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I'm trying to prove the following identity:

$$ \begin{equation} \label{eq:1} x\frac{d}{dx}\delta(x)=-\delta(x) \end{equation} $$

I integrated both sides with respect to $x$ over the limits of $(-\infty,\infty)$ and saw that they both evaluated to $-1$.

I'm not sure if that is the right approach because the worked solution I saw multiplied both sides of the above relation by an arbitrary function $f(x)$.

Is there any reason for multiplying both sides by an arbitrary $f(x)$? I'm a little helpless here because the approach that I took seemed more direct.

GanTheMan
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    https://math.stackexchange.com/questions/2118286/differential-of-dirac-delta-function – user326159 Apr 26 '20 at 09:50
  • Having two functions integrate to the same thing is not enough to prove that they are equivalent. You could prove a variety of absurd things with this fact, like $\sin x = \cos x = e^x = \cdots$ with the right bounds. – Ninad Munshi Apr 26 '20 at 10:35
  • @NinadMunshi Thanks for the clarification. I guess it's simpler things like what you described that really contribute to a better foundation! – GanTheMan Apr 27 '20 at 03:13

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By definitions, for any $\varphi \in C_c^\infty(\mathbb R),$ $$ \langle x \delta', \varphi \rangle = \langle \delta', x\varphi \rangle = - \langle \delta, (x\varphi)' \rangle = - \langle \delta, \varphi + x\varphi' \rangle = - (\varphi + x\varphi')|_{x=0} = -\varphi(0) = \langle -\delta, \varphi \rangle . $$ Thus, $x \delta' = -\delta.$

md2perpe
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  • Doesn't $\delta(x)=\infty$ at $x=0$? How were you able to get −⟨,+′⟩=−(+′) when =0? – GanTheMan Apr 27 '20 at 03:45
  • The definition of $\delta$ is that $\langle \delta, \varphi \rangle = \varphi(0)$ for every $\varphi \in C_c^\infty(\mathbb R).$ – md2perpe Apr 27 '20 at 05:49
  • Okay. I'm not completely familiar with inner products (I am still learning linear algebra). But am I correct to say that $⟨,⟩=(0)$ is analogous to $$\int_{-\infty}^{\infty} f(x) \cdot \delta (x) dx$$? – GanTheMan Apr 27 '20 at 07:49
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    $\delta$ is not really a function so integrals involving it are not defined. Instead it's a distribution and the application of a distribution on a test function is often written as an inner production (which it's not). But it's anyway common to treat it as a function and use it inside integrals. In this case: $$\langle \delta, \varphi \rangle = \int_{-\infty}^{\infty} \delta(x),\varphi(x),dx = \varphi(0).$$ – md2perpe Apr 27 '20 at 11:28
  • Thanks a lot for clarifying my question! – GanTheMan Apr 28 '20 at 04:43