Can one do without Riesz Representation? 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.
 A: 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.
A: 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?
A: 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.
A: 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.
