# Scaling property of Dirac delta function is not intuitive!

It is known that the Dirac delta function scales as follows:$$\delta(kx)=\frac{1}{|k|}\delta(x)$$ I have studied the proof for it, considering Dirac delta function as a limit of the sequence of zero-centred normal distributions (as given here).

However, when intuitively thought about it, this does not seem correct. Since $$\delta(x)$$ is zero everywhere except at $$x=0$$, $$\delta(kx)$$ should also be zero for any non-zero value of $$x$$ (given $$k\in R-\{0\}$$). Also for $$x=0, kx=0$$, and, thus, $$\delta(kx)=\delta(x)$$.

From the above logic it is evident that the scaling property should be the following.$$\delta(kx)=\delta(x)\forall x\in R, k\neq 0$$ However, as we know this is not true, can you point out where I am going wrong in thinking like this. Please note that I do not require some other kind of proof (until necessary), just a flaw in this kind of thinking.

• When you plug in specific numbers, as in "for $x=0$, $kx=0$ and, thus, $\delta(kx)=\delta(x)$", you're treating $\delta$ as if it were a function. It's not a function, and treating it as one can lead to error. Sep 4, 2020 at 16:02
• As a comment, perhaps the more intuitive way for a physicist is just to write $\delta_0 = H'$, the derivative of the heaviside function. Since $H(kx) = H(x)$ when $k>0$, you easily deduce the result. Sep 4, 2020 at 22:12
• Notice that using your intuitive reasoning also $\delta(x/k)=\delta(x)$ which leads to a contradiction with $\delta(kx)=\delta(x)$ unless $k=1$. Mar 4, 2021 at 19:30
• Jan 29 at 2:23

However, when intuitively thought about it, this does not seem correct. Since $$\delta(x)$$ is zero everywhere except at $$x=0$$, $$\delta(kx)$$ should also be zero for any non-zero value of $$x$$ (given $$k\in R-\{0\}$$). Also for $$x=0, kx=0$$, and, thus, $$\delta(kx)=\delta(x)$$.

Also $$C \, \delta(x)$$ is zero everywhere except at $$x=0.$$ Why do you think that $$\delta(kx)$$ must be $$\delta(x)$$ and not $$C\,\delta(x)$$ for some $$C\neq 1$$?

You know that $$\int_{-\infty}^{\infty} \delta(x) \, dx = 1.$$ But if $$k>0$$ we have $$\int_{-\infty}^{\infty} \delta(kx) \, dx = \left\{ x=\frac{y}{k} \right\} = \int_{-\infty}^{\infty} \delta(y) \, \frac{dy}{k} = \frac{1}{k} \int_{-\infty}^{\infty} \delta(y) \, dy = \frac{1}{k}$$ Therefore $$\delta(kx)$$ can not equal $$\delta(x)$$ but rather equals $$\frac{1}{k} \delta(x).$$

You can also look at ordinary functions that that approximate $$\delta,$$ e.g. $$d_\epsilon(x) = \begin{cases} \frac{1}{2\epsilon}\text{ if -\epsilon You have $$\int_{-\infty}^{\infty} d_\epsilon(x) \, dx = 1.$$ But if you scale it in the $$x$$ direction you get another integral, $$\int_{-\infty}^{\infty} d_\epsilon(kx) \, dx = \left\{ x=\frac{y}{k},\ k>0 \right\} = \int_{-\infty}^{\infty} d_\epsilon(y) \, \frac{dy}{k} = \frac{1}{k} \int_{-\infty}^{\infty} d_\epsilon(y) \, dy = \frac{1}{k}.$$

• "Why do you think that δ(kx) must be δ(x) and not Cδ(x) for some C≠1?" I think so beacuse for x=0, δ(kx)=δ(x)=δ(0); but Cδ(x)=Cδ(0). Sep 4, 2020 at 16:28
• @BigBrother. On the other hand, $\delta(0) = \infty = C \infty = C \, \delta(0).$ As you hopefully understand, one cannot handle $\delta$ as a pointwisely defined function. Sep 4, 2020 at 16:32
• @md2perpe one can: mathoverflow.net/questions/439535/… Jan 29 at 2:24
• Why the substitution $x\to y/k$ is permitted here? It seems ad-hoc and unjustified. Jan 29 at 2:26
• Substitution does not come automatically from the definition of distributions, but there is a natural definition based on how it works for integrals of ordinary functions. My last equation, where I integrate an approximation to the identity therefore works as a justification. Jan 29 at 8:59

Since $$\delta$$ is a distribution, you need to phrase everything in that language. You can't just go around evaluating it. If you're viewing $$\delta(x)"="\infty$$ at $$x=0$$, then of course you cannot distinguish between $$\delta(0),$$ $$|k|\delta(0),$$ etc. This is the reason that you're having trouble with intuition. First, I'll give a more formal explanation to the general reader, then I'll address "intuition" a little bit more.

