I am stumped at showing exercise 3.14 on Brezis' Functional Analysis:
Let $E$ be a reflexive Banach space and $I$ a set of indices. Consider a collection $(f_{i})_{i \in I}$ in the dual space $E'$ and a collection $(\alpha_{i})_{i \in I}$ in $\mathbb{R}$. Let $M>0$.
Show that the following properties are equivalent:
- There exists some $a \in E$ with $|a| \le M$ such that $\langle f_{i}, a\rangle=\alpha_{i}$ for every $i \in I$.
- One has $|\sum_{i \in J} \beta_{i} \alpha_{i}| \leq M\|\sum_{i \in J} \beta_{i} f_{i}\|$ for every collection $(\beta_{i})_{i \in J}$ in $\mathbb{R}$ with $J \subset I, J$ finite.
The first direction 1 implies 2 is easy through direct computation. The converse direction from 2 to 1 is less obvious to me and I am not sure how to proceed. Why does reflexivity help in this case?
Best Answer
Let $Y$ denote the linear span of the elements $(f_i)_{i\in I}.$ Let $\varphi$ be the linear functional on $Y$ defined by $$ \varphi\left (\sum_{i\in J}\beta_if_i\right )=\sum_{i\in J}\beta_i\alpha_i\quad J\underset{\rm finite}{\subset}I$$ The functional is well defined due to the inequality of $2.$ Moreover $\|\varphi\|_{Y'}\le M.$ By the Hahn-Banach theorem the functional $\varphi$ can be extended to a linear functional $\Phi$ on $X'$ such that $\|\Phi\|_{X''}=\varphi\|_{Y'}\le M.$ As $X$ is reflexive there is $a\in X$ for which $\Phi(x')=x'(a)$ for any $x'\in X'$ and $\|a\|_X=\|\Phi\|_{X''}\le M.$ Thus $$\sum_{i\in J}\beta_i\alpha_i=\varphi\left (\sum_{i\in J}\beta_if_i\right )=\Phi\left (\sum_{i\in J}\beta_if_i\right )=\sum_{i\in J}\beta_if_i(a)$$ In particular for $J=\{i\}$ and $\beta_i=1$ we get $f_i(a)=\alpha_i.$