3
$\begingroup$

Can anyone give a brief proof or a reference of a proof for the following property of Hilbert spaces?

If $H$ is a Hilbert space and $M$ is a closed subspace of $H$ then $H/M$ is a Hilbert space.

Definitions:

  • A Hilbert space $H$ is a real or complex inner product space that is also a complete metric space with respect to the distance function induced by the inner product.

  • A closed subspace of $H$ is a linear subspace of $H$ that is closed with respect to the topology induced by the norm (which is induced by the inner product).

I know that $H/M$ denotes the quotient set. But I don't know how to prove the Hilbert space structure on this set.

$\endgroup$
5
$\begingroup$

I will show you that if $X$ is Banach and Y is a closed subspace then $X/Y$ is Banach. From this hopefully you can extrapolate to your case. Now, you just need to define an inner product on your factor space.

To that end, let $X$ be Banach and $Y$ a closed subspace, and let $\{X_n\} \subset X/Y$ be an absolutely convergent series in $X/Y$, i.e, $\sum_{n \in \mathbb{N}}{\|X_n\|_{X/Y}} < \infty$. Recall that our quotient norm is defined as $$\|x\|_{X/Y}=\inf_{u \in x+Y}{\|u\|_X}=\inf_{y \in Y}{\|x+y\|_X}.$$ Then by definition of the quotient norm for any $X_n \in X/Y$ there exists $x_n \in X_n$ such that $$\|x_n\|_{X} \leq \|X_n\|_{X/Y}+2^{-n}.$$ Then we have that $\sum_{n \in \mathbb{N}}{\|x_n\|_{X}} < \infty$ and hence is absolutely convergent and therefore $\exists x \in X$ such that $x=\sum_{n \in \mathbb{N}}{x_n}$. Note this follows from the fact that $X$ is a Banach space.

Now, define $X=x+Y$ and $S_k=\sum_{n=1}^{k}{x_n}+Y$. Now, note that $$\|U-V\|_{X/Y}=\inf_{k \in (u-v)+Y}{\|k\|_{X}} = \inf_{y \in Y}{\|(u-v)+y\|_{X}} \leq \inf_{y \in Y}{ \|u-v\|+\|y\|}$$ and lastly, we note that $y=0 \in Y$ and hence we have the right most term is equal to $\|u-v\|_{X}$. Hence, we get that $$\|X-S_k\|_{X/Y} \leq \|x-s_k\|_{X} \underset{k \to \infty}{\longrightarrow} 0,$$ where $s_k=\sum_{n=1}^{k}{x_n}$. Hence, we have that our series in our factor space is convergent and thus $X/Y$ is Banach

Per my comment below:

You can also do this by noting that Given a closed subspace $Y \subset X$, $X/Y \simeq Y^{\perp}$ and the result follows trivially.

$\endgroup$
  • $\begingroup$ Thanks for the help, yes, I know, under the hypothesis it is easy to see that $H/M$ is Banach, but, one question, the norm about the quotient defines an internal product? $\endgroup$ – ChuckTesta May 29 '17 at 14:42
  • $\begingroup$ You tell me. Do you know how to check? Another way you could do this is recall the decomposition theorem, namely given a closed subspace of H $ = Y \oplus Y^{\perp}$ then you can conclude that $H/Y \simeq Y^{\perp}$. Then it follows trivially. $\endgroup$ – ADA May 29 '17 at 14:49
  • $\begingroup$ you're right, it's more easy from the theorem of the descomposition, thanks a lot. $\endgroup$ – ChuckTesta May 29 '17 at 14:53
3
$\begingroup$

First of all, note that $H/M$ is a normed linear space, with the norm $$\|[x]\|_{H/M}=\inf_{y\in M}\|x-y\|_H,$$ where $\|\cdot\|_H$ is the norm induced by the scalar product on $H$. To show that $H/M$ is a Hilbert space, we need to show two things: that $H/M$ is a complete metric space (under the metric induced by $\|\cdot\|_{H/M}$) and that $\|\cdot\|_{H/M}$ is induced by an inner product.

We can do this by the following three steps:

  1. Show that $H/M$ is isometrically isomorphic to $M^\perp$ via the map $\psi\colon H/M\to M^\perp$ defined by $\psi\colon [x]\mapsto P^\perp x$, where $P^\perp$ denotes the orthogonal projection of $H$ onto $M^\perp$.

  2. Note that $M^\perp$ is a closed subspace of $H$. Since $H$ is a complete metric space, and closed subsets of complete metric spaces are complete themselves, this shows that $M^\perp$ is a complete metric space. Show that this, together with the isometry, means that $H/M$ is complete.

  3. Recall that a norm on a linear space is induced by an inner product if and only if the norm is subject to the parallelogram law. Note that since $\|\cdot\|_H$ is induced by an inner product, it must satisfy the parallelogram law. Use the isometry to conclude that $\|\cdot\|_{H/M}$ also satisfies the parallelogram law, and thus is induced by an inner product, just like we wanted to show.

$\endgroup$

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.