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.


  • 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.


2 Answers 2


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.


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.

  • $\begingroup$ 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, 2017 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, 2017 at 14:49
  • $\begingroup$ you're right, it's more easy from the theorem of the descomposition, thanks a lot. $\endgroup$
    – ChuckTesta
    May 29, 2017 at 14:53

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