# Order of divergence in square of Dirac delta distribution

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?

• Cauchy-Schwarz inequality applied to $1 = \int_{0}^{\epsilon} f'(x) \, dx$ is probably one way of justifying the divergence. Jun 1, 2017 at 18:48
• I'm not sure how to do that, unfortunately. But, I think I've found a way. Thanks. Jun 2, 2017 at 14:44
• What I mean is $$1 = \left( \int_{0}^{\epsilon} 1 \cdot f'(x) \, dx \right)^2 \leq \left( \int_{0}^{\epsilon} f'(x)^2 \, dx \right)\left( \int_{0}^{\epsilon} 1 \, dx \right)$$ by the Cauchy-Schwarz inequality. So we have $$\frac{1}{\epsilon} \leq \int_{0}^{\epsilon} f'(x)^2 \, dx.$$ This proof works for any absolutely continuous function interpolating $0$ for $x < 0$ and $1$ for $x > \epsilon$. Jun 2, 2017 at 14:50
• Great. But it doesn't really show how fast the divergence really is. It could be that $$\int_{0}^{\epsilon} [f'(x)]^{2} \mathrm{d}x \sim \frac{1}{\epsilon^{2}} \quad \text{as} \quad \epsilon \to 0.$$ Jun 2, 2017 at 14:59
• Yes, I was thinking that $f$ is constructed via dilation. Jun 2, 2017 at 15:17

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 $$\lim_{\epsilon \to 0} \int_{0}^{\epsilon} [f'_{\epsilon}(x)]^{2} \mathrm{d}x = \lim_{\epsilon \to 0} \frac{r}{\epsilon} K_{r}^{2}.$$