# Delta function question

Given the functions $$f(x)= \delta (x-a)$$ $$g(x)= \frac{1}{a} \delta \left(x- \frac{1}{a}\right)$$ for a real constant $a\gt0$, is there a relationship between $f$ and $g$?

I believe that $g\left(\frac{1}{x}\right)=af(x)$, but I'm not $100\%$ sure. I have used the properties of $\delta (f(x))$ with $f(x)= \frac{1}{x}-\frac{1}{a}$, which has the only root $x=a$, but I do not know what else to do in order to prove it.

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$\delta (-a)=0= \delta (\infty-a)$ and $f(a)= \infty = g(1/a)$ –  Jose Garcia Aug 27 '12 at 10:51

$$g\left(\frac1x\right)=\frac1a\delta\left(\frac1x-\frac1a\right)=\frac1a\delta(x-a)\left|\left[\frac{\mathrm d}{\mathrm dx}\left(\frac1x-\frac1a\right)\right]_{x=a}\right|^{-1}=a\delta(x-a)=af(x)\;.$$

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I don't see, what motivates the second $=$? –  draks ... Aug 27 '12 at 11:56
@draks: en.wikipedia.org/wiki/… –  joriki Aug 27 '12 at 11:58
Interesting, thanks, +1... –  draks ... Aug 27 '12 at 12:16