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In more detail, can one establish that the continuous linear dual of a Hilbert space is again a Hilbert space without appealing to the Riesz Representation Theorem?

For me, the Riesz Representation Theorem is the result that every continuous linear functional on a Hilbert space is of the form $v \mapsto \langle v, u \rangle$ for some $u$ in the Hilbert space.

Whilst I have no particular quarrel with the Riesz Representation Theorem itself, I wonder if it's possible to do without it. My motivation is fairly flimsy, but consider the situation where you have an arbitrary inner product space, $V$. Then its dual is a Hilbert space. However, to use Riesz Representation to prove that, you first have to complete $V$ to a Hilbert space and then apply Riesz. Completing metric spaces, and in particular showing that the completion of an inner product space is a Hilbert space, seems like a lot of just hard slog to me (and hard to motivate to students in particular) so I wondered if one could avoid it by proving directly that the dual was a Hilbert space.

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Surely you mean the continuous dual? –  Theo Johnson-Freyd Nov 12 '09 at 1:52
Really?!? If I take the continuous functions on an interval, then its dual space contains the delta functions, and I have no idea what the inner product of a delta function with itself is supposed to be. –  Theo Johnson-Freyd Nov 12 '09 at 1:53
The first two times I wrote "continuous linear dual", I guess I figured that would be enough to just write "dual" in the rest. To answer your second comment, delta functions are not in the continuous linear dual of the space of continuous functions on an interval with the $L^2$ topology. –  Andrew Stacey Nov 12 '09 at 8:30
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4 Answers

It's been a while since I was made to look at the proof rather than just quote it, but IIRC the gist of the RRT for Hilbert spaces, is the bijection between closed hyperplanes (=closed codimension $1$ subspaces) in a Hilbert space $H$ and the lines orthogonal to each, and the fact that this can be set up so as to be conjugate-linear. This in turn is based -- I think -- on the fact that for each $x \in H$ and each closed subspace $V$ there is a unique point in $V$ closest to $x$.

If you wanted to look at the (continuous) dual of an inner product space $E$, then the above reasoning suggests to me that completion of $E$ is going to enter the picture somehow. For if $\psi$ is a continuous linear functional on $E$, we want to consider $\ker \psi$ and then associate to it a choice of normal vector, but I'm not sure we can show that a suitable choice exists without using completeness.

(There exist plenty of codimension 1 dense subspaces in incomplete i.p. spaces, of course: equip $C[0,1]$ with the inner product given by integrating along $[0,1]$, i.e. the $L^2[0,1]$ inner product, and consider the subspace of $C[0,1]$ consisting of all those functions in it which vanish at $0$. So in the setting above, the continuity of ψ has to get used in the proof that the dual of $E$ is a Hilbert space.)

I take your point that perhaps there is a way to show that the dual of $E$ is a Hilbert space, which doesn't start by completing $E$. But one may end up constructing some kind of abstract completion anyway.

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At the risk of further muddying the waters, one application of this approach would be that it gave one a construction of the completion that didn't involve Cauchy sequences: simply take the bidual. –  Andrew Stacey Nov 11 '09 at 19:51
The idea of "taking the bidual" is actually a good approach to take the completion of normed space: if X is a normed space, then X* and also X** are Banach spaces, and X embeds isometrically into X**. The completion of X is just the closure of X in X**. I'm certain I've seen this in books, although I couldn't give you an example right now. –  Matthew Daws Nov 11 '09 at 21:31
Yes, this works in general. Though for an inner product space it only establishes that the completion is a Banach space, not a Hilbert space. To show that the norm comes from an inner product on the completion needs a continuity argument (or needs the result that I'm asking for). –  Andrew Stacey Nov 12 '09 at 8:32
If you wish, you could recover the inner product on the completion using the polarization identity. Or more directly, given two sequences $(u_n)$ and $(v_n)$ in the original space converging to $u$ and $v$ respectively, define $\langle u,v\rangle=\lim\langle u_n,v_n\rangle$ (the latter is easily seen to be a Cauchy sequence of complex numbers. –  Harald Hanche-Olsen Nov 12 '09 at 12:29
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I haven't thought this through. But the continuous linear dual is a Banach space, and if you could show that it satisfied the parallelogram rule then the norm would come from an inner product, yes?

