Functional Analysis, Sobolev Spaces and Partial Differential Equations.

## Chapter 1 The Hahn-Banach Theorems.

### 1.1 The Analytic Form of the Hahn-Banach Theorem

Suppose $$E$$ is a vector space, a functional is a function defined on $$E$$ or on some subspace of $$E$$ with values in $$\mathbb{R}$$. $$E^*$$ denotes the dual space of $$E$$, that is, the space of all continuous linear functionals on $$E$$. The dual norm on $$E^*$$ is defined by

$\| f \|_{E^*} = \sup_{\|x\| \leq 1 \\ x \in E} \vert f(x) \vert = \sup_{\|x\| \leq 1 \\ x \in E} f(x).$

Given $$f \in E^*$$ and $$x \in E$$ we often write $$\langle f, x \rangle$$ instead of $$f(x)$$; it reads the scalar product for the duality $$E^*, E$$.

(Corollary 1.3.)

Proof. Note that $$G = \{x_0\}$$ and $$f_0$$ is equivalent to $$g$$ on $$G$$, hence,

\begin{aligned} t \| x_0 \|^2 &= \sup_{x_0 \in G} t \| x_0 \|^2 = \sup_{x_0 \in G} g(t x_0) = \sup_{x_0 \in G} g\left(t \|x_0\| \frac{x_0}{\| x_0 \|}\right) \\ &= t \|x_0\| \sup_{x_0 \in G} g\left(\frac{x_0}{\| x_0 \|}\right) = t \|x_0\| \| g \|_{G^*}, \end{aligned}

which leads to $$\| g \|_{G^*} = \| x_0 \|$$.

\begin{aligned} \langle f_0, x_0 \rangle = \sup_{x_0 \in G} \langle f_0, x_0 \rangle &= \sup_{x_0 \in G} \| x_0 \| \left\langle f_0, \frac{x_0}{\| x_0 \|} \right\rangle \\ &= \| x_0 \| \sup_{x_0 \in G} \left\langle f_0, \frac{x_0}{\| x_0 \|} \right\rangle \\ &= \| x_0 \| \| g \|_{G^*} = \| x_0 \|^2. \end{aligned}

(Corollary 1.4.)

Proof.

$\sup_{f \in E^* \\ \|f\| \leq 1} \vert \langle f, x \rangle \vert = \sup_{f \in E^* \\ \|f\| \leq 1} \|x\| \left\vert \left\langle f, x / {\|x\|} \right\rangle \right\vert \leq \|x\| \sup_{f \in E^* \\ \|f\| \leq 1} \| f \| \leq \| x\|.$

### 1.3 The Bidual $$E^{**}$$.

$$J: E \longrightarrow E^{**}$$, given $$x \in E$$ then $$Jx \in E^{**}$$. It is defined as follows: given $$x \in E$$, the map $$f \mapsto \langle f, x \rangle$$ is a continuous linear functional on $$E^*$$; thus it is an element of $$E^**$$, which is denoted by $$Jx$$.

$\langle J x, f\rangle_{E^{\star \star}, E^{\star}}=\langle f, x\rangle_{E^{\star}, E} \quad \forall x \in E, \quad \forall f \in E^{\star}$

Remark. We need both $$f$$ and $$x$$ to define bidual norm.

### 1.4 Theory of Conjugate Convex Functions

For fixed $$x \in E$$, the function $$f \mapsto \langle f, x \rangle - \varphi(x)$$ is convex where $$f \in E^*$$. Let $$T(f) = \langle f, x \rangle - \varphi(x)$$,

\begin{aligned} T(t f_1 + (1-t) f_2) &= \langle t f_1, x \rangle - \varphi(x) + \langle (1-t) f_2, x \rangle - \varphi(x)\\ &= t \langle f_1, x \rangle - \varphi(x) + (1-t) \langle f_2, x \rangle - \varphi(x)\\ &= t T(f_1) + (1 - t) T(f_2), \end{aligned}

which is ture for all $$f_1, f_2 \in E^*, t \in (0, 1)$$.

Proposition 1.10.

\begin{aligned} \Phi([x, \lambda]) &= \Phi([x, 0] + [0, \lambda])\\ &= \Phi([x, 0]) + \lambda \Phi([0, 1])\\ &= \langle f, x \rangle + \lambda k. \end{aligned}

Theorem 1.11.

Step 1:

\begin{aligned} \left\langle \frac{f}{k + \epsilon}, x \right\rangle + \varphi(x) &\geq \frac{\alpha}{k + \epsilon} \Rightarrow \\ \left\langle -\frac{f}{k + \epsilon}, x \right\rangle - \varphi(x) &\leq -\frac{\alpha}{k + \epsilon} \Rightarrow \\ \varphi^*\left(- \frac{f}{k+\epsilon}\right) &\leq - \frac{\alpha}{k + \epsilon}. \end{aligned}

Step 2:

\begin{aligned} (\bar{\varphi})^*(f) &= \sup_{x \in E}\left\{ \langle f, x \rangle - \bar{\varphi}(x) \right\}\\ &= \sup_{x \in E}\left\{ \langle f + f_0, x \rangle - \bar{\varphi}(x) - \varphi^*(f_0)\right\}\\ &= \varphi^*(f + f_0) - \varphi^*(f_0). \end{aligned} \begin{aligned} (\bar{\varphi})^{**}(x) &= \sup_{f \in E^*}\left\{ \langle f, x \rangle - \bar{\varphi}^*(f) \right\}\\ &= \sup_{f \in E^*}\left\{ \langle f + f_0, x \rangle - \varphi^*(f+f_0) - \langle f_0, x \rangle + \varphi^*(f_0)\right\}\\ &= \varphi^{**}(x) - \langle f_0, x \rangle + \varphi^*(f_0). \end{aligned}

Example 1.

\begin{aligned} \varphi^*(f) &= \sup_{x \in E}\left\{ \langle f, x \rangle - \| x \| \right\} \\ &= \sup_{x \in E}\left\{ \|x\| \left(\left\langle f, \frac{x}{\| x \|} \right\rangle - 1\right) \right\}. \end{aligned}