On the derivative of Dirac delta

Let $$\delta(x)$$ be the Dirac's delta, i.e. the "strange object" characterized by the two properties $$\delta(x)=0$$ for all $$x\neq 0$$ and $$\int \delta(x) f(x)\text{ d}x= f(0)$$ ($$\int f$$ is the integral over $$\mathbb{R}$$). Now, let's define the derivative of $$\delta(x)$$ as

$$\begin{equation*} \dot{\delta}(x)\triangleq \lim_{\tau \to 0} \frac{\delta(x+\tau)-\delta(x)}{\tau} \end{equation*}$$ For simplicity, I'm considering the univariate case $$x\in\mathbb{R}$$. Now let's try to see what happens when we compute a weighted integral of $$f(x)$$ involving $$\dot{\delta}(x)$$. I would write naively \begin{equation*} \begin{aligned} \int \dot{\delta}(x) f(x)\text{ d}x&= \int \left(\lim_{\tau \to 0} \frac{\delta(x+\tau)-\delta(x)}{\tau}\right)f(x)\text{ d}x\\ &= \lim_{\tau \to 0} \frac{1}{\tau}\left[\int \delta(x+\tau)\,f(x)\text{ d}x - \int \delta(x)\,f(x)\text{ d}x\right]\\ &= \lim_{\tau \to 0} \frac{1}{\tau}\left[f(0-\tau) - f(0)\right]=-\dot{f}(0)\\ \end{aligned} \end{equation*} provided that we can push the limit sign outside the integral (if I'm not wrong, here Lebesgue help us to see when this can be done); the difference between the two integrals is not in the form $$\pm \infty \mp\infty$$; the derivative of $$f(x)$$ exists in $$x=0$$.

If I'm not missing anything else, we can say that, just like $$\int \delta(x) f(x) \text{d}x=f(0)$$, we have $$\int \dot{\delta}(x) f(x)\text{ d}x=-\dot{f}(0)$$.

Question

Since the $$\delta(x)$$ is a "strange object", I'm not 100% sure about my conclusions. So, my question is: am I right? If not, where I'm wrong?

• You are wrong by a sign: $\lim_{\tau \to 0} \frac{1}{\tau}\left[f(0-\tau) - f(0)\right]=-\dot{f}(0)$, hence $\int \dot{\delta}(x) f(x)\text{ d}x=-\dot{f}(0)$. Dec 26, 2023 at 3:04
• Yes, though in general manipulation of weird things like this is easy to mess up. See en.wikipedia.org/wiki/Dirac_delta_function#Derivatives
– Eric
Dec 26, 2023 at 3:39
• You have missed one important properrty of Dirac Delta: $\delta(a x)=\frac1{|a|}\delta(x)$. This thing exactly makes it the most strange Dec 26, 2023 at 8:52
• where do we use $\delta(ax)$? Dec 26, 2023 at 9:54

Your argument shows that the derivative of the delta distribution, let's denote it by $$\delta',$$ is given by a limit of difference quotients, i.e. your $$\dot{\delta}.$$
1. For a distribution $$\phi,$$ its derivative $$\phi'$$ is defined by the relation $$\int \phi' f:=-\int \phi f'$$ for any test function $$f.$$ The motivation behind this definition is that when $$\phi$$ is a (continuously differentiable) function, this identity is just integration by parts.
2. The limit $$\lim_{n\to \infty} \phi_n$$ of a sequence of distributions $$\phi_n$$ is characterized by $$\int(\lim_{n\to \infty} \phi_n) f=\lim_{n\to \infty}\int \phi_n f$$ for any test function $$f.$$ This can either be taken as a definition of convergence for distributions or proven as a theorem.
3. For a distribution $$\phi$$ and $$\tau\in \mathbb{R}$$ there is its shift $$\phi(\cdot+\tau),$$ which is again a distribution given by the identity $$\int \phi(\cdot+\tau) f=\int \phi f(\cdot-\tau)$$ for any test function $$f.$$ Similarly to 1., for functions this is just a change of variables.
Using these facts, your proof shows $$\delta'=\lim_{\tau\to 0} \frac{\delta(\cdot+\tau)-\delta}{\tau},$$ an equality of two distributions, which is exactly the result we would expect from the corresponding equality for functions.
• (+1) You might consider correcting the OP's assertions about the "properties" that characterized the Dirac Delta. Namely, $\delta(x)=0$ for all $x\ne0$ is not at all correct. And $\int_{\mathbb{R}}\delta(x)f(x)\,dx=f(0)$ is only a notation, and not an integral (unless the Dirac Delta is regarded as a measure, in which case its derivative is not). Dec 26, 2023 at 15:43