Prove that in a complex Hilbert space $\|Tx\|\le\|T\|^\frac{1}{2}\langle Tx,x\rangle^\frac{1}{2}$

Question

Let $H$ be complex Hilbert space. Prove that $\|Tx\|\le\|T\|^\frac{1}{2}\langle Tx,x\rangle^\frac{1}{2}$.

I am trying to prove this result by using the following result.

Let $T:H\to H$ be a bounded positive self-adjoint linear operator on a complex Hilbert space $H$ and with the positive square root of $T$, then $|\langle Tx,y\rangle|\le \langle Tx,x\rangle^\frac{1}{2}\langle Ty,y\rangle^\frac{1}{2}.$

In this, if we substitute $y=Tx$, then $|\langle Tx,Tx\rangle|\le \langle Tx,x\rangle^\frac{1}{2}\langle TTx,Tx\rangle^\frac{1}{2}$.

But I am not getting the result. Any help? Thank you all.

Answer 1

@daw The square root is not needed. We have $$\displaylines{\|Tx\|=\sup_{\|y\|\le 1}|\langle Tx,y\rangle |\le\langle Tx,x\rangle^{1/2} \sup_{\|y\|\le 1}\langle Ty,y\rangle ^{1/2}\\ =\langle Tx,x\rangle^{1/2} \left (\sup_{\|y\|\le 1}\langle Ty,y\rangle \right )^{1/2}\le \langle Tx,x\rangle^{1/2}\|T\|^{1/2} }$$

Answer 2

You haven't made use of the given fact that $T$ has a positive self-adjoint square root $T^{1/2}$.

Note that $\langle Tx, x \rangle^{1/2} = \langle T^{1/2} x, T^{1/2} x \rangle^{1/2} = \|T^{1/2} x\|$.

We also have $\|T\|^{1/2} = \|T^{1/2}\|$ from self-adjoint-ness of $T$.

Finally, note that $\|Tx\| = \|T^{1/2} T^{1/2} x\| \le \|T^{1/2}\| \|T^{1/2} x\|$.