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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?
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?