I'm looking for either a reference or a guildeline on how to complete a pre-hilbert space(vector space with inner product). Many sources I've read normally just state it's possible and I'm interested in how.

I do know that if $(X,d)$ is a metric space, I can isometrically embed $X$ into the Banach space $l^\infty(X)=\{f:X\to\mathbb{C}|$f$ $ is bounded$\}$. To do this, we fix any $y\in X$ and we map $x\mapsto\phi_x$, where $\phi_x(t)=d(t,x)-d(t,y)$.

Now I don't think this process will work when $X$ also happens to be a pre-hilbert space. In particular, my function from $X\to l^\infty (X)$ won't be linear. Is it a similar approach though?

  • $\begingroup$ Cf section completion in Complete metric space on Wikipedia. $\endgroup$ – user251257 Dec 5 '16 at 23:37
  • $\begingroup$ Kreyszig Introductory Functional Analysis is a good reference. $\endgroup$ – Chee Han Dec 6 '16 at 0:01
  • $\begingroup$ I don't understand your example showing that any metric space $(X,d)$ can be embedded into $\ell^\infty(X)$. If $X=\mathbb{R}$ with the usual metric, and $x=y$, then $\phi_x(t)=|t-x|$ is not bounded. $\endgroup$ – Aweygan Dec 6 '16 at 6:46
  • $\begingroup$ Sorry, I had a typo. The reverse triangle inequality will give you boundedness. $\endgroup$ – Jake Dec 6 '16 at 13:23

Now I don't think this process will work when $X$ also happens to be a pre-hilbert space. In particular, my function from $X \to l^{\infty}(X)$ won't be linear.

That means making it work will be more cumbersome, but one can still make it work. If we have an isometric embedding $J \colon X \to Y$ where $Y$ is a complete metric space, we can extend addition and scalar multiplication of $X$ to $Z := \overline{J(X)}$ so that $Z$ becomes a Hilbert space. The addition on $X$ is transported to $J(X)$ via

$$\alpha_0 (u,v) := J(J^{-1}(u) + J^{-1}(v)).$$

Since the addition $X\times X \to X$ is uniformly continuous and $J$ is isometric, $\alpha_0$ is uniformly continuous, and hence has a uniformly continuous extension $\alpha \colon Z \times Z \to Z$. One then uses the continuity of $\alpha$ to show that $(Z,\alpha)$ is an abelian group. In a similar vein, for each $z \in \mathbb{C}$, the multiplication with $z$ is uniformly continuous on $X$, and thus has a uniformly continuous extension $\mu_z \colon Z \to Z$. One then verifies (using continuity) that $\odot \colon (z,u) \mapsto \mu_z(u)$ is a scalar multiplication compatible with $\alpha$, so that $(Z,\alpha,\odot)$ becomes a complex vector space. Then one checks that the uniformly continuous extension of the norm on $X$ yields a norm on $Z$ that satisfies the parallelogram identity, so $Z$ is a Hilbert space, and $J \colon (X, +, \,\cdot\,) \hookrightarrow (Z, \alpha, \odot)$ is an isometric linear embedding with dense image.

That is, however, not the most convenient way to obtain a completion of $X$ if $X$ is a normed space.

For normed spaces, the most convenient way to obtain a completion is to consider the canonical embedding $\Phi \colon X \hookrightarrow X'';\, \Phi(x) \colon \lambda \mapsto \lambda(x)$, where $Y'$ is the topological dual of a topological vector space, i.e. the space of continuous linear functionals on $Y$. Since all dual spaces of normed spaces are Banach spaces, $X''$ is complete, and since $\Phi$ is isometric, $\tilde{X} = \overline{\Phi(X)} \subset X''$ is a completion of $X$. In the case where $X$ is a (Hausdorff) pre-Hilbert space, it's easy to see that $\tilde{X}$ is a Hilbert space. The parallelogram identity of the norm on $X$ extends to $\tilde{X}$ by continuity. In this case, we always have $\tilde{X} = X''$, but for general normed spaces, it can be a proper subspace.

For metrisable topological vector spaces, one can look at the construction of the completion of a metric space using Cauchy sequences, and check that the construction is compatible with the linear structure of the space, so that it naturally leads to a complete metrisable TVS.

For general (Hausdorff) topological vector spaces, one follows the programme at the top of this answer, one considers the completion of $X$ as a uniform space, and extends the vector space structure to the completion as sketched above.


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.