# Process to complete a pre Hilbert space?

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?

• Cf section completion in Complete metric space on Wikipedia. – user251257 Dec 5 '16 at 23:37
• Kreyszig Introductory Functional Analysis is a good reference. – Chee Han Dec 6 '16 at 0:01
• 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. – Aweygan Dec 6 '16 at 6:46
• Sorry, I had a typo. The reverse triangle inequality will give you boundedness. – 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.