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.
Best Answer
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.