Observe that for any $$\varphi\in C_c^\infty$$ and $$f\in L^1_{loc}$$, $$\int\limits_{-\infty}^\infty f(kx)\varphi(x)\, dx=\begin{cases}\frac{1}{k}\int\limits_{-\infty}^\infty f(y)\varphi(y/k)\, dy && k>0\\ \frac{1}{k}\int\limits_{\infty}^{-\infty} f(y)\varphi(y/k)\, dy && k<0 \end{cases}$$ That is,

$$\int\limits_{-\infty}^\infty f(kx)\varphi(x)\, dx=\frac{1}{|k|}\int\limits_{-\infty}^\infty f(y)\varphi(y/k)\, dy.$$ This motivates the definition for general distributions, such as the Dirac delta. Alternatively, check out a dense subspace and extend by density.

If you want something less rigorous, see e.g. https://proofwiki.org/wiki/Scaling_Property_of_Dirac_Delta_Function

The Dirac delta will not follow standard intuition due to the fact that it's not a "simple" object, like a function from $$\mathbb{R}$$ to $$\mathbb{R}$$. For this reason, I'd argue that a derivation like this, or the one that you linked, give the desired intuition. It probably makes the most sense to think about it from the perspective of a regularizing sequence like you linked and observing the property from there. This gets much more at how $$\delta$$ behaves.

• What you wrote is correct but I seriously doubt that the person asking the question has the mathematical level to understand your answer. Sep 4, 2020 at 15:56
• This is potentially true, but I think that it could be helpful to a reader other than just the OP who looks for a question like this :)
– cmk
Sep 4, 2020 at 15:57
• "This motivates the definition for general distributions" - but this does not work even for general functions. And the distributions are a superset of functions, no? Jan 29 at 2:30
• @Anixx fixed, thanks for the catch
– cmk
Feb 9 at 1:09

A bit belatedly: in addition to other good answers, indeed this is a good time to see that it is simply not sustainable (and not useful!) to insist on a notion of "function" for $$\delta$$ that pretends that it is describable by pointwise values.

Yes, this is possibly heretical after all the drill about "what functions are", in very-basic calculus and analysis. But, perhaps unsurprisingly, those definitions are excellent in_a_certain_regime ... and this was a big step forward 100+ years ago! ... but inadequate for talking about Heaviside's, Dirac's, and others' generalized functions. L. Schwartz' rigorization of most of those ideas accomplished many things, but it did not manage to save the idea that a function is described by pointwise values.

There are at least two styles of "intuition" that seem to suggest a way to get correct outcomes. (After all, we are talking about computations that refer to real things! ...) One is thinking of $$\delta$$ as a ("weak") limit of "spike" functions (which do have pointwise values), to prove its properties. Another is to repackage that weak-limit game (which is arguably not physically explanatory to many...?) as something about infinitesimals, and the behavior of $$\delta$$ "infinitesimally near $$0$$".

From an optimistic viewpoint, these two ideas are "the same", for me.

Formally, they are certainly different. The infinitesimal approach has some annoying technical limitations, but it does approximately restore the idea that even generalized functions have pointwise values... but with complications in the notion of "pointwise" and "values". :)

• "it is simply not sustainable (and not useful!) to insist on a notion of "function" for δ that pretends that it is describable by pointwise values." - why? It is absolutely sustainable and useful: mathoverflow.net/questions/439535/… Jan 29 at 2:33

I would like to share my understanding which helps me memorize the scaling property of 𝛿(t).

Considering the physical meaning of 𝛿(t) is to introduce a model that can model the reaction toward some stimulus, like adding a voltage on a capacitor, when the slope is infinitely large and the action time is infinitely small.

So we can easily know that no matter the function is 𝛿(at) or 𝛿(t), the systems always give reaction to the stimulus at the same time. So, the range of t doesn't change. 𝛿(at) and 𝛿(t) have the same range of t.

𝛿(t) can be shaped as a rectangle or something else, it doesn't matter which shape you choose. Nevertheless, let's suppose its shape is a rectangle. Since the range of time doesn't change, the bottom line of 𝛿(at) will expand a times. Since the total area is constant, the height of this rectangle should be 1/a. Consequently, 𝛿(at) = (1/a)*𝛿(t)