (u (t))$ . Now, note that

\begin{aligned} | w (t) | & = | Re (u (t)) | \\ \leq | u (t) | \\ \leq | | t | | \\ ⟹ w (t_{1}) - w (t_{2}) & = w (t_{1} - t_{2}) \forall t_{1}, t_{2} \in M \\ \leq | w (t_{1} - t_{2}) | \\ \leq | | t_{1} - t_{2} | | \\ \leq | | t_{1} - x | | + | | t_{2} - x | | \\ ⟹ w (t_{1}) - | | t_{1} - x | | & \leq w (t_{2}) + | | t_{2} - x | | \forall t_{1}, t_{2} \in M \\ ⟹ sup_{t \in M} w (t) - | | t - x | | & \leq w (t_{2}) + | | t_{2} - x | | \forall t_{2} \in M \\ ⟹ sup_{t \in M} w (t) - | | t - x | | & \leq inf_{t \in M} w (t) + | | t - x | | \end{aligned}

So we can choose an $α$ between these two quantities such that for all $t \in M$ we have

\begin{aligned} w (t) - | | t - x | | & \leq α & \leq w (t) + | | t - x | | \\ ⟹ - | | t - x | | & \leq α - w (t) & \leq | | t - x | | \\ ⟹ | w (t) - α | & \leq | | t - x | | \end{aligned}

Using the same process for the imaginary part of $u (t)$ and repeating the argument with $i x$ , we can find a $λ$ 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 + a x) = u (t) + a λ

on all of $M + C x$ . Thus we are done.

\begin{matrix} ◼ \end{matrix}

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