If $X$ is a complex inner product space and $T, S \in B(X)$ (are bounded operators), it holds
$S=T \iff \left<Sx, x\right>=\left<Tx, x\right> \ \ \ \forall x \in X $
Can you prove this claim from Lemma 5.29 of "Linear functiona analysis" by Rynne and Youngson ? Thanks.
I noticed that this is equal to say that $\left<Fx, x\right>=0$ then $F=0$
It easily follows from the polarization identity: if $B(x,y)$ is a map $B:H \times H \rightarrow \mathbb C$ that is linear in the first argument and antilinear in the second, then $$ B(x,y) = \sum_{k = 0}^3 i^k Q(x +i^k y),$$ where $Q(x) = B(x,x)$. It can be proved by direct verification.
The direct corollary is: if $B$ is a map with the foregoing properties and $Q(x) = 0$ for all $x$, then $B = 0$.
Now if $T:H \rightarrow H$ is a linear operator, then $B(x,y) = \langle Tx,y\rangle$ satisfies the properties. Thus, if $\langle Tx,x\rangle = 0$ for all $x$, then $\langle Tx,y\rangle = 0$ for all $x,y$. it easily follows that $T = 0$.