There are similar question but the characterization of the space $E$ that I have gives me problem in computing the actual norm.
Let $E=\{u \in C[0,1]\ ,\ u(0) = 0\}$ with the usual $\parallel \cdot\parallel_\infty$-norm.
Consider the linear functional $f$ defined as: $$f(u)=\int_0^1u(t)dt, \ \forall u\in E\ .$$ $\cdot$Show that $f ∈E^⋆$ and compute $∥f∥_{E^⋆}$.
I can show that $f ∈E^⋆$ but I have troubles computing the actual norm of $f$.
$f ∈E^⋆$ iff $\ \ \ $ $f\ : \ E$ -> $\mathbb{R}$ and $f$ is linear and continuous. Linearity follows easily from the properties of the integral. In order to show continuity, we show that $f$ is bounded since for linear operator boundedness is equivalent to continuity.
$|f(u)| = |\int_0^1 u(t)dt\ |\ \leq\ \int_0^1 |u(t)|dt\ \leq ||u||_\infty$, so taking the $\sup_{||u||\leq 1}$ we get that $||f||\leq 1$. Hence I would guess that: $$||f||=1$$ since we have that: $\forall u \in E, |f(u)|\leq||f||\cdot||u||_\infty$.
Now I should find some $u\in E$ such that $||u||_\infty = 1$ and $|f(u)|=1$ and the trick would be done, but I can not find such function.
Any suggestion?
It is not necessary to find $u\in E$ such that $|f(u)|/||u||_\infty=1$. It is enough to find a sequence $u_n$ such that $|f(u_n)|/||u_n||_\infty\to1$.
For this take the function(s) $u_n(x)=1$ for $x\geq1/n$ and $u_n(x)=nx$ for $x\in[0,1/n]$.
We can see that for this sequence
$$|f(u_n)|/||u_n||_{\infty}=|\int_{0}^{1}u_n(x)dx|\geq|\int_{1/n}^{1}u_n(x)dx|=1-\frac{1}{n}\to1.$$