If $E$ is a Hilbert separable space, and $T:E\to E$ a compact operator. Is it true that there is an $x\in E$ with $||x||\leq 1$ so that $||T||=||T(x)||$?

Hints? Thank you

  • $\begingroup$ The image of the unit ball in $E$ under $T$ is, in fact, compact (see this); it thus has an element of maximal norm. $\endgroup$ – David Mitra Jun 30 '19 at 13:06
  • $\begingroup$ So it is not necessary that $E$ is Hilbert and separable? $\endgroup$ – JN_2605 Jun 30 '19 at 13:10
  • $\begingroup$ @JN_2605 You only need to have $T(\mathcal{B})$ compact and $T$ continuous with $\mathcal{B}$ the closed unit ball in $E$. $\endgroup$ – Raito Jun 30 '19 at 14:44

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