I have a problem understanding the solution of Ex 3.27.4 in Brezis's book of Functional Analysis. Below is the Ex 3.27.4:
Let $E$ be a separable Banach space with norm $\|\cdot\|$ . The dual norm on $E^{\star}$ is also denoted by $\|\cdot\|$. The purpose of this exercise is to construct an equivalent norm on $E$ that is strictly convex and whose dual norm is also strictly convex.
Let $\left(a_{n}\right) \subset B_{E}$ be a dense subset of $B_{E}$ with respect to the strong topology. Let $\left(b_{n}\right) \subset B_{E^{\star}}$ be a countable subset of $B_{E^{\star}}$ that is dense in $B_{E^{\star}}$ for the weak topology $\sigma\left(E^{\star}, E\right)$. Why does such a set exist? Given $f \in E^{\star}$, set $$ \|f\|_{1}=\left\{\|f\|^{2}+\sum_{n=1}^{\infty} \frac{1}{2^{n}}\left|\left\langle f, a_{n}\right\rangle\right|^{2}\right\}^{1 / 2} $$
- Prove that $\|\cdot\|_1$ is a norm equivalent to $\|\cdot\|$.
- Prove that $\|\cdot\|_1$ is strictly convex. Given $x \in E$, set $$ \|x\|_{2}=\left\{\|x\|_{1}^{2}+\sum_{n=1}^{\infty} \frac{1}{2^{n}}\left|\left\langle b_{n}, x\right\rangle\right|^{2}\right\}^{1 / 2} $$ where $\|x\|_{1}=\sup _{\|f\|_{1} \leq 1}\langle f, x\rangle$.
- Prove that $\|\cdot\|_2$ is a strictly convex norm that is equivalent to $\|\cdot\|$.
- Prove that the dual norm of $\|\cdot\|_2$ is also strictly convex. [Hint: Use the result of Exercise 1.23, question 3.]
Below is the solution. Could you elaborate on why the minimization has a solution, i.e., $$ \color{blue}{\inf} _{h \in E^{\star}}\left\{\|f-h\|_{1}+[h]^{2}\right\}=\color{blue}{\min} _{h \in E^{\star}}\left\{\|f-h\|_{1}^{2}+[h]^{2}\right\} . $$
$B_{E^{\star}}$ is compact and metrizable for $\sigma\left(E^{\star}, E\right)$. Hence there exists a countable subset of $B_{E^{\star}}$ that is dense for $\sigma\left(E^{\star}, E\right)$.
- Clearly $\|f\| \leq\|f\|_{1} \leq \sqrt{2}\|f\| \forall f \in E^{\star}$.
- Set $|f|^{2}=\sum_{n=1}^{\infty} \frac{1}{2^{n}}\left|\left\langle f, a_{n}\right\rangle\right|^{2}$. Note that the norm || is associated to a scalar product (why?), and thus it is strictly convex, i.e., the function $f \mapsto|f|^{2}$ is strictly convex. More precisely, we have $\forall t \in[0,1], \forall f, g \in E^{\star}$, (S1) $|t f+(1-t) g|^{2}+t(1-t)|f-g|^{2}=t|f|^{2}+(1-t)|g|^{2} .$ Consequently, the function $f \mapsto\|f\|^{2}+|f|^{2}$ is also strictly convex.
- Same method as in question 2. Note that if $\left\langle b_{n}, x\right\rangle=0 \forall n$, then $x=0$ (why?).
- Given $x \in E$ set $[x]=\left\{\sum_{n=1}^{\infty} \frac{1}{2^{n}}\left|\left\langle b_{n}, x\right\rangle\right|^{2}\right\}^{1 / 2}$, and let $[f]$ denote the dual norm of [] on $E^{\star}$. Note that $[f]$ also satisfies the identity (S1). Indeed, we have $$ \begin{aligned} \frac{1}{2}[t f+(1-t) g]^{2} &=\sup _{x \in E}\left\{\langle t f+(1-t) g, x\rangle-\frac{1}{2}[x]^{2}\right\} \\ \frac{1}{2}[f-g]^{2} &=\sup _{y \in E}\left\{\langle f-g, y\rangle-\frac{1}{2}[y]^{2}\right\} \end{aligned} $$ and thus $$ \begin{aligned} &\frac{1}{2}[t f+(1-t) g]^{2}+\frac{1}{2} t(1-t)[f-g]^{2} \\ &\quad=\sup _{x, y}\left\{\langle t f+(1-t) g, x\rangle+t(1-t)\langle f-g, y\rangle-\frac{1}{2}[x]^{2}-\frac{1}{2} t(1-t)[y]^{2}\right\} . \end{aligned} $$ We conclude that (S1) holds by a change of variables $x=t \xi+(1-t) \eta$ and $y=\xi-\eta$. Applying question 3 of Exercise 1.23, we see that $$ \|f\|_{2}^{2}=\color{blue}{\inf} _{h \in E^{\star}}\left\{\|f-h\|_{1}+[h]^{2}\right\}=\color{blue}{\min} _{h \in E^{\star}}\left\{\|f-h\|_{1}^{2}+[h]^{2}\right\} . $$
Below is Question 3 of Exercise 1.23
Let $(E, |\cdot|)$ be a normed linear space and $(E', \| \cdot \|)$ its dual. Let $\varphi, \psi:E \to (-\infty, +\infty]$ such that $\varphi, \psi \not\equiv +\infty$. The conjugate $\varphi':E' \to (-\infty, +\infty]$ of $\varphi$ is defined as $$ \varphi'(f) := \sup_{x\in E} [ \langle f, x \rangle - \varphi(x)], \quad \forall f \in E'. $$
The conjugate $\psi'$ of $\psi$ is defined similarly. The inf-convolution of $\varphi$ and $\psi$ is defined by $$ (\varphi \nabla \psi)(x) := \inf_{y\in E} [\varphi(x-y) + \psi(y)], \quad \forall x\in E. $$
Assume $\varphi, \psi$ are convex, and there exists $x_0 \in E$ such that $\varphi(x_0), \psi(x_0) \neq +\infty$ and that $\varphi$ is continuous at $x_0$. Then $$ (\varphi+\psi)' = ( \varphi' \nabla \psi') \quad \text{on} \,\, E'. $$