Let $(\mathcal{H}, \langle \cdot, \cdot \rangle)$ denote a seperable Hilbert space with orthonormal basis $(e_n)_{n \in \mathbb{N}}$ with induced norm $\Vert \cdot\Vert$.
Let $T:\mathcal{H} \rightarrow \mathcal{H}$ denote a linear, continuous map and let $T^*:\mathcal{H} \rightarrow \mathcal{H}$ denote the adjoint map i.e. $T^{*}$ satisfies that $$\langle T(x), y \rangle = \langle x, T^*(y) \rangle.$$
Consider now the norm $$ \Vert T\Vert_{\infty} := \sup\{\Vert T(x) \Vert : x\in \mathcal{H}, \Vert x \Vert \leq 1 \}. $$
I want to prove two properties in this setup:
(1) $\quad$$\Vert T^* \circ T \Vert_{\infty} = (\Vert T\Vert_{\infty})^2$
(2) $\quad$$\sum_{n=1}^\infty \Vert T(e_n) \Vert^2 = \sum_{n=1}^\infty \Vert T^*(e_n) \Vert^2$
I have already proven some properties that may be useful in doing so. Here is a list:
(A) $\quad$ $T^{*}$ is linear and continuous
(B) $\quad$ $(T^*)^* = T$
(C) $\quad$ $\Vert T\Vert_{\infty} = \Vert T^*\Vert_{\infty} $
(D) $\quad$ $\Vert T \circ T \Vert_{\infty} \leq (\Vert T\Vert_{\infty} )^2$
Thoughts so far:
(1) Using property (C) and (D), we can easily arrive at $\Vert T \circ T^* \Vert_{\infty} \leq (\Vert T\Vert_{\infty} )^2$. I have trouble showing the other inequality however. I was think of maybe utilizing property (B) somehow. But maybe there is a different route entirely, perhaps one where we just show the equality directly?
(2) Here it suffices to show that $\Vert T(e_n) \Vert^2 = \Vert T^*(e_n) \Vert^2$ for all $n$. However whenever I try to write out the norms, my manipulations are not fruitful:
$$ \Vert T(e_n) \Vert^2 = \langle T(e_n), T(e_n) \rangle = \langle T(e_n), T(e_n) \rangle = \langle e_n, T^{*}(T(e_n)) \rangle $$ or (using (B)) $$ \Vert T(e_n) \Vert^2 = \langle T(e_n), T(e_n) \rangle = \langle (T^{*})^*(e_n), T(e_n) \rangle = \langle e_n, T^{*}(T(e_n)) \rangle $$ Can anyone help me?
Update: I have managed to prove part (2) using Parsevals equation (for Fourier coefficients) and Tonellis theorem. So only (1) remains!
In order to show the reverse inequality in $(1)$, let $f\in \mathcal{H}$ with $||f||\leq 1$, then
\begin{align} ||Tf||^2=\langle Tf, Tf\rangle=\langle T^*T f,f\rangle\leq ||T^*Tf||\cdot ||f||\leq ||T^*T||_{\infty}, \end{align} where we used Cauchy-Schwartz in the first inequality. Taking the supremum over all $f$ implies \begin{align} ||T||_{\infty}^2\leq ||T^*T||_{\infty}. \end{align}