# Approximating derivative of Dirac delta function using mollifiers

Given $\rho_\epsilon (x)$ a delta approximating sequence, what is the limit in distribution of the function $$\frac{\rho_\epsilon (x) - \rho_\epsilon (x-\epsilon)}{\epsilon} \text{?}$$ Intuitively it should converge to the derivative of the Dirac delta in zero $\delta'_0$, though I'm having trouble showing this rigorously. I assumed $\rho_\epsilon (x) = \rho(\epsilon x)/\epsilon$ for some mollifier $\rho$.

The book I'm using suggests that I use the fundamental theorem of calculus to represent the derivative of $\rho_\epsilon$ first, though this only makes calculations messier and I can't see how this is supposed to help.

The usual way of doing this is to change variables so that the difference quotient falls on the test function. (This is a "discrete" version of integration by parts.) Suppose $\varphi$ is a smooth test function. Then \begin{align*} \int_{-\infty}^{\infty} \left(\frac{\rho_{\epsilon}(x) - \rho_{\epsilon}(x - \epsilon)}{\epsilon}\right) \varphi(x) \, dx &= - \int_{-\infty}^{\infty} \rho_{\epsilon}(y) \left(\frac{\varphi(y + \epsilon) - \varphi(y)}{\epsilon} \right) \, dy \\ &= - \int_{-\infty}^{\infty} \rho_{\epsilon}(y) \varphi'(y_{\epsilon}) \, dy, \end{align*} where $y_{\epsilon}$ is a point on the interval between $y$ and $y + \epsilon$. Send $\epsilon \to 0$ to find $$\lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \left(\frac{\rho_{\epsilon}(x) - \rho_{\epsilon}(x - \epsilon)}{\epsilon}\right) \varphi(x) \, dx = - \varphi'(0).$$ This proves that $$\frac{\rho_{\epsilon}(\cdot) - \rho_{\epsilon}(\cdot - \epsilon)}{\epsilon} \to \delta_{0}'$$ in the sense of distributions.