Brezis' Functional analysis 3.27 item 2.

Question

I´m work in the following exercise of Brezis, but a i can´t understand some things.

Let $E$ be a Banach space whit norm $\|,\|$. The norm in $E^{\star}$ is also denoted by $\|,\|$. The purpose this exercise is to construct a equivalent norm on $E$ that is strictly convex and whose dual norm in the dual is also strictly convex. Let $(a_{n})\subset B_{E}$ be a dense subset of $B_{E}$ whit respect to the strong topology.Let $(b_{n})\subset B_{E^{\star}}$ be a countable subset in $B_{E^{\star}}$ that is dense in $B_{E^{\star}}$ for the weak$^\star$ topology $\sigma(E^\star, E)$. Why does such set exists?(1) Given $f \in E^{\star}$ set: \begin{equation}\label{nor1} \|f\|_{1} = \left\lbrace \|f\|^{2} + \sum_{n=1}^{\infty}\frac{1}{2^{n}}|\left< f, a_{n}\right>|^{2} \right\rbrace^{\frac{1}{2}} \end{equation} Given $x\in E$, set: \begin{equation}\label{nor2} \|x\|_{2} = \left\lbrace \|x\|_{1}^{2} + \sum_{n=1}^{\infty}\frac{1}{2^{n}}|\left< b_{n}, x \right>|^{2} \right\rbrace^{\frac{1}{2}} \end{equation} Onde $\|x\|_{1} = sup_{\|f\|_{1}\leq 1}\left< f, x \right>$ .

Prove $\|,\|_{1}$ is stricly convex.

My attempt: Set $|f|^{2} = \sum_{n=1}^{\infty}|\left <f, a_n\right >|^{2}$ which satisfies the parallelogram law, so $|f|^{2}$ arise from inner product, then is strictly convex and the function $f \mapsto |f|^{2}$ is strictly convex. more precisely, we have for any $t \in [0,1]$ and $f, g \in E^{\star}$: \begin{equation}\label{equs1} |tf + (1-t)g|^{2} + t(1-t)|f-g|^{2} = t|f|^{2} + (1-t)|g|^{2}. \hspace{3mm} (2) \end{equation}

My doubts are:

(1) Why does such set exists? I think is due to existence of $(a_n)$ using the Hahn-Banach theorem, but i´m not sure. I can´t support it.

(2) Why is this equality true? Thanks.

Answer 1

$E$ is assumed to be a separable Banach space. ($(a_n)$ would not exist otherwise). The closed unit ball of the dual space is compact (by Banach Alaoglu Thoerem) and it is metrizable by separability of $E$. ($d(f,g)=\sum \frac {|f(a_n)-g(a_n)|} {2^{n}|f(a_n)-g(a_n)|}$ metrizes weak* topology on the ball). Hence, this ball is separable.

The identity you have stated holds in any inner product space. Just expand the norm squares and simplify.