If the scalar product are equal then the operators are equal.

Question

I want to show the following:

Let H be a $\mathbb C$ -hilbert space and $S,T\in L(X)$

If $\langle Sx,x \rangle = \langle Tx,x \rangle$ for all $x\in H$, then $S=T$

Any hints for me?

Answer 1

Hints:

1. It suffices to show that $$\langle Tx,x \rangle=0, \qquad x \in H,$$ implies $T=0$.
2. Conclude from $\langle T(x+y),(x+y) \rangle=0$ that $$\langle Tx,y \rangle + \langle Ty,x \rangle = 0. \tag{1}$$
3. Replace $y$ in $(1)$ by $\imath y$.
4. Combine both equations (from step 2,3) to obtain $$\langle Tx,y \rangle=0.$$
5. Conclude.

Reference: Walter Rudin, Funtional Analysis, Theorem 12.7.

Answer 2

Soppuse $x,y\in H$, then $x+y, x+iy\in H$. Thus since $\langle x,T(x)‎\rangle=\langle x,S(x)‎\rangle$, we have $\langle x+y,T(x+y)‎\rangle+‎\langle x+iy,T(x+iy)‎\rangle =‎ ‎\langle x+y,s(x+y)‎\rangle +‎\langle x+iy,s(x+iy)‎\rangle ‎‎$‎,

so we have $2\langle x,T(y)‎\rangle =‎ 2‎\langle x,s(y)‎\rangle$ for each $x,y\in H$. Thus $\langle x,T(y)-S(y)‎\rangle=0$ for each $x,y\in H$. let $x=T(y)-S(y)$, then $\langle T(y)-S(y),T(y)-S(y)‎\rangle=0$, therefore $T(y)=S(y)$ for each $y\in H$, so $T=S$