TITLE: Necessity of completeness of the inner product space in Riesz representation theorem
QUESTION [14 upvotes]: I wanted to find a counter example to show that the completeness of the inner product space is necessary in Riesz representation theorem. Please give an example of a bounded linear functional $T$ on an incomplete inner product space $X$ which do not have any inner product representation i.e. there does not exist any $z$ in $X$ s.t. $T(x)= \langle ,z\rangle$ for all $x$ in $X$.

REPLY [2 votes]: Let $X$ be an inner product space and let $X'$ denote its continuous dual. Define $\Phi:X\to X'$ as $z\mapsto f_z$ where $f_z(x)=<x,z>$ for $x\in X$. Clearly $\Phi$ is an anti-linear isometry (we assume that the inner product is linear in the first variable). The Riesz theorem says that $\Phi$ is onto if $X$ is complete. You want to see if the converse is true, i.e. whether $\Phi$ is not onto whenever $X$ is not complete. One way to easily see this is by observing that the continuous dual of any normed linear space is always complete. Hence if $\Phi$ is onto, then it would become a bijective isometry of the underlying metric spaces, forcing $X$ to be complete. Formally, we can prove as follows.
If $X$ is not complete, then there exists a Cauchy sequence $(z_n)$ of (unit) vectors in $X$ which does not converge. Consider the corresponding sequence $(f_{z_n})$ in $X'$ which is also Cauchy, because $\Phi$ is anti-linear and isometry $(\Vert f_{z_n}-f_{z_m}\Vert=\Vert z_n-z_m\Vert)$. Thus $X'$ being complete, there exists $f\in X'$ such that $(f_{z_n})\to f$. We claim that there exists no $z\in X$ such that $f=f_z$. Assume, for arriving at a contradiction, that $f=f_z$ for some $z\in X$. As $\Vert z_n-z\Vert=\Vert f_{z_n}-f_z\Vert$ and the right hand side goes to $0$ as $n$ tends to infinity, we conclude that the sequence $(z_n)$ converges to $z$ which gives a contradiction to where we started.
For an explicit example, consider $f:X=c_{00}\subseteq\ell^2\to \mathbb{C}$ definded as $f(x)=\sum\frac{x_n}{n}$ where $x=(x_n)$.