Positive operator and inner product We have a positive and self-adjoint linear operator $T: H\to H$. Moreover, There is $x\in H$ such that $\langle Tx, x\rangle=0.$ How we can prove that $Tx=0?$
 A: By definition $T$ being positive means that
\begin{equation}
\langle Ty, y \rangle \geq 0
\end{equation}
for all $y \in H$. In particular
\begin{align}
\langle T(x+y), x+y \rangle &\geq 0 \\
\langle T(x-y), x-y \rangle &\geq 0.
\end{align}
Expanding the above inner products, using that $\langle Tx, x\rangle = 0$ and that $T$ is self-adjoint we get
\begin{equation}
|\langle Tx, y\rangle| \leq \frac 12 \langle Ty, y\rangle.
\end{equation}
Now, we can replace $x$ by $\lambda x$ for any $\lambda \in \mathbb{R}$ above and the argument doesn't change. Therefore we must have $\langle Tx, y \rangle = 0$ for all $y \in H$. Therefore $Tx = 0$ as required. 
A: Here are a couple  arguments. 


*

*If you know that positive operators always have a positive square root, you can write $$0=\langle Tx,x\rangle=\langle T^{1/2}x,T^{1/2}x\rangle=\|T^{1/2}x\|^2.$$ Then $T^{1/2}x=0$, and $Tx=T^{1/2}(T^{1/2}x)=0$. 

*Because $\langle Ty,y\rangle\geq0$ for all $y$, the sequilinear form $[y,z]:=\langle Ty,z\rangle$ is positive. So Cauchy Schwarz applies! Then
$$
\|Tx\|^2=\langle Tx,Tx\rangle=[x,Tx]\leq [x,x]^{1/2}\,[Tx,Tx]^{1/2}=\langle Tx,x\rangle^{1/2}\,[Tx,Tx]^{1/2}=0. 
$$
So $Tx=0$. 
