# Is the solution of this distribution identity unique?

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.

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

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 $$F = \delta' + C \delta$$.

• 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. Commented Sep 17, 2021 at 11:37
• @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"? Commented Sep 17, 2021 at 12:23
• 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 Commented Sep 17, 2021 at 12:29
• Thank you, @Arctic Char! Commented Sep 17, 2021 at 12:33