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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.

Michael Hardy
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Conjecture
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1 Answers1

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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

Tsemo Aristide
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  • Thank you! I looked at this proof in this other question https://math.stackexchange.com/questions/3104664/proving-james-theorem. But It looks like it uses Alaoglu theorem for the unit ball in the space itself and not the dual. Is it a valid proof? Also is the compactness of $T$ superfluous in my question? – Conjecture Dec 07 '19 at 03:29