($\Rightarrow$) This implication is immediate (is valid for real spaces).
($\Leftarrow$) (is not valid for real spaces) Suppose $\langle Tx,x \rangle=\langle Sx,x \rangle$ for all $x\in X$. And i dont know what to do
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@SeverinSchraven I think I'm making a mistake in the calculations. Im find $\langle x, Tx\rangle=\langle x, Sx\rangle.$ ='( – Rodolfo Cardoso Apr 03 '21 at 18:01
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Let $X=\mathbb{R}^2$ and let $$T(x,y)=(-y,x)$$ $$S(x,y) =(-2y,2x)$$ then $$\left<T(x,y),(x,y)\right>=-yx+xy =0$$ $$\left<S(x,y) ,(x,y)\right>=-2yx +2xy=0$$ hence $$\left<T(x,y) ,(x,y)\right>=\left<S(x,y),(x,y)\right>$$ for $(x,y)\in \mathbb{R}^2 $ but $$S\neq T.$$
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Neither your $S$ nor your $T$ appear to be bounded, so how does this answer the question? – postmortes Apr 14 '21 at 06:12
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Suppose that $\langle Tx,x \rangle=\langle Sx,x \rangle$ for all $x\in X$. Hence $$ \langle (T-S)x,x \rangle=0, \forall x\in X.$$ How to $X$ is a complex IPS we had $(T-S)\equiv 0$. Because, if $\langle Tz,z \rangle=0$ for all $z\in X$, then $T\equiv 0$ how to showing in Prove that if $\langle Tx,x\rangle =0$ for all $x \in X$, then $T = 0$.