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.
Best Answer
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. $$ d_\epsilon(x) = \begin{cases} \frac{1}{2\epsilon}\text{ if $-\epsilon<x<\epsilon$}\\ 0\text{ otherwise} \end{cases}$$ You have $\int_{-\infty}^{\infty} d_\epsilon(x) \, dx = 1.$ But if you scale it in the $x$ direction you get another integral, $$ \int_{-\infty}^{\infty} d_\epsilon(kx) \, dx = \left\{ x=\frac{y}{k},\ k>0 \right\} = \int_{-\infty}^{\infty} d_\epsilon(y) \, \frac{dy}{k} = \frac{1}{k} \int_{-\infty}^{\infty} d_\epsilon(y) \, dy = \frac{1}{k}. $$