Is it true that exists $v_0 \in H$ such that $\lVert T \rVert =\frac{\lvert T(v_0)\rvert }{\lVert v_0 \rVert}$?

Question

Let $H$ be an complex infinite dimensional Hilbert space.

Let $T$ be a continuous linear functional on $H$.

Is it true that exists $v_0 \in H$ such that $\lVert T \rVert =\frac{\lvert T(v_0)\rvert }{\lVert v_0 \rVert}$ ?

Answer 1

Although your original question has already been answered, I will give a slight generalization:

If $X$ is any reflexive Banach space, then for each $T\in X'$ we find $v_0\in X$ with $\|v_0\| \leq 1$ and $\|T\| = |T(v_0)|$.

Proof: Since $X$ is reflexive, the closed unit ball $\overline B(0,1)$ of $X$ is weakly compact. Now $T$ is one of the generators of the weak topology on $X$, so $T$ and then also $|T|$ is weakly continuous. Putting both together, $|T|$ must assume its supremum on $\overline B(0,1)$ in some point $v_0$. But $\|T\|$ is exactly the supremum of $|T|$ on $\overline B(0,1)$.