Dual of an inner product space (not necessarily Hilbert space) is Hilbert space?

Question

I know that if H is an Hilbert space, then H' is also a Hilbert space. Now if X is just an inner product space (specifically not a Hilbert space), does it follow that X' is Hilbert? And if it is not the case, what's the counterexample?

I have proved that if X is real, then it is indeed true (using parallelogram equality). What's about the complex case?

The dual of any normed linear space (real or complex) is complete. — Kavi Rama Murthy, Apr 06 '24 at 08:11
Related: if $X$ and $Y$ are normed spaces and $Y$ is complete, then the space $\mathcal{B}(X,Y)$ of bounded linear operators from $X$ to $Y$, normed by the operator norm, is complete, also this question and answer — leslie townes, Apr 06 '24 at 08:20
The dual space is naturally isometrically isomorphic to the dual space of the completion, which is a Hilbert space. So, being isomorphic to the dual space of a Hilbert space, it must be a Hilbert space itself. — David Gao, Apr 06 '24 at 08:22
If $X'$ is Hilbert, then $X''$ will also be Hilbert (by the first sentence of your post). And not only that, $X'$ is isometrically isomorphic to $X''$ due to the Riesz Representation Theorem. And there is a canonical embedding of $X$ and $X''$ by the map $x\mapsto \text{ev}{x}$ where $\text{ev}{x}(f)=f(x)$ for all $f\in X'$. But as $X'$ and $X''$ are Hilbert, this embedding is an isometric isomorphism. Hence $X$ itself is also complete — Mr. Gandalf Sauron, Apr 06 '24 at 08:22
By the way, I’m not sure what you think the difference between real and complex cases is. The parallelogram law characterizes norms induced by inner products in both the real and the complex cases. — David Gao, Apr 06 '24 at 08:23
Also, the Parallelogram identity is for identifying iff the norm is given by an inner product. In particular, you don't get information about the completeness of the norm by it. In this question, the main thing to prove is that $X$ itself is complete. — Mr. Gandalf Sauron, Apr 06 '24 at 08:24
@Mr.GandalfSauron That doesn’t make any sense. The embedding is certainly isometric, but there’s no reason for it to be surjective. In fact, any normed space embeds naturally (and isometrically) into its double dual, and its double dual is always complete. That certainly does not mean all normed spaces are complete. — David Gao, Apr 06 '24 at 08:26

Dual of an inner product space (not necessarily Hilbert space) is Hilbert space?

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