I´m trying to proof the following Statement.

$$\delta(kx)=\frac{1}{|k|} \delta(x).$$

I already tried to proof and I got this.

$$u=kx \Rightarrow x=\frac{u}{k},dx=\frac{1}{k} du \\ \int_{-\infty}^\infty f(x) \delta(kx) dx = \int_{-\infty}^\infty f \left ( \frac{u}{k} \right ) \delta(u) \frac{du}{k}.$$

But I dont know how to proceed from here.

I know that I must catch the value of u. But my doubt is what value?

  • 4
    $\begingroup$ What are you trying to prove? $\endgroup$ – Kaster Oct 19 '15 at 0:22
  • 1
    $\begingroup$ Assuming you are trying $\delta(kx)=\frac{1}{|k|} \delta(x)$, you have made a single error thus far: if $k<0$, then the substitution reverses the order of the limits. Reversing it back gives a minus sign, which is where the absolute value comes from. Once you've fixed that, you apply the definition of $\delta$ to finish the proof. $\endgroup$ – Ian Oct 19 '15 at 0:27
  • $\begingroup$ Ian. The problem is the application of the definition of Delta. $\endgroup$ – Vinicius L. Beserra Oct 19 '15 at 0:27
  • $\begingroup$ Please use MathJax next time. I've done it for you here. $\endgroup$ – Ian Oct 19 '15 at 0:30
  • $\begingroup$ There is no difficulty there: $\int_{-\infty}^\infty \frac{f(u/k)}{|k|} \delta(u) du = \frac{f(0)}{|k|}$ by definition. If you are using pretty much any other definition, then your very first task should be to prove this, since this is "really" the definition. $\endgroup$ – Ian Oct 19 '15 at 0:30

When $T$ is a distribution and $k$ is a positive number, how should we define the rescaled distribution $T_k$? When $T$ is represented by a function $g$, we just want $g(kx)$. In terms of the integral against a test function $f$, this gives $$ \int_{\mathbb{R}^n} g(kx) f(x) \,dx =k^{-n} \int_{\mathbb{R}^n} g(y) f(x/y) \,dx $$ So, for a general distribution $T$ we define the rescaled distribution $T_k$ as $$ T_k(f) = k^{-n} T(f(x/k)) $$

In the specific case of Dirac delta at $0$, the expression $k^{-n} T(f(x/k))$ evaluates to $ k^{-n}f(0)$. (You work in one dimension, $n=1$.)


I worked this out referring to what you have already done and what Ian suggested. Also refer to https://proofwiki.org/wiki/Scaling_Property_of_Dirac_Delta_Function .

To prove $\delta(kx)=\frac{1}{|k|} \delta(x)$ , instead prove ${|k|}\delta(kx)= \delta(x)$, where k is a nonzero real constant.

Use the definition of $\delta(x)$ consiting of the first condition:

$$\delta \left({x}\right) = \begin{cases} +\infty & : x = 0 \\ 0 & : x \neq 0 \end{cases}$$ and the second condition: $$I=\int_{-\infty}^{+\infty} \delta \left({x}\right) dx = 1$$

Then show $\delta \left({kx}\right)$ satifies the second condition

$$I=\int_{-\infty}^{+\infty}|k| \delta \left({kx}\right) dx $$

$$=|k|\int_{-\infty}^{+\infty} \delta \left({kx}\right) dx = 1$$

since the first condition is obviously satisfied by $\delta \left({kx}\right)$ which is equal to $\infty$ for $x=0$ and is otherwise equal to zero.

Proceeding to find the integrals, $I$: let $y=kx$, then $dy=kdx$ and $dx=\frac{dy}{k}$.


$$I=|k|\int_{-\infty}^{+\infty} \delta \left({kx}\right) dx =|k|\int_{-\infty}^{+\infty} \delta \left({y}\right) \frac{dy}{k}$$ for $k>0$ $$I=\frac{|k|}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =\frac{k}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =1$$ and for $k<0$, noting that $y=kx$ will change the signs of the limits of integration, and also that reversing the limits multiplies the integrand by $-1$ $$I=\frac{|k|}{k}\int_{+\infty}^{-\infty} \delta \left({y}\right) dy =\frac{-|k|}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =\frac{k}{k}\int_{-\infty}^{+\infty} \delta \left({y}\right) dy =1$$ which finishes the proof.


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