When is the completion of a space of functions a space of functions? If $V$ is a $\mathbb C$-vector space of functions $f: X \to \mathbb C$ on some common domain $X$ and $\tau$ is a Hausdorff, locally convex topology on $V$, when may the completion of $(V,\tau)$ also be realized as a space of functions on $X$? Are there any general conditions that can be imposed on $V$ and $\tau$ that will ensure this?
Since completions are, of course, determined only up to isomorphism, the question is perhaps in too imprecise a form to provide a meaningful answer, but it's motivated by the difference between basic examples from functional analysis. Compare, for instance, the difference between (1) the space of polynomials on $[0,1]$ having completion $C[0,1]$ (a space of functions on $[0,1]$) under the uniform norm and (2) the vector space of integrable simple functions on $[0,1]$ having $L^p[0,1]$ as its completion with respect to $\|\cdot\|_p$. More dramatic examples than (2) appear to be common. One from Sobolev theory I stumbled on recently can be found in Theorem 3.1 and the subsequent Remark in this article.
Thank you for any help or references you can provide.
 A: The key issue is the continuity of point evaluation. For every $x\in X$ we have the evaluation functional $\phi_x : V\to\mathbb{C}$ defined by $\phi_x(f)=f(x)$.  If these functionals are continuous in $\tau$, they extends to the completion $\overline{V}$ and therefore allow us to interpret the elements of $\overline{V}$ as functions on $X$. On the other hand, if the evaluation functionals are not continuous, we are unlikely to have such an interpretation.  Consider your examples: 


*

*Polynomials with the uniform norm on $[0,1]$. The evaluation functional at any point is bounded with norm $1$. The completion is the space of continuous functions. 

*Polynomials (or continuous functions) with the $L^p$ norm on $[0,1]$, $1\le p<\infty$. The evaluation functional at any point $x_0$ is discontinuous, because the sequence $f_n(x)=\max(0,1-n|x-x_0|)$ converges to $0$ in the norm while $f_n(x_0)=1$. The completion $(L^p)$ ends up being the space of some equivalence classes of functions, but not of actual functions on $[0,1]$. 


Similarly with Sobolev spaces: point evaluation is bounded on $W^{k,p}$ if and only if $kp>n$ where $n$ is the dimension (exception: it's also bounded on $W^{1,1}$ in one dimension). These are precisely the Sobolev spaces whose elements can be naturally identified with functions, rather than equivalence classes.
