TITLE: Defining the value of a distribution at a point
QUESTION [9 upvotes]: Let $\omega\in D'(\mathbb R^n)$ be a distribution and $p\in \mathbb R^n$. If there is an open set $U\subset \mathbb R^n$ containing $p$ such that $\omega|_U$ is  given by a continuous function $f\in C(U)$, then for every  $\phi\in C^\infty_c(\mathbb R^n)$ with $\int_{\mathbb R^n}\phi(x)d x=1$ we can define a Dirac sequence $\{\phi^p_j\}_{j\in \mathbb N}\subset D(\mathbb R^n)$ by $\phi^p_j(x):=j^n\phi(j(x-p))$ which fulfills
$$
\omega(\phi^p_j)\to f(p)\quad \text{ as }j\to \infty.
$$
This shows that we can recover the value $\omega(p)\equiv f(p)$  of the distribution $\omega$ at the point $p$ via a limit of such Dirac sequences.
Now, suppose that for some $\omega\in D'(\mathbb R^n)$ and $p\in \mathbb R^n$ we just know that
$
\lim_{j\to \infty}\omega(\phi^p_j)
$ 
exists for every $\phi\in C^\infty_c(\mathbb R^n)$ with $\int_{\mathbb R^n}\phi(x)d x=1$ and is independent of $\phi$. In view of the above it then seems reasonable to define $\omega(p):=\lim_{j\to \infty}\omega(\phi^p_j)$ and to say that $\omega$ has a well-defined value at the point $p$.
Q: Is this definition useful in any sense? I have the feeling that it might be fundamentally flawed. In that case, I'd find it interesting to know what's the greatest generality in which one can make sense of "the value of a distribution at a point".
Additional thoughts after 1st edit: Some "consistency checks" for the definition would in my opinion be the following: 

If the value of $\omega$ exists at every point in some open set $U\subset \mathbb R^n$ and the function $f$ defined on $U$ by these values is continuous, then $\omega|_U$ is given by $f$.
If the value of $\omega$ exists at Lebesgue-almost  every point in some open set $U\subset \mathbb R^n$ and the values define a function $f\in L^1_{\mathrm{loc}}(U)$, then $\omega|_U$ is given by $f$.

I believe that at least property 1 should be true and I'll check it once I find the time.
2nd edit: My question is related to this MO question which corresponds to the case $f\equiv 0$.

REPLY [7 votes]: As indicated above, the concept of the limit resp. value of a distribution at a point was studied intensively over 50 years ago.  Here is a very  elementary and natural definition due to Sebastião e Silva (it is  definition 6.9 in his paper „On integrals and orders of growth of distributions“. (I will not give a reference since it can be found online just by googling the title).
A distribution $s$ on an interval $I$ is said to be continuous at a point $c$ if there is a natural number $p$ and a continuous function $F$ on $I$ so that $f=D^pF$ (distributional derivative) and $\dfrac {F(x)}{(x-c)^p}$ converges in the usual sense as $x$ goes to $c$  Then we write $f(c)$ for $p!$ times this limit and call it the value of the distribution at $c$. As an example, he shows that $\cos \frac 1 x$ has the value $0$ at $0$.

REPLY [3 votes]: This is not an answer, and maybe even marginally off-topic, but I'd like to point out the following example which might be useful to keep in mind when trying to define the value of a distribution at a point (and which is too long to fit in a comment):
Let $g\colon\mathbb{R}\to\mathbb{R}$ be $g(x) = x^2\sin(\frac{1}{x})$ (obviously extended by $g(0)=0$).  This is a continuous, in fact, even, differentiable, function on $\mathbb{R}$, so we can unproblematically identify it with a distribution, call it $T$.  Now since $g$ is differentiable, we probably want to identify its derivative $g'$, as a real function, with the derivative $T'$ of the corresponding distribution $T$, so we might want to conclude that the value at $0$ of the distribution $T'$ should be (well-defined and equal to) $g'(0) = 0$.  But since $g'$ is not continuous at $0$, it is not easy to come up with a justification for why $T'$ takes that value at that point.