Consider the following problem, item 2. of the exercise 1.2 in Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations:
Let $E$ be a vector space of dimension $n$ and let $(e_i)_{1 \leq i \leq n}$ be a basis of $E$. Given $x \in E$ write $x = \sum_1^n x_i e_i$ with $x_i \in \Bbb{R}$. Given $f \in E^*$, define $f_i = \langle f, e_i \rangle$. Consider on $E$ the norm $||x||_\infty = \max_{1 \leq i \leq n}|x_i|$. (a) Compute $||f||_{E^*}$ explicitly, in terms of the $f_i$. (b) Determine explicitly the duality map for every $x \in E$.
The duality map is defined as follows:
The duality map is the set $F(x) = \{f \in E^* \ : \ ||f|| = ||x|| \text{ and } \langle f, x \rangle = ||x||^2\}$.
My partial solution:
(a) By definition,
\begin{align*}
||f|| & = \sup_{||x||_\infty \leq 1} \langle f, x \rangle \\
& = \sup_{||x||_\infty \leq 1} \sum_1^n x_i f_i \\
& = \sum_1^n |f_i|,
\end{align*}
that is, the sup is attained when $|x_i| = 1$ for every $i$.
(b) Given $x \in E$, let $I = \{1 \leq i \leq n \ : \ |x_i| = ||x||_\infty\}$. We will show that $f \in F(x)$ if and only if
(i) $f_i = 0 \ \forall i \notin I$,
(ii) $f_i x_i \geq 0$ for all $i \in I$ and $\sum_I |f_i| = ||x||_\infty$.
First, assume that (i) and (ii) hold. That $||f|| = ||x||_\infty$ is trivial by the previous item. Now, notice that
$$
\langle f, x \rangle = \sum_I f_i x_i = \sum_I |f_i| \ |x_i| = ||x||_\infty \sum_I |f_i| = ||x||_\infty^2.
$$
My question is:
How to prove the converse statement?
Best Answer
(a) It's basically correct. However, to attain the supremum, choose $x_i:=1$ if $f_i\ge 0$ and $x_i=-1$ if $f_i<0$.
(b) If either (i) or (ii) doesn't hold and $\|x\|=\|f\|=\sum_i|f_i|$, then $$\langle f,x\rangle\ =\ \sum_i f_ix_i\ <\ \sum_i |f_i|\,\|x\|\ =\ \|x\|^2\,.$$