I’m following Brezis’ book on functional analysis and I'm stuck with the following exercise:
Let $E$ be a vector space of dimension $n$ with the $1$-norm, that is, if $(e_i)_{1\leq i \leq n}$ is a basis, then for every $x\in E$ we define $||x||= \sum_{i=1}^{n}|x_i|$ where $x=\sum_{i=1}^n x_ie_i$. Show that for every $x\in E$, the duality map $F(x)$ is exactly the set of linear functionals $f\in E^*$ such that
$$
f_i=
\begin{cases}
(\mathrm{sign}(x_i))||x||_1 &\text{if}\, x_i\neq 0\\
\text{anything in the interval} [-||x||_1, ||x||_1] &\text{if}\, x_i=0
\end{cases}
$$
where $f_i= f(e_i)$.
We know from the previous exercise in the book that the norm in the dual space $E^*$ is given by $||f||= \max|f_i| $.
I can’t really show any progress since I’ve tried all I can think of with no luck. I know the definition of the duality map and have seen the solution to the exercise but I can’t see how to actually obtain it myself. Any help would be much appreciated.
Best Answer
Recall that $f \in F(x)$ means that $\|f\| = \|x\|$ and $\langle f,x \rangle = \|x\|^2$. The first condition states $$\max_{1 \le i \le n} |f_i|= \|x\|$$ which gives you $-\|x\| \le f_i \le \|x\|$ for all $i$.
The second condition states $$\sum_{i=1}^n f_i x_i = \|x\|^2$$ and in particular $$ \sum_{i=1}^n \frac{f_i}{\|x\|} x_i = \|x\| = \sum_{i=1}^n |x_i|.$$ Since $$ \frac{f_i}{\|x\|} x_i \le |x_i|$$ for all $i$, to obtain equality in the sum over all indices it must be the case that $\displaystyle \frac{f_i}{\|x\|} x_i = |x_i|$ for all $i$. If $x_i = 0$ this adds no restriction to to $f_i$. If $x_i \not= 0$ it forces $f_i = \mathrm{sgn}(x_i) \|x\|$.