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Let $H(x)$ be the Heaviside function defined by

\begin{cases} 1 & \text{if } x\geq0\\ 0 & \text{if } x<0 \end{cases}

I know that

  1. $H'(x)=\delta(x)$. The derivative of the Heaviside function is the delta function.
  2. $\delta(x)=\delta(-x)$. The delta function is symetric.

Does

  1. $H(x)=H(-x)$?
  2. $H(x)=-H(x)$?

It appears that

$$ -\delta(x)\delta(-y)=\delta(x)\delta(y)$$

and

$$ -\delta(-x)\delta(y)=\delta(x)\delta(y)$$

Do both of these properties follow from the definition of the Heaviside function?

Axion004
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    $\delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps . –  Mar 21 '19 at 19:18

1 Answers1

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No, $H(x)=1-H(-x)$ for $x\ne 0$. Integrating something even gives something odd plus an integration constant.

J.G.
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  • And indeed, differentiating that using $H'(x) = \delta(x)$ gives the correct $\delta(x) = \delta(-x)$. – eyeballfrog Mar 21 '19 at 19:22
  • I see that, can this justify that $-\delta(x)\delta(-y)=\delta(x)\delta(y)$? – Axion004 Mar 21 '19 at 19:46
  • @Axion004 Since $\delta$ is even, the leftmost $-$ sign shouldn't be there. – J.G. Mar 21 '19 at 19:56
  • That is the part which is confusing to me. As $\delta(-y)=\delta(y)$, we have that $-\delta(x)\delta(-y)=-\delta(x)\delta(y)=\delta(x)\delta(y)$. I don't see why $-\delta(x)\delta(y)=\delta(x)\delta(y)$. – Axion004 Mar 21 '19 at 20:41
  • @Axion004 Where did you get the "it appears that" equations from? – J.G. Mar 21 '19 at 20:44
  • I saw that $-H'(x)H'(-y) =\delta(x)\delta(y)$ and concluded that $-H'(x)H'(-y) =-\delta(x)\delta(-y)=\delta(x)\delta(y)$. – Axion004 Mar 21 '19 at 20:48
  • @Axion004 Ah, I see; I think you went from the first equation to the second by replacing $y$ with $-y$, but forgot that $H^\prime (-y)=H^\prime(y)$ rather than $-H^\prime(y)$. – J.G. Mar 21 '19 at 20:51
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    Yes, got it. Thanks! – Axion004 Mar 21 '19 at 21:14