Norm of adjoint proof check

Asked Nov 26 '20 at 22:47

Active Nov 26 '20 at 22:47

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Let ${\cal H},{\cal K}$ be Hilbert spaces. Let $T\in B({\cal H},{\cal K})$. If ${\cal H}={\cal K}$, I know that $||T||=||T^*||$. So I was wondering if the same is true when ${\cal H}$ is not necessarily ${\cal K}$.

attempt of proof

If $T=0$, then equality holds. Assume $T\neq 0$. Observe: $$||T^*k||^2=\langle T^*k,T^*k\rangle=\langle TT^*k,k\rangle\leq ||TT^*||\cdot ||k||^2$$ for all $k\in {\cal K}$. Then $$||T^*||^2\leq ||TT^*||\leq ||T||\cdot||T^*||.$$ Cancelling $||T^*||$, we get $||T^*||\leq ||T||$. By a symmetric argument, $||T||\leq ||T^*||$.

asked Nov 26 '20 at 22:47

chhro

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Yes, it seems ok to me. – Berci Nov 26 '20 at 23:06
Thanks for the input! – chhro Nov 26 '20 at 23:08
@Berci you should make that an answer. – supinf Nov 27 '20 at 07:52
Actually, $|TT^*|=|T|^2$ would follow, and that made me uncertain, since I don't remember such a general result. Nevertheless the proof still seems valid. – Berci Nov 27 '20 at 08:00
@Berci, that's one of my concerns as well that's why I asked. – chhro Nov 28 '20 at 00:10
@Berci, another way: Consider $S=\begin{bmatrix} 0& 0\T& 0\end{bmatrix}\in B({\cal H}\oplus {\cal K})$. Then $||T||=||S||=||S^||=||T^||$. – chhro Nov 28 '20 at 00:29

Norm of adjoint proof check

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