Exercise 1.4 in Brezis's Functional Analysis, Sobolev Spaces and partial Differential Equations reads as follows:
Let $E = c_0$ be the space of real sequences tending to $0$ with norm $||u|| = \sup_k |u_k|$. In $E$ define for $u = (u_1, u_2, \ldots)$ $$ f(u) = \sum_1^\infty \frac{1}{2^n} u_n. $$ 1. Check that $f$ is a continuous linear functional on $E$ and compute $||f||_{E^*}$. 2. Can one find some $u \in E$ such that $||u|| = 1$ and $f(u) = ||f||_{E^*}$?
I was able to find a solution for this problem, but I lack confidence. Is the following correct?
- Linearity is trivial. Now, note that $$ |f(u)| = \left|\sum_1^\infty \frac{1}{2^n} u_n\right| \leq \sum_1^\infty \frac{1}{2^n} |u_n|. $$ The norm on $c_0$, according to Section 11.3, is $$ ||u|| = \sup_n |x_n|. $$ We know that $x_n \to 0$, so by the definition of convergence we have that $|u_n| \to 0$. Then $(|u_n|)_n$ is bounded by $\sup_n |u_n|$. Hence $$ |f(u)| \leq \left(\sup_n |u_n|\right) \left(\sum_1^\infty \frac{1}{2^n}\right) = ||u||. $$ We conclude that $f$ is bounded and hence continuous.
Now, we have that $$ ||f|| = \sup_{||u|| \leq 1} |f(u)| \leq 1, $$ by the above result. Fix $\varepsilon > 0$. Then there is some $k \in \Bbb{N}$ such that $\sum_1^k 1/(2^n) > 1 - \varepsilon$. But this is just $f(u_k)$, where $$ u_k = (1, 1, \ldots, 1, 0, 0 \ldots), $$ where the $1$s go until the $k$-th position. That means that for arbitrarily small $\varepsilon > 0$ there are elements in $B_{c_0}(0)$ such that $|f(u)| > 1 - \varepsilon$. Hence $$ 1 \leq \sup_{||x|| \leq 1} |f(u)| \leq 1 $$ and therefore $$ ||f|| = 1. $$
- If there was such an $u$, we would have $$ f(u) - 1 = 0 \implies \sum_1^\infty \frac{1}{2^n}(u_n - 1) = 0, $$ which in turn implies that $u_n = 1$ for every $n$. But this is absurd, since $u_n \to 0$.
Thanks in advance and kind regards.