$\|T\| \leq\|T\|_{0}^{1 / 2}\left\|T^{*}\right\|_{0}^{1 / 2}$ for compact operators on Hilbert spaces.

Let $H$ be a Hilbert space over $\mathbb{K}$ and $T$ be a compact operator on $H$. Let $\|\cdot\|_{0}$ be a norm on $H$ and $$ \|T\|_{0}=\sup \left\{\|T(x)\|_{0}: x \in H,\|x\|_{0} \leq 1\right\} $$ Then $\|T\| \leq\|T\|_{0}^{1 / 2}\left\|T^{*}\right\|_{0}^{1 / 2}$. In particular, if $\|\cdot\|_{0}$ is an norm on $\mathbb{K}^{n}$ and $M$ is an $n \times n$ matrix, then $\|M\|_{2} \leq\|M\|_{0}^{1 / 2}\left\|\overline{M}^{t}\right\|_{0}^{1 / 2}$.

My attempt : Notice that $\|T\| = \sup \left\{\|T(x)\|: x \in H, \langle x,x\rangle \leq 1\right\}$, where the norm $\|\cdot \|$ is induced from the inner product. $\| \cdot\|_{0}$ may not be induced from an inner product. The second part of the problem, can be solved by simply substituting $T = M$. If we for a moment, assume that it is true for self-adjoint compact operators, i.e $\|T\| \leq\|T\|_{0}$ holds true for self-adjoint compact operators, then $\|T^* T\| = \| T\|^2 \leq\left\|T^{*}T\right\|_{0} \le \|T\|_{0}\left\|T^{*}\right\|_{0}$, from which we obtain the result. How do I show $\|T\| \leq\|T\|_{0}$ holds true for self-adjoint compact operators ?

Solution 1:

Hint: By definition of adjoint, we have $\|Tx\|^2=\langle Tx,Tx \rangle =\langle x,T^*Tx\rangle=|\langle x,T^*Tx\rangle|.$ Consider Cauchy-Schwarz inequality.