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## Main Question or Discussion Point

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 [itex] x\delta (x) =0[/itex]. Let [itex] f[/itex] be an arbitrary (integrable) function. Then, [tex] \int_{-\infty} ^{\infty} (xf(x)) \delta (x) \; dx=0f(0)=0[/tex] where we have used the filtering property of [itex] \delta[/itex]-function. As [itex] f[/itex] was chosen arbitrarily we must conclude that [itex] x\delta (x) [/itex].

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

Would this be correct at all?

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

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

Would this be correct at all?