Defining the value of a distribution at a point

Question

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:

1. 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 $f$.

2. 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 $f\equiv 0$.

Answer 1

The definition of the value of a distribution at a point you describe in your question does not seem flawed to me since, at least from the point of view of the independence on $\delta$-sequences, follows the path traced years ago by Stanisław Łojasiewicz in the paper [1], so I describe his approach to the problem below.

Łojasiewicz analyzes the problem for functions of one variable, i.e. $n=1$: by using the definition of change of variables in a distribution (see for example [2], §1.9 pp. 21-22) and considering the change of variable $y=x_0+\lambda x$, for $ x,x_0,\lambda \in\Bbb R$, i.e. $$ \begin{split} T(x_0+\lambda x)&\triangleq \langle T(x_0+\lambda x),\varphi(x)\rangle\\ &=\left\langle T(y),\frac{\varphi\big(\lambda^{-1} (y-x_0)\big)}{\lambda}\right\rangle \end{split} \quad \varphi\in\mathscr{D}(\Bbb R) $$ he defines the limit of a distribution at a point $x_o$ as ([1], §1 p. 2-3) $$ \lim_{x\to x_0} T\triangleq \lim_{\lambda\to 0} T(x_0+\lambda x) \label{1}\tag{1} $$ and proves that

• $\lim_{x\to x_0} T=\lim_{x\to x_0^+} T=\lim_{x\to x_0^-} T$
• by using an earlier result of Zieleźny, if the limit \eqref{1} exists, it is necessarily a constant $C\in \Bbb C$, or more precisely a constant distribution $C$.
• a necessary and sufficient condition for the limit \eqref{1} to exist is (see [1], §2, theorem 2.2, pp. 5-7) that $T=f^{(n)}$, where $f\in C^0(\Bbb R)$ and $$ \lim_{x\to x_0}\frac{f(x)}{(x-x_0)^n}=\frac{C}{n!}. $$ Then Łojasiewicz assumes \eqref{1} as the definition of the value of a distribution at a point: note that this definition does not rely on any particular test function (or sequence of such) $\varphi\in\mathscr{D}(\Bbb R)$, as stated above. Now a few observations:

1. Łojasiewicz ([1], §1 p. 1) states that the case $n>1$ will be analyzed in a subsequent paper which to my knowledge has never been published. However (but this only my opinion), a generalization of \eqref{1} could perhaps be tried by using the Stoltz condition as described, for example, in the textbook of Griffith Bailey Price (1984) Multivariable Analysis, Springer-Verlag.
2. Łojasiewicz gives another necessary and sufficient condition for the limit \eqref{1} to exists, in terms of Denjoy differentials ([1], §2, corollary to theorem 2.2, p. 7).
3. The term $\lambda^{-1}$, more or less intrinsically used in \eqref{1}, suggests the possible use of the Mellin transform: this suggestion was followed by Bogdan Ziemian in [3], §12 pp. 41-42. He defines the (generalized) spectral value of a function/distribution at a point and proves ([3], §12 p. 43) that it coincides with Łojasiewicz point value \eqref{1} when the latter exists (by using the necessary and sufficient condition above): however, the construction of Ziemian does not apply to all distributions.

References

[1] Stanisław Łojasiewicz (1957-1958), "Sur la valeur et la limite d'une distribution en un point" (French), Studia Mathematica, Vol. 16, Issue 1, pp. 1-36, MR0087905 Zbl 0086.09405.

[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

[3] Bogdan Ziemian (1988), "Taylor formula for distributions", Rozprawy Matematyczne 264, pp. 56, ISBN 83-01-07898-7, ISSN 0012-3862, MR0931848, Zbl 0685.46025.

Answer 2

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 $0$.

Answer 3

It's not a bad definition and I think it is better to think of it as a particular case of the "restriction problem", i.e., trying to define the restriction $\omega|_{\Gamma}$ of $\omega$ to some subset $\Gamma\subset\mathbb{R}^n$. When one succeeds the result is called a trace theorem. This usually requires some quantitative regularity hypothesis on $\omega$, e.g., being in a Sobolev space $H^s$ with $s>$something.

A particularly important case is when $\Gamma$ is an affine subspace, or say for simplicity a linear subspace like $\Gamma =\mathbb{R}^m\times\{0\}^{n-m}\subset\mathbb{R}^n$. A rather standard approach is to start with $\omega\in\mathcal{D}'(\mathbb{R}^n)$. The convolution $\omega\ast \phi_j^0$ is in the space of $C^{\infty}$ functions $\mathcal{E}(\mathbb{R}^n)\subset \mathcal{D}'(\mathbb{R}^n)$ and converges to $\omega$ in the topology of $\mathcal{D}'(\mathbb{R}^n)$ (the strong topology). The ordinary restriction $\omega\ast \phi_j^0|_{\Gamma}$ makes sense and you can ask if the limit $\lim_{j\rightarrow\infty}\omega\ast \phi_j^0|_{\Gamma}$ exists inside $\mathcal{D}'(\mathbb{R}^m)$.

Your particular case $p=0$ corresponds to mine with $m=0$.

Another problem of this kind is pointwise multiplication. If $\omega_1(x)$ and $\omega_2(x)$ are two distributions, then there is no problem defining $\omega_1(x_1)\omega_2(x_2)$ (tensor product), but the issue is how to restrict to the diagonal $\Gamma=\{x_1=x_2\}$.

Finally, note that all of these problems become much more interesting for random distributions, because it's like magic: you can sometimes do the (deterministically) impossible.

---

Small addendum: Suppose that for some reason one has a trace theorem but only for large enough $m$ and one cannot do the $m=0$ or the point restriction case. Then one can still do the following "stabilization" trick: change $\omega$ to $\omega\otimes 1$ where one tensors with the constant function equal to one seen as a distribution in say $p$ new variables. If you can restrict it from $\mathbb{R}^{n+p}$ to a subspace of dimension $p$, then you will have your point evaluation after factoring out the $\otimes 1$. The last step of course needs your restriction construction to be invariant/covariant by translation along $\Gamma$.

Answer 4

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.