Differentiating the delta function

Question

I want to proof that:

$$\lim_{a \to 0}\frac{\delta(x+2a)-\delta(x+a)}{a} = \delta'(x)$$

I tried with the definition of differential but I could not end the proof. Any idea?

Answer 1

So here is a proof: Let $\varphi \in C_c^{\infty}(\mathbb{R})$ be any test function. Then

$$ \int_{\mathbb{R}} \left( \frac{\delta(x+2a)-\delta(x+a)}{a} \right) \varphi(x) \, \mathrm{d}x = \frac{\varphi(-2a) - \varphi(-a)}{a}. $$

Now letting $a \to 0$, either by writing $\frac{\varphi(-2a) - \varphi(-a)}{a} = \frac{\varphi(-2a) - \varphi(0)}{a} + \frac{\varphi(0) - \varphi(-a)}{a}$ or applying the L'Hôpital's rule, we can check that

\begin{align*} \lim_{a \to 0} \frac{\varphi(-2a) - \varphi(-a)}{a} &= -\varphi'(0) = \int_{\mathbb{R}} \delta'(x)\varphi(x)\, \mathrm{d}x. \end{align*}

Therefore

$$ \lim_{a \to 0} \frac{\delta(x+2a)-\delta(x+a)}{a} = \delta'(x) $$

in the space $\mathcal{D}(\mathbb{R})$ of distributions.