Let $H$ be hilbert and $T$ a BLO, such that $T:H\rightarrow H$. Prove that $\langle T(x),x \rangle = 0$ implies $T = 0$. Let $H$ be hilbert and $T$ a BLO, such that $T:H\rightarrow H$. Prove that $\langle T(x),x \rangle = 0$ implies $T = 0$. Any hints to tackle this problem? i tried writing x as $x = u + v$ where $u \in Y$ and $v \in Y^T$ for some closed linear subspace of $H$, but i did not see somethins smart. Is it maybe smart to try using the contra positive?

since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\\ \end{eqnarray} Then $$ \langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad (1) $$ so we replace $y$ by $iy$ in the last equality we get : $$ -i\langle T x,y\rangle +i\langle Ty,x\rangle=0 \qquad (2) $$ multiplying $(2)$ by $i$ and add to $(1)$ we get $$ \langle Tx,y\rangle=0 \qquad \forall x,y\in H $$ then we put $y=Tx$ we get $\|Tx\|^2=0$ for all $x\in H$ so $T=0$.