$TT^*$ norm is equal to $||T||^2$ if $T:\mathcal{H}\to \mathcal{B}$ where $\mathcal{H}$ is hilbert and $\mathcal{B}$ is banach. I am aware that $$||T||^2_{\mathrm{op}}=||TT^*||_{\mathrm{op}}.$$ If $T | Step-by-Step Solution

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$TT^*$ norm is equal to $||T||^2$ if $T:\mathcal{H}\to \mathcal{B}$ where $\mathcal{H}$ is hilbert and $\mathcal{B}$ is banach. I am aware that $$||T||^2_{\mathrm{op}}=||TT^*||_{\mathrm{op}}.$$ If $T$ is from an Hilbert space to an Hilbert space I am interested in the following: Let $\mathcal{H}$ be an Hilbert space and $\mathcal{B}$ be a Banach space. If $T:\mathcal{H}\to \mathcal{B}$ is a bounded linear opeartor then $$||T||^2_{\mathrm{op}}=||TT^*||_{\mathrm{op}}.$$ Clearly $$||T||^2_{\mathrm{op}}=||T||_{\mathrm{op}}\cdot ||T||_{\mathrm{op}}=||T^*||_{\mathrm{op}}\cdot ||T||_{\mathrm{op}} \ge||TT^*||_{\mathrm{op}}$$ So I need help with the other direction... Is my attempt any good? By defintion of $TT^*:\mathcal{B}^*\to \mathcal{B}$. Let $f\in \mathcal{B}^*$ be with norm $1$ we have that $$||TT^*||_{\mathrm{op}}\ge||f||\cdot ||TT^*f|| \ge|f(TT^*f)|=|\underbrace{f\circ T}_{=T^*f} \circ T^* f| =|T^*f \circ T^*f|=|\langle T^*f,T^*f \rangle|$$ where we used the notation $\langle x,f \rangle=f(x)$. Notice that $T^*f\in \mathcal{H}$ there we have that $$|\langle T^*f,T^*f \rangle|=||T^*f||^2$$ thus, we obtain $$||TT^*||_{\mathrm{op}}\ge ||T^*||^2=||T||^2.$$ I am not really sure that I can do the abuse of notation $\langle T^*f,T^*f \rangle|=||T^*f||^2$ just because $T^*f \in \mathcal{H}$. Moreover, I am not really sure what I am trying to show is true... any help would be appreciated!