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.)


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

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.

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