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How does one prove that \begin{equation} x T = 0 \Longleftrightarrow T = c \delta_0, \end{equation} with $T \in \mathcal{D}'(\mathbb{R})$?

I see why we have "$\Longleftarrow$ " ($T = c\delta_0 \Leftrightarrow xT = xc\delta_0 \Leftrightarrow \langle xT, \phi \rangle = \langle c\delta_0, x\phi \rangle$ for all $\phi \in \mathcal{D}(\mathbb{R})$ and $\langle \delta_0, x\phi \rangle = 0$ therefore $xT = 0$).

  • There is a slight complication here: defining the product of distributions isn't so well-defined. – Cameron L. Williams Mar 26 '17 at 18:19
  • For $g \in C^\infty(\mathbb{R})$, one may multiply a distribution by $g$ I believe? –  Mar 26 '17 at 18:20
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    Yeah, it just may make the analysis a little murky, is what I meant. I'll see if I can answer the question at hand. – Cameron L. Williams Mar 26 '17 at 18:21
  • Another method would be to prove that the support of $T$ is a singleton ${0}$, and then use a theorem that distributions with singleton support are linear combinations of delta-distribution and its derivatives. – TZakrevskiy Mar 27 '17 at 11:40

1 Answers1

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There may be some details to work out, but here's a rough sketch. Let us apply Fourier transform techniques.

\begin{align} 0 &= \langle xT, \varphi\rangle \\ &= \langle T,x\varphi\rangle \\ &= \langle \mathcal{F}T, \mathcal{F}(x\varphi)\rangle \\ &= \langle \mathcal{F}T, (\mathcal{F}\varphi)'\rangle \\ &= -\langle (\mathcal{F}T)',\mathcal{F}\varphi\rangle \end{align}

Thus the distribution $\mathcal{F}T$ has derivative zero. This implies that it is constant and the inverse Fourier transform of a constant distribution is a multiple of the Dirac delta at $0$.

Here are some details for showing that if the derivative of a distribution is zero, it is constant.

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  • I think that you'd need to also add the fact that such a distribution is compactly supported, hence the Fourier transform is well-defined. – Marko Karbevski Jun 25 '17 at 18:10