4
$\begingroup$

Let $U$ be a linear isometry between Hilbert spaces. Why does the fact that the range of $U$ is dense imply that the range of $U$ is closed?

I am trying to understand the proof of theorem 5.4 in Conway's A Course in Functional Analysis.

$\endgroup$
5
$\begingroup$

In fact, the range of a linear isometry $U \colon H \rightarrow H'$ between Hilbert spaces must always be closed. If the range of $U$ is also dense then $U(H) = H'$ so $U$ is one-to-one and onto.

The reason is that the range $U(H)$ of $U$ is complete and a complete subspace of a normed space must be closed. To see that $U(H)$ is complete, let $Ux_n$ be a Cauchy sequence in $U(H)$ so $\| Ux_n - Ux_m \|_{H'} \rightarrow 0$. Since $U$ is an isometry, $\| x_n - x_m \|_{H} = \| Ux_n - Ux_m \|_{H'}$ and so $(x_n)$ is Cauchy in $H$. Since $H$ is complete, $x_n \rightarrow x$ for some $x \in H$ but then $Ux_n \rightarrow Ux$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.