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Introduction

This question is about distributions as they are used in physics (a.k.a. generalized functions). They are written like functions from $\mathbb{R}$ to $\mathbb{R}$, but are understood as formal objects that are defined by their integral when multiplied with a test function (most likely, this means a function in $C_c^\infty(\mathbb{R})$). For example, the delta distribution $\delta(x)$ is defined by $$\int_\mathbb{R} \delta(x) f(x) \, dx = f(0) \qquad \forall \text{ test functions } f.$$

Using integration by parts, one can give a meaning to the derivative of a distribution. For instance, the derivative of the delta distribution is defined by $$\int_\mathbb{R} \delta'(x) f(x) \, dx = - \int_\mathbb{R} \delta(x) f'(x) \, dx = -f'(0) \qquad \forall \text{ test functions } f.$$

Problem

It is known that $$ x \, \delta'(x) = - \delta(x), $$ meaning $$\int_\mathbb{R} x \, \delta'(x) f(x) \, dx = \int_\mathbb{R} \delta(x) f(x) \, dx = f(0) \qquad \forall \text{ test functions } f.$$

Now, assume that we have given a distribution $F$ with $$ x \, F(x) = - \delta(x). $$ Can we conclude then that $$ F = \delta' \quad ?$$

Solution approach

Since $$\int_\mathbb{R} x \, \delta'(x) f(x) \, dx = \int_\mathbb{R} x \, F(x) f(x) \, dx \qquad \forall \text{ test functions } f,$$ $\delta'$ and $F$ act identically on all test functions of the form $x \, f(x)$, where $f$ is a test function itself. However, it is not clear if any test function can be written in that form.

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  • $\begingroup$ Also $F=-\delta'+C\delta,$ where $C$ is a constant, satisfies $xF=\delta.$ $\endgroup$
    – md2perpe
    Commented Sep 17, 2021 at 11:03

1 Answer 1

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Not true. Note $F = \delta' + C\delta$ also satisfies $xF = -\delta$ for any constant $C$.

In general, given $xF = -\delta$, we have

$$ \tag{1} F(x\varphi (x)) = -\varphi(0)$$ for all test function $\varphi$.

Now for any test function $\varphi$ with $\varphi (0) = 0$, write

$$ \varphi (x) = xg(x),$$ where $$ g(x) =\int_0^1 \varphi'(sx) ds.$$ Since $\varphi$ is a test function, so is $g$ and thus $$ F(\varphi (x)) = F(x g(x)) = -g(0) = -\varphi'(0).$$

In general, let $T$ be a test function so that $T(0) = 1$ and $T'(0) = 0$. Then for all test function $\varphi$, $$ F(\varphi) = F( \varphi - \varphi (0) T) + \varphi (0) F(T)= -\varphi'(0) + C\varphi (0),$$

where $C = F(T)$. Thus $F = \delta' + C \delta$.

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    $\begingroup$ Note to Larss96: Here the action of a distribution $F$ on a test function $f$ is written as $F(f).$ With integral notation this corresponds to $\int F(x)\,f(x)\,dx.$ It is also common to use a pairing $\langle F, f \rangle$ which makes it look like an inner product. $\endgroup$
    – md2perpe
    Commented Sep 17, 2021 at 11:37
  • $\begingroup$ @md2perpe: Could you explain why "$\varphi$ is a test function" implies "$g: x \mapsto \frac{1}{x} \, \int_0^x \varphi'(s) \, ds$ is a test function"? $\endgroup$
    – Larss 96
    Commented Sep 17, 2021 at 12:23
  • $\begingroup$ I think it is pretty clear that $x\mapsto \int_0^1 \varphi ' (sx) dx$ is smooth in $x$ right? Using $\varphi = xg$, one also see that $g$ has compact support since $\varphi$ has. @Larss96 $\endgroup$ Commented Sep 17, 2021 at 12:29
  • $\begingroup$ Thank you, @Arctic Char! $\endgroup$
    – Larss 96
    Commented Sep 17, 2021 at 12:33

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