Existence of $v$,$\lvert\lvert v \rvert\rvert = 1$, such that $\langle Tv, Tv \rangle = \lvert\lvert T \rvert \rvert^2$

Question

Let $T : H \rightarrow H$ be a compact linear operator on a Hilbert space which is NOT necessarily self-adjoint. How would you prove that there exists $v \in H$, $\lvert\lvert v \rvert\rvert = 1$, such that $\langle Tv, Tv \rangle = \lvert\lvert T \rvert \rvert^2$ ?

I don't know how I should take on this problem. Of course, I can use the compacity assumption right at the beginning. But then, I'm lost.

Answer 1

Apparently my original answer (below) overlooked an important detail.

With that being said, we can reduce this question to the case of self-adjoint compact operators if we note that $$ \langle Tv,Tv \rangle = \langle v, T^*Tv \rangle = \|\sqrt{T^*T}v\|^2 $$ and that $T^*T$ is self-adjoint and compact.

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Hint: Consider the function $v \mapsto \|v\|^2$ as a continuous function restricted to the image of the closed unit ball under $T$.

Answer 2

By definition of the operator norm, there is for each $n$ a vector $v_n$ with $\|v_n\|\le 1$ such that $$ \|Tv \|\ge \|T\|-1/n. $$ Hilbert spaces are reflexive, after extracting a subsequence if necessary we have $v_n\rightharpoonup v$ and (by compactness) $Tv_n\to Tv$. By weak lower semicontiuity of norms, $\|v\|\le1$. This shows $\|v\|\le1$ and $\|Tv\|=\|T\|$. If $T\ne0$ this implies $v\ne0$ and scaling argument shows $\|v\|=1$.