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

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.

• What‘s the definition of the operator norm? – Fakemistake Jan 22 '18 at 16:35
• @Fakemistake supremum of $\lvert\lvert Tv \rvert\rvert$ over elements $v$ of norm less or equal to $1$ – Desura Jan 22 '18 at 16:35

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.

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

• I'm sorry, I still cannot see how to do it. $T(\overline{B_1(0)})$ is only relatively compact so (for example) I cannot immediately conclude that the max of the function $v \mapsto \lvert\lvert v \rvert\rvert^2$ is attained. – Desura Jan 22 '18 at 17:41
• @Desura ah, that is a problem. I'll delete this answer if I can't think of a fix. – Ben Grossmann Jan 22 '18 at 18:06
• @Desura does that answer let you get the rest of the way? If so, I'd like to know how – Ben Grossmann Jan 22 '18 at 18:16
• @Omnomnomnom Do you need to take a detour through functional calculus to define the square root? – Cameron Williams Jan 22 '18 at 18:19
• @CameronWilliams it suffices to define the square root by power series. Pedersen's Analysis Now uses something like that in his proof of the polar decomposition – Ben Grossmann Jan 22 '18 at 18:38

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 $\|v\|=1$.