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.

with the $L^2$ topology. $\endgroup$ – Andrew Stacey Nov 12 '09 at 8:30