If $|\omega(\xi)|\le \lambda \|x\xi\|$ for all $\xi \in H$, is $\omega(\xi) = \langle x\xi, \eta\rangle$ for some $\eta \in H$?

Question

Let $H$ be a Hilbert space. Let $\omega \in H^*$ be a bounded functional, $x\in B(H)$ and $\lambda \ge 0$ satisfy $$|\omega(\xi)| \le \lambda \|x\xi\|$$ for all $\xi \in H$.

Does there exist $\eta \in H$ such that $\omega(\xi) = \langle x \xi, \eta\rangle$ for all $\xi \in H$? If necessary, I can assume that $x$ is a positive operator.

This looks an awful lot like the Riesz representation theorem, but not quite. Any hints?

I can show that $$\omega(\xi) = \langle \xi, \eta\rangle$$ where $\eta \in [xH]$. Perhaps $x^*\eta = \eta?$ Then I would be done.

Answer 1

You can define a linear functional $\nu$ on the range of the operator $x$ via $$ \nu(x \xi) := \omega(\xi). $$ It can be checked that this is well defined and continuous. Via Hahn-Banach, we can extend $\nu$ to the entire space $H$. Then, the Riesz representative $\eta \in H$ of this extension satisfies the desired $$ \langle x \xi, \eta\rangle = \nu(x \xi) = \omega(\xi). $$