I am solving the following problem.
Show that if $T$ is bounded linear, positive, self-adjoint operator on Hilbert space $H$. Then $\left\|T{x}\right\| \leq\|T\|^{\frac{1}{2}}\langle T x, x\rangle^{\frac{1}{2}}$. Thus $\langle T x, x\rangle=0 \Leftrightarrow x \in \ker T$
Can anyone help me?
First, observe that $$ \|T^{1/2}x\|^2=\langle T^{1/2}x,T^{1/2}x\rangle=\langle Tx,x\rangle\le\|Tx\|\|x\|\le \|T\|\|x\|^2, $$ that is, $$ \|T^{1/2}\|\le \|T\|^{1/2}. $$ Now just observe that $$ \|Tx\|=\|T^{1/2}T^{1/2}x\|\le \|T^{1/2}\|\langle T^{1/2}x,T^{1/2}x\rangle^{1/2}\le \|T\|^{1/2}\langle T^{1/2}x,T^{1/2}x\rangle^{1/2} $$ $$ =\|T\|^{1/2}\langle Tx,x\rangle^{1/2} $$ where the fact proved above has been used in the second inequality above.
The meaning of this is that for positive selfadjoint operators being in their kernel is the same as being in the kernel of the associated quadratic form.