# Proving properties of the Dirac delta function

I've been thinking about the properties of the Dirac delta function recently, and having been trying to prove them. I'm not a pure mathematician but come from a physics background, so the following aren't rigorous to the extent of a full proof, but are they correct enough?

First I aim to prove that $x\delta (x) =0$. Let $f$ be an arbitrary (integrable) function. Then, $$\int_{-\infty} ^{\infty} (xf(x)) \delta (x) \; dx=0f(0)=0$$ where we have used the filtering property of $\delta$-function. As $f$ was chosen arbitrarily we must conclude that $x\delta (x)$.

Next, I aim to prove that $\delta (x^{2} - a^{2}) =\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a))$. Consider the following, $$\int_{-\infty} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx$$ Now, let $x=\sqrt{u}$ then $dx=\frac{1}{2\sqrt{u}} du$ and so $$\int_{0} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (u - a^{2})\frac{f(\sqrt{u})} {2 \sqrt{u}} \;dx =\frac{f(\vert a\vert)} {2 \vert a\vert}$$ Similarly, let $x=-\sqrt{u}$ then $dx=-\frac{1}{2\sqrt{u}} du$ and so $$\int_{-\infty} ^{0}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (u - a^{2})\frac{f(-\sqrt{u})} {2 \sqrt{u}} \;dx =\frac{f(-\vert a\vert)} {2 \vert a\vert}$$ and hence $$\int_{-\infty} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx+\int_{-\infty} ^{0}\delta (x^{2} - a^{2})f(x)\;dx=\frac{f(\vert a\vert)} {2 \vert a\vert}+\frac{f(-\vert a\vert)} {2 \vert a\vert}\\ \qquad\qquad\qquad\qquad\quad\;\;\;=\frac{1} {2\vert a\vert}(f(a)+f(-a))=\int_{-\infty} ^{\infty}\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a))f(x) \;dx$$ Thus implying that $\delta (x^{2} - a^{2}) =\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a))$.

Would this be correct at all?