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So, I have a Fourier analisys course right now and got a problem to solve. Here's how it sounds

Given that $\delta(x)$ is a Dirac delta function, $\phi(x)$ is a test function, by using properties of a test function prove that $\delta(kx) = \frac{\delta(x)}{|k|}$. (Hint: $\langle \delta(x), \phi(x) \rangle = \phi(0)$)

And, here's how I tried to prove that. I assumed that hint is not for nothing there, so I decided that it should be a decent place to start.

$$ \langle \delta(x), \phi(x) \rangle = \phi(0) $$ $$ \langle \delta(kx), \phi(kx) \rangle = \phi(0) $$ $$ \int_{-\infty}^{\infty} \delta(kx) \phi(kx) d(kx) = \phi(0) $$ $$ u = kx \implies du = d(kx) = kdx $$ $$ \int_{-\infty}^{\infty} k \delta(kx) \phi(kx) dx = \phi(0) $$ $$ \int_{-\infty}^{\infty} \delta(kx) \phi(kx) dx = \frac{\phi(0)}{k} $$ And... I am not sure what exactly I achieved here. $ \delta(kx) $ is still $ \delta(kx) $, so I assume I made no progress by going this way, and starting with a hint was not a good idea. So I started over with a slightly different approach.

$$ \langle \delta(kx), \phi(x) \rangle = $$ $$ = \int_{-\infty}^{\infty} \delta(kx) \phi(x) dx = * $$ $$ u = kx \implies du = kdx $$ $$ * = \int_{-\infty}^{\infty} \frac{1}{k} \delta(u) \phi(\frac{u}{k}) du = ??? $$ A-a-and, I'm in another dead end again. I feel that I need to somehow utilize the hint $\langle \delta(x), \phi(x) \rangle = \phi(0)$ and it should somehow lead me to $ \int_{-\infty}^{\infty} \frac{1}{k} \delta(u) \phi(\frac{u}{k}) du = \frac{\phi(0)}{k} $ which I'm not even sure is true or not, but even if it is, it still does not put me any closer to the statement I need to prove and absolutely does not use any properties of a test function.

I understand that this question is probably a duplicate, and I already checked a lot of answers like this one, but, sadly, they don't really help with my problem because they don't use the definition and properties of a test function, but the definition of a Dirac delta itself, which I believe is not what this question about. Or they actually do, but I can not understand in what way.

So, please, I clearly do not understand something about the question and would be very thankful if someone helped me solve this problem.

Yuki Endo
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    In your equation $*$ you should have $\frac{1}{|k|}$ in the integrand. It's correct as is if $k$ is positive. If $k$ is negative, the limits of integration are switched, so you must negate them which then multiplies by your $k$. After that, it's not clear why you're confused about that being equal to $\frac{\phi(0)}{|k|}$. That's just from the definition $\langle \delta , \phi \rangle = \phi(0)$ – Dionel Jaime Jan 17 '22 at 21:21
  • @DionelJaime well, you're absolutely correct, I overlooked that part, but still, I don't really understand how do I get from $ \int_{-\infty}^{\infty} \frac{1}{|k|} \delta(u) \phi(\frac{u}{k}) du $ to $ \delta(kx) = \frac{\delta(x)}{|k|} $ and how exactly does this solution uses test function properties? – Yuki Endo Jan 17 '22 at 22:06
  • Do you have a property similar to "If $f$ is a functional and $\langle f, \phi\rangle = 0$ for every test function $\phi$, then $f=0$"? – Brian Moehring Jan 17 '22 at 23:31
  • I'm not entirely sure what's confusing you, but maybe this will help. Set $F(u) = \phi(u/k)$. Then your integral is $\frac{1}{|k|}\int_{-\infty}^{\infty} \delta(u) F(u) du$. By your hint, this is equal to $\frac{F(0)}{|k|} = \frac{\phi(0)}{|k|}$. – Dionel Jaime Jan 18 '22 at 02:57
  • So what this means is that $\langle \delta(kx) , \phi(x) \rangle = \frac{1}{|k|} \langle \delta(x), \phi(x) \rangle $. Or in other words $\delta(kx) = \frac{\delta(x)}{|k|}$ – Dionel Jaime Jan 18 '22 at 03:02
  • I'm also just rolling with your notation. If I felt like putting in the effort, I would go on a rant about how I actually think putting an argument for $\delta$ (i.e pretending $\delta$ is some function that takes in values $x$ ) is counter productive and will lead to a lot of confusion when one learns this from a mathematical point of view. And writing $\langle \delta(x), \phi(x) \rangle$ as opposed to $\langle \delta ,\phi \rangle$ is even extra confusing. Paul mentions this in his answer (though I don't think he has an issue with it). But anyway, I hope my answer makes more sense lol – Dionel Jaime Jan 18 '22 at 03:04
  • @DionelJaime thank you very much, now I think I understand what happens here. So basically, $ \int_{-\infty}^{\infty} \delta(kx) \phi(x)dx = \int_{-\infty}^{\infty} \frac{\delta(x)}{|k|} \phi(\frac{x}{k})dx = \int_{-\infty}^{\infty} \frac{\delta(x)}{|k|} \Phi(x)dx $ and since $ \Phi(x) $ is also a test function that satisfies the condition in the hint, it is safe to substitute it with $\phi(x)$, so it becomes $ \int_{-\infty}^{\infty} \delta(kx) \phi(x)dx = \int_{-\infty}^{\infty} \frac{\delta(x)}{|k|} \phi(x)dx $, is this correct? – Yuki Endo Jan 19 '22 at 08:35

1 Answers1

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Yes, this is a chronically possibly-counterintuitive aspect of $\delta$. But there is a two-fold way to understand it.

First, somewhat the intention of $\delta$, is as some kind of function on $\mathbb R$ (for example) that "can be integrated against" very nice functions $\varphi$, and $\int \delta(x)f(x)dx=f(0)$. Whether or not this is literally correct/rigorous, it certainly suggests the heuristic that, for $k>0$, $$ \int \delta(kx)\,f(x)\;dx \;=\; \int \delta(x)\,f(x/k)\;d(x/k) \;=\; {1\over k} \int \delta(x)\,f(x/k)\;dx \;=\; {1\over k} f(0/k) \;=\; {1\over k}f(0) $$ (Yes, there is a little fooling around to see that it's $|k|$ for $k<0$, and not just $k$, but this is a subsidiary point... $\delta$ is even...)

A disquieting aspect of that argument is that $\delta$ is not literally a function, and that integral cannot be literal. But the heuristic is what we want to be correct. The change of variables is certainly correct for any even function in place of $\delta$... and we do want $\delta$ to be some kind of limit of classical, nice functions, so we insist that the conclusion is correct (for consistency with classical functions, and expressing $\delta$ as a weak limit of them!).

So, second, on $\mathbb R^n$ (just to see where the dimension enters), we define the dilation $u(kx)$ with $k>0$ of a distribution $u$ (even though it's not a pointwise-valued function... so writing this as though $u$ had an argument is misleading) for $k>0$ by $u(kx)(f)=k^{-n}u(f(x/k))$. More precisely, letting $T_ku$ be that dilation, without mentioning pointwise values, $(T_ku)(f)=k^{-n}\cdot u(T_{k^{-1}}f)$.

paul garrett
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