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This is the route I would start down. I have no idea how easy this would be, though. –  Andrew Stacey Nov 11 '09 at 19:52
Good point; hadn't thought of this, although one now has the problem of finding appropriate "norming vectors" for given functionals, in order to verify the parallelogram rule. But perhaps, since we don't have to show that maxima are attained, we can get round incompleteness of the original i.p. space... I'll have to think about this some more. –  Yemon Choi Nov 11 '09 at 19:58
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Well, there is a way, but I don't know if I like it: Let $V$ be a complex inner product space, and for any $f\in V^*$ and $\epsilon>0$ let $$V^f(\epsilon)=\{v\in V:\|v\|=\|f\|, f(v)\ge\|f\|^2(1-\epsilon)\}$$

Now, as $\epsilon\to0$ we expect the functionals induced by $v\in V^f(\epsilon)$ to converge to $f$. More precisely, let $\bar v$ be the functional $\bar v(u)=\langle u,v\rangle$. It's easy to estimate $|f(u)-\bar v(u)|$ when $u$ is parallel to $v$. It's just a tiny bit harder to do so when $u\perp v$: Assume first that $\|u\|=\|v\|=\|f\|$ and consider $f(\overline{f(u)}u+\overline{f(v)}v)$, compute the norm using Pythagoras and employ the definition of $\|f\|$ to find $$\|f(u)\|^2\le\|f\|^4-|f(v)|^2\le\|f\|^4(2\epsilon-\epsilon^2)=\|f\|^2\|u\|^2(2\epsilon-\epsilon^2)$$

I left out some details, but the end result is that $\|\bar v-f\|\to0$ when $v\in V^f(\epsilon)$ and $\epsilon\to0$. Now we can finally define the inner product on $H^*$ by $$\langle f,g\rangle=\lim\langle w,v\rangle$$ where $v\in V^f(\epsilon)$, $w\in V^g(\epsilon)$ and $\epsilon\to0$.

Pretty? I don't think so, but it seems to answer your question. As a bonus, the Riesz representation theorem is now of course just around the corner.

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I wouldn't expect a "pretty" answer since it's going to involve trying to get at the norm of the functionals by evaluating them on vectors, we can only do that approximately. –  Andrew Stacey Nov 13 '09 at 12:35
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You might do it by coordinatizing everything. That is, if you're happy using an orthonormal basis $(e_i)_{i\in I}$ for the Hilbert space (where the index set may well be uncountable), for then if $f$ is a linear functional it is not hard to show that $f(x)=\sum\langle x,e_i\rangle f(e_i)$ and hence $\|f\|^2=\sum|f(e_i)|^2$, and you're essentially done.

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Not recognising '\sb' is extremely annoying as this would be a neat solution to the problem. But underscores are usually okay in the real markdown, but the preview version isn't as smart as the one on the server. –  Andrew Stacey Nov 11 '09 at 21:09
I thought about that a few minutes ago as I was preparing a lecture on this (which is what sparked the question!), but couldn't see easily how to get from $f(x) = \lim \langle x, \sum_{j=1}^n f(e_i)e_i \rangle$ to showing that $\sum |f(e_i)|^2$ converges. But as you surmise, I'm happier using orthonormal bases than using completeness. Hmm, maybe I ought to add this to my answer to "When to choose a basis"... –  Andrew Stacey Nov 11 '09 at 21:15
Use the Cauchy–Schwarz inequality for sums! $f(x)=|\sum f(e_i)\langle x,e_i\rangle|^2\le(\sum|\langle x,e_i\rangle|^2)(\sum|f(e_i)|^2)=\|x\|^2\sum|f(e_i)|^2$ and more to the point, use your knowledge of when the inequality becomes equality (with finite sums to play it safe) and conclude that $\|f\|^2=\sum|f(e_i)|^2$. (Gee, writing this with no preview is scary.) –  Harald Hanche-Olsen Nov 12 '09 at 1:15
Ditto Andrew's comment re \sp. Real markdown (serverside) does not try to process underscores that are in the middle of strings, usually, although preview (clientside) does. Also, markdown (eitherside) will convert \_ into _ (in case that doesn't render --- no preview in comments --- I mean it converse backslash-underscore into underscore) before sending the text to jsMath. –  Theo Johnson-Freyd Nov 12 '09 at 1:58
(Ah, no, it messed up. Let's try again: I mean that markdown takes \\_ and makes it into \_.) –  Theo Johnson-Freyd Nov 12 '09 at 2:00
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