Does every inner product space have Hilbert completion which is a Hilbert space? If so, how can we define the inner product in the Hilbert space?


Yes, note that an inner product induces a norm by $\|x\|:=\sqrt{\langle x,x\rangle}$ and a norm is induced by an inner product iff it satisfies the the parallelogram law: $$2\|x\|^2+2\|y\|^2 \overset!= \|x+y\|^2+\|x-y\|^2\qquad \text{ for all $x,y\in V$}.$$

So if $V$ is equipped with an inner product let $\overline{V}$ denote the completion of $V$ wrt the norm induced by the inner product. If $x,y\in \overline V$ and $x_n,y_n\in V$ with $x_n\to x$, $y_n\to y$ you have that: $$2\|x\|^2+ 2\|y\|^2= \lim_n (2\|x_n\|^2+2\|y_n\|^2) =\lim_n( \|x_n+y_n\|^2+\|x_n-y_n\|^2)\\ = \|x+y\|^2+\|x-y\|^2$$ hence the norm on $\overline V$ still obeys the parallelogram law and as such $\overline V$ is a Hilbert space completion of $V$.

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    $\begingroup$ And a phrase I like that succinctly describes this episode as well as many others, is "extension by continuity". $\endgroup$ – paul garrett Jun 7 '20 at 19:41
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    $\begingroup$ What is the meaning of the $\overset!=$ operator in your first equation? $\endgroup$ – Carmeister Jun 8 '20 at 1:02
  • $\begingroup$ @Carmeister it has no formal meaning, it just means something like "this is the important equality", "pay attention to this" or, in this case, "this is what must hold". $\endgroup$ – s.harp Jun 8 '20 at 12:15

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