Let $A$ be a bounded operator and $V$ a linear isometry, both defined on a complex Hilbert space $H$ (infinite dimensional). I could easily prove that $\|VA\|=\|A\|$. But, I just couldn't prove that $\|AV\|=\|A\|$. Is this last property correct, and if yes could you supply a proof?

  • $\begingroup$ Thanks Jonas...I am relieved! I just could not do it and to be honest I statrt to find counterexamples but this result appeared without proof in Kubrusly book. Now, I think he means $\|BV^*\|=\|B\|$. $\endgroup$ – user220398 Mar 2 '15 at 18:18

You were correct to doubt that. It is false in general that $\|AV\|=\|A\|$ when $V$ is an isometry. If $V$ is any nonunitary isometry, there is a nonzero $A\in B(H)$ such that $AV=0$, because the range of $V$ is a proper closed subspace of $H$, so $A$ can be chosen to be $0$ on this range but nonzero elsewhere. In particular, $I-VV^*$ is the projection onto the orthogonal complement of the range of $V$, and $(I-VV^*)V=0$, while $\|I-VV^*\|=1$ unless $V$ is unitary.

For a very concrete example, on $\ell^2$ take $V(x_0,x_1,x_2,\ldots)=(0,x_0,x_1,\ldots)$, and $A(x_0,x_1,x_2,\ldots)=(x_0,0,0,\ldots)$.

You mention that perhaps your source meant to say that $\|AV^*\|=\|A\|$ when $V$ is an isometry on $H$ and $A$ is a bounded operator on $H$. This is true and easy to show using the fact that $\|X^*\|=\|X\|$ for all $X\in B(H)$ and the result you already proved.


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