# How to extend evaluation at a point from continuous maps to square-integrable ones?

Consider the Hilbert space $L^2[0, 1]$ of square integrable functions on $[0, 1]$. Saying more carefully, there is one subtle detail: this space is defined as factor space by the functions which are have norm zero, in particular if we change value of $f(x)$ at one point, then in this factor space the two functions will be still the same. So formally speaking the value at points of elements from $L^2[0 1]$ is not defined.

Take point $x$ in $[0, 1]$. A continuous function $f$ can be evaluated at $x$: $ev_x (f) = f(x)$.

Question: Can the linear functional $ev_x$ be extended from continuous functions to $L^2[0, 1]$? In other words, does there exists a continuous linear functional $\Psi:L^2[0, 1]\to \mathbb{R}$, such that it coincides with $ev_x$ on the subspace of continuous functions?

Another way to Reformulation: let $i: C[0, 1] \to L^2[0, 1]$ be the inclusion and consider the dual map $i : (L^2[0,1])^{ * } \to (C[0,1])^{ * }$. The dual spaces involved are known: for $C[0, 1]$ it is space of functions of bounded variation, whereas $(L^2[0, 1])^{*} = L^2[0, 1]$. So we can ask whether $ev_x$ lies in the image of $i^{ * }$.

Motivation: in physics "coherent states" defined by this idea, (as far as I understand), e.g. approach by Rawnsley. But they are defined using "holomorphic polarization", where of course we can take value of holomorphic function at a point. So I am puzzled wether we really need holomorphic polarization.

-
Of course not: $ev_x$ is not continuous on $C([0,1])$ wrto the $L^2$ norm. There are continuous functions $f$ with arbitrarily small $L^2$ norm and such that $f(x)=1$. – Pietro Majer Feb 2 '12 at 12:52
Heh, you are right! Please write it as answer I will accept it. – Alexander Chervov Feb 2 '12 at 12:58

We can't extend the evaluation at $x$ to a continuous linear functional on $L^2[0,1]$, because it is not continuous on $C([0,1])$ wrto the $L^2$ norm. For instance $f_n(t):=(1-nt)_+$ has $f_n(0)=1$ and $\|f_n\|_2\to 0$.