Let $E, F$ be Banach spaces and $T ∈ \mathcal L(E, F )$ a linear continuous compact operator.
If $E$ is reflexive prove that $\exists x ∈ E$ such that $\|x\| ≤ 1$ and $\|T (x)\| = \|T\|$.
We know that for any $x ∈ E$ such that $\|x\| ≤ 1$, we have $\|T(x)\|\le \|T\|$.
Let $M:=\sup_{x ∈ E, \|x\| ≤ 1} \|T(x)\|$
Let $A:=\{T(x) ∈ F,x ∈ E,\|x\| ≤ 1\}$ and let $(T(x_n))$ be a sequence converging to $M$ with $x_n\in A$.
I need to show that $M=\|T\|$ and we have $M\le\|T\|$ but I don't know how to go further by extracting sub-sequence as $E$ is reflexive and $(x_n)$ bounded, or if there is a better way to tackle this problem.
Thank you for your help.
The fact that there exists an element $x$ such that $\|x\|=1, \|T(x)\|=\|T\|$ is an application of the theorem of James.
https://en.wikipedia.org/wiki/James%27s_theorem