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.



How to extend evaluation at a point from continuous maps to square-integrable ones?
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 $\|f\_n\|\_2\to 0$.


