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I am trying to show

$$\delta(2|x|-3)=|x|\delta(x^2-\frac{9}{4}).$$

The composition property states $\delta(g(x))=\sum\frac{\delta(x-a_i)}{|g'(a_i)|}$ where $a_i$'s are the roots of $g(x)=0$.

If I start with the RHS, the roots of $x^2-9/4$ are $\pm 3/2$ with corresponding derivatives $\pm 3.$ This gives $$|x| \delta(x^2-\frac{9}{4})=\frac{|x|}{3}\delta(x-\frac{3}{2})+\frac{|x|}{3}\delta(x+\frac{3}{2}).\tag{*}$$

If I start with the LHS $2|x|-3$, has two roots $\pm 3/2$ and corresponding derivatives $\pm 2$. This implies $$\delta(2|x|-3)=\frac{1}{2}\delta(x+\frac{3}{2})+\frac{1}{2}\delta(x-\frac{3}{2}).\tag{**}$$

Now I am stuck. How do I show that (*) and (**) are equal?

Is it ok to use the composition property for $2|x|-3$ even though is not continuously differentiable everywhere?

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2 Answers 2

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Simply use the definition: $$\int f(x)\delta(x-x_0)=\int f(x_0)\delta(x-x_0)=f(x_0)$$

Therefore $$|x|\delta(x-x_0)=|x_0|\delta(x-x_0)$$

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Define $f(y) :=y^2-\frac{9}{4}$. Then $f(|x|) =|x|^2-\frac{9}{4}$ and $f^{\prime}(|x|) =2|x|$. So

$$|x|\delta(x^2-\frac{9}{4})~=~|x|\delta(f(|x|)) ~=~ \frac{|x|}{|f^{\prime}(|x|)|}\delta(|x|-\frac{3}{2}) ~=~ \frac{1}{2}\delta(|x|-\frac{3}{2}) ~=~ \delta(2|x|-3).$$

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