Order of divergence in square of Dirac delta distribution

Question

This question is related to the square of Dirac delta distribution. In one of the answers to this question, it was argued that if we represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$, we obtain $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [\delta_{\epsilon}(x)]^{2} \mathrm{d}{x} = \lim_{\epsilon \to 0} \frac{1}{\epsilon}.$$

So, my question is, given a smooth function $f$ satisfying the following properties (so that $\lim_{\epsilon \to 0} f'(x) = \delta(x)$) $$f(x) = \begin{cases} 1 &\text{for } x \geq \epsilon \\ h(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases}$$ where $h(x)$ is some function interpolating the two asymptotic values, can we show that $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [f'(x)]^{2} \mathrm{d}{x} = \lim_{\epsilon \to 0} \frac{1}{\epsilon}$$ is true without specifying the form of $h(x)$ explicitly?

Answer 1

I think I've figured out how to show that this is the case (assuming the family of smooth functions considered is constructed via uniform dilation).

Consider a family of smooth functions $$ f_{\epsilon}(x) = \begin{cases} 1 &\text{for } x \geq \epsilon \\ g_{\epsilon}(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases} \implies f'_{\epsilon}(x) = \begin{cases} 0 &\text{for } x \geq \epsilon \\ g'_{\epsilon}(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases} $$ where $g_{\epsilon}(x)$ is some smooth interpolating function and prime indicates derivative with respect to the argument. When we fix $ \epsilon = r $, we have $$ f_{r}(x) = \begin{cases} 1 &\text{for } x \geq r \\ g_{r}(x) &\text{for } 0 < x < r \\ 0 &\text{for } x \leq 0 \end{cases} \implies f'_{r}(x) = \begin{cases} 0 &\text{for } x \geq r \\ g'_{r}(x) &\text{for } 0 < x < r \\ 0 &\text{for } x \leq 0 \end{cases} $$ and $$ \int_{-\infty}^{+\infty} \left[ f'_{r}(x) \right]^{p} \mathrm{d}x = \int_{0}^{r} \left[ g'_{r}(x) \right]^{p} \mathrm{d}x \equiv K_{r}^{p} < \infty$$ for any non-negative integer $p$ since $ g'_{r}(x) $ is necessarily bounded. Otherwise, it is not smooth. Note that some care is required if one is to extend $p$ to rational or real numbers. Note also that, by definition, there is no dependence on $\epsilon$ in $K_{r}^{p}$ as we have already fixed $\epsilon = r$.

Now, we know that $f_{\epsilon}(x)$ is just a strectched version of $f_{r}(x)$. So, we can write $$ f_{\epsilon}(x) = f_{r}(sx) $$ for some stretching function $s$ which, in general, depends on $x$ and $\epsilon$. In our case, however, we would like the stretching to be uniform for all $x$. Thus, we require that $s$ to be independent of $x$. Given the criteria, we have, in particular, that $$ f_{\epsilon}(\epsilon) = f_{r}(s\epsilon) = f_{r}(r). $$ It follows that $s = r/\epsilon$ and $$ f'_{\epsilon}(x) = \frac{r}{\epsilon} f'_{r} \left( \frac{rx}{\epsilon} \right). $$

Thus, \begin{align} \int_{0}^{\epsilon} [g'_{\epsilon}(x)]^{p} \mathrm{d}x = \int_{-\infty}^{+\infty} [f'_{\epsilon}(x)]^{p} \mathrm{d}x &= \left( \frac{r}{\epsilon} \right)^{p} \int_{-\infty}^{+\infty} \left[ f'_{r}\left(\frac{rx}{\epsilon}\right) \right]^{p} \mathrm{d}x \\ &= \left( \frac{r}{\epsilon} \right)^{p} \int_{0}^{\epsilon} \left[ g'_{r}\left(\frac{rx}{\epsilon}\right) \right]^{p} \mathrm{d}x \\ &= \left( \frac{r}{\epsilon} \right)^{p-1} \int_{0}^{r} [g'_{r}(y)]^{p} \mathrm{d}y \\ &= \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p} \end{align} or, written compactly \begin{align} K_{\epsilon}^{p} &= \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p} \\ \implies \lim_{\epsilon \to 0} K_{\epsilon}^{p} &= \lim_{\epsilon \to 0} \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p}. \end{align} In particular, we have the desired result (except for some pre-factor independent of $\epsilon$) when $ p = 2$, that is $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [f'_{\epsilon}(x)]^{2} \mathrm{d}x = \lim_{\epsilon \to 0} \frac{r}{\epsilon} K_{r}^{2}. $$