I am to prove that $\delta(ax) =\frac{\delta(x)}{|a|} $

The following is my reasoning:

I begin with, $$\int_{-\infty}^{\infty} \delta(x') f(x') dx'=f(0) \tag{1} $$ Where $\delta$ is the Dirac delta function and where $f$ is a single variable function. I make the substitution $x'=|a|x$ to get, $$\int_{-\infty}^{\infty} \delta(ax) f(|a|x) |a|d(x)\tag{2}$$ Since $\delta$ is an even function. Eqs. (1) and (2) give the same result. So, on equating them we see, $$\int_{-\infty}^{\infty} \delta(x') dx'=|a| \int_{-\infty}^{\infty} \delta(ax) d(x)\tag{3}$$ Where $f(x')$and $ f(|a|x)$ are treated as constants and are equal in an infinitesimal interval $(-\epsilon/a, \epsilon/a) $ about zero. Here, $\epsilon$ is an infinitesimal. Eq.(3) can be re-written as $$\int_{-\epsilon/a}^{\epsilon/a} \frac{\delta(x')}{|a|} dx'=\int_{-\epsilon/a}^{\epsilon/a} \delta(ax) d(x)$$ The Delta function is zero everywhere else. Eq.(5) implies that $\delta(x) =\frac{\delta(x)}{|a|} $. This concludes the proof.

I would like to know if there are any flaws in reasoning.

  • $\begingroup$ Mathematicians would have easier to follow if you had used the normal notation for function application $f(x)$ instead of $\langle x | f \rangle$. When you also have $|a|$ inside those brackets it gets really confusing. $\endgroup$
    – md2perpe
    Jun 26 '18 at 9:46

Let me first answer this question physics-style and then reanswer this rigorously for any math students who may happen to see this.

Physics Answer

The basic conceptual problem is that the only way to prove an equality of generalised functions $\phi$ and $\psi$ is to show that $$ \int_{-\infty}^\infty \phi(x)\,f(x)\,dx = \int_{-\infty}^\infty \psi(x)\,f(x)\,dx $$ for all suitable ordinary functions $f$—anything else is completely unreliable. In this case, this would mean showing that $$ \int_{-\infty}^\infty \delta(ax)\,f(x)\,dx = \int_{-\infty} \frac{1}{\lvert a \rvert}\delta(x)\,f(x)\,dx $$ for all suitable ordinary functions $f$.

So, we need to proceed more carefully. Fix $a \neq 0$, and let $f$ be any suitable ordinary function. Let $x^\prime = ax$. Since $f(x) = f(a^{-1}x^\prime)$, so that $$ f(x)\rvert_{x = 0} = f(0) = f(a^{-1}0) = f(a^{-1}x^\prime)\vert_{x^\prime=0}, $$ it follows that $$ \begin{align} \int_{-\infty}^\infty \delta(x)\,f(x)\,dx &= f(0)\\ &= f(a^{-1}0)\\ &= \int_{-\infty}^\infty \delta(x^\prime) f(a^{-1}x^\prime)\,dx^\prime\\ &= \int_{-\infty}^\infty \delta(ax)\,f(x)\lvert a \rvert dx\\ &= \int_{-\infty}^\infty \left(\lvert a \rvert \delta(ax) \right) f(x) \, dx; \end{align} $$ since $f$ was arbitary, you can conclude that $$ \delta(x) = \lvert a \rvert \delta(ax), $$ which gives you what you want.

Mathematics Answer

Recall that the Dirac delta is the distribution (generalised function) $\delta \in \mathcal{D}(\mathbb{R}) := C_c^\infty(\mathbb{R})^\prime$ defined by $$ \forall f \in C_c^\infty(\mathbb{R}), \quad \delta(f) := f(0). $$ For any $a \in \mathbb{R} \setminus \{0\}$, define a linear transformation $S_a : C_c^\infty(\mathbb{R}) \to C_c^\infty(\mathbb{R})$ by $$ \forall f \in C_c^\infty(\mathbb{R}), \; \forall x \in \mathbb{R}, \quad (S_a f)(x) := f(ax). $$ For any $a \neq 0$, one can use the change of coordinates $x^\prime = ax$ to show that for all $f,g \in C_c^\infty(\mathbb{R})$, $$ \begin{align} \int_{-\infty}^\infty (S_a f)(x)\, g(x)\,dx &= \int_{-\infty} f(ax)\,g(x)\,dx\\ &= \int_{-\infty}^\infty f(x^\prime) g(a^{-1}x^\prime)\frac{1}{\lvert a \rvert}dx^\prime\\ &= \int_{-\infty} f(x) \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}g\right)(x) \,dx, \end{align} $$ i.e., that $S_a^\ast = \lvert a \rvert^{-1} S_{a^{-1}}$ with respect to the $L^2$ inner product on $\mathbb{R}$, so, by duality, we're obliged to define $S_a\delta$ (i.e., $\delta(ax)$) to be the distribution $$ S_a\delta := \delta \circ \frac{1}{\lvert a \rvert}S_{a^{-1}} \in \mathcal{D}(\mathbb{R}). $$ But now, for any $f \in C_c^\infty(\mathbb{R})$, $$ (S_a\delta)(f) = \delta\left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right) = \left(\frac{1}{\lvert a \rvert}S_{a^{-1}}f\right)(0) = \frac{1}{\lvert a \rvert}f(a^{-1}0) = \frac{1}{\lvert a \rvert}f(0) = \left(\frac{1}{\lvert a \rvert}\delta\right)(f), $$ so that, indeed, $$ S_a\delta = \frac{1}{\lvert a \rvert}\delta, $$ which is the rigorous statement of the identity $\delta(ax) = \frac{1}{\lvert a \rvert}\delta(x)$.

  • $\begingroup$ I realized the fatal flaw. I corrected it. $\endgroup$
    – R004
    Jun 26 '18 at 10:43
  • $\begingroup$ My edits came in the moment you posted. I'll go through your answer now. $\endgroup$
    – R004
    Jun 26 '18 at 10:44
  • $\begingroup$ There's a more fundamental conceptual issue---please see the comments in the edited answer about equality of generalised functions. $\endgroup$ Jun 26 '18 at 10:48
  • $\begingroup$ This is an excellent answer(I could follow the physics answer). Clears it for me. +1 $\endgroup$
    – R004
    Jun 26 '18 at 10:59

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