Does every inner product space have Hilbert completion?

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 $$V$$.
• What is the meaning of the $\overset!=$ operator in your first equation? – Carmeister Jun 8 '20 at 1:02