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This question has been motivated by the post making sense of distributions on the diagonal.

Let $T$ be a tempered distribution on $\mathbb{R}^2$ and $\eta$ be a given mollifier on $\mathbb{R}$. For $f \in \mathcal{S}(\mathbb{R})$, let \begin{equation} \eta_{f,\epsilon}(x,y):= \frac{1}{\epsilon}f\Bigl(\frac{x+y}{2}\Bigr) \eta\Bigl(\frac{x-y}{\epsilon}\Bigr) \end{equation} and assume that the limit \begin{equation} \lim\limits_{\epsilon \to 0^+} T(\eta_{f,\epsilon}) \end{equation} exists for all $f \in \mathcal{S}(\mathbb{R})$.

Then, my issue is

For any other mollifier $\phi$ on $\mathbb{R}$, do we have $\lim\limits_{\epsilon \to 0^+} T(\phi_{f,\epsilon})=\lim\limits_{\epsilon \to 0^+} T(\eta_{f,\epsilon})$?

For a Schwartz function $F \in \mathcal{S}(\mathbb{R}^2)$ understood as a tempered distribution, a direct computation yields that \begin{equation} \lim\limits_{\epsilon \to 0^+} F(\phi_{f,\epsilon})=\lim\limits_{\epsilon \to 0^+} F(\eta_{f,\epsilon}) = \int_{\mathbb{R}} F(x,x) f(x) dx \end{equation}

Also, we know that the tempered dsitribution $T$ above can be approximated by Schwartz functions in the weak$^*$ topology. That is, we can find a sequence $F_n$ in $\mathcal{S}(\mathbb{R}^2)$ such that \begin{equation} F_n(g) \to T(g) \text{ as } n \to \infty \end{equation} for each $g \in \mathcal{S}(\mathbb{R}^2)$.

Therefore, my issue is reduced to finding a sequence $F_n$ approximating above $T$ such that \begin{equation} \lim\limits_{\epsilon \to 0^+} \lim\limits_{n \to \infty} F_n(\eta_{f,\epsilon}) = \lim\limits_{n \to \infty} \lim\limits_{\epsilon \to 0^+} F_n(\eta_{f,\epsilon}) \end{equation} and \begin{equation} \lim\limits_{\epsilon \to 0^+} \lim\limits_{n \to \infty} F_n(\phi_{f,\epsilon}) = \lim\limits_{n \to \infty} \lim\limits_{\epsilon \to 0^+} F_n(\phi_{f,\epsilon}) \end{equation} at the same time, for each $f \in \mathcal{S}(\mathbb{R})$.

However, I cannot see how to justify interchange of limits as written above. I guess I need some sort of "uniformity" in approximating $T$ by $F_n$.

Could anyone please help me?

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1 Answer 1

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For the first "highlighted issue": Not generally. For example, let $T$ be defined by $T\,g = \int_{-\infty}^{+\infty}g\,(s,s)\,{\rm d}s$. Taking $\eta$ with $\eta_1(0)=0$ we get $\lim_{\varepsilon\to 0^+}T\,(\eta_{f,\varepsilon}) = 0$ independently of $f$ whereas for $\phi$ with $\phi_1(0) > 0$ and for $f$ with $\int_{\,\mathbb R}f > 0$ we have $\lim_{\epsilon\to 0^+}T\,(\phi_{f,\varepsilon}) = +\infty$. This $T$ actually is the distribution that a theoretical physicist would write "$\delta(x-y)$".

For the problem of restricting a general (tempered) distribution to a submanifold that is not an open subset, you should "take seriously" some comments that are already given to some of your earlier questions, i.e. consider the singular support and wave front sets.

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    $\begingroup$ Certainly it is possible. For example on $\mathbb R$ take any smooth compactly supported $\varphi$ with integral equal to one and support included in $\mathbb R_+$ and define $\varphi_\varepsilon(s)=\varepsilon^{-1}\varphi(\varepsilon^{-1}s)$. $\endgroup$
    – TaQ
    Commented Mar 14 at 18:39
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    $\begingroup$ "any compactly supported function with unit integral is a mollifier". Add smooth there! This is precisely the case! Now you should carefully make some explicit computations before continuing this extended discussion! $\endgroup$
    – TaQ
    Commented Mar 14 at 18:45
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    $\begingroup$ Just compute and do not argue! $\endgroup$
    – TaQ
    Commented Mar 14 at 18:48
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    $\begingroup$ Your limit above is wrong. Compute it explicitely, noting that in the integral one takes the values of $f$ only near zero. Use the standard $\varepsilon,\delta$-argument, noting that $f$ is continuous at zero. $\endgroup$
    – TaQ
    Commented Mar 14 at 18:56
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    $\begingroup$ Ok, I see that I made a mistake. Sorry about this. $\endgroup$
    – Isaac
    Commented Mar 14 at 19:00

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