Compact operator in Hilbert separable space attains its norm

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

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