Derivative of a distribution in $\mathcal{D}'(\mathbb{R})$

Question

How do I show that the Newton's quotient is still valid for distribution? By this, I mean that if $u \in \mathcal{D}'(\mathbb{R})$ and if $u_r$ is the translation $\langle u_r,\phi \rangle = \langle u, \phi_{r}\rangle$ of $u$, in which $\phi_{-r}(x) = \phi(x+r)$ and $\phi \in \mathcal{C}_0^\infty$, then $$\lim_{r \rightarrow 0} \frac{1}{r}(\langle u_r,\phi \rangle - \langle u,\phi \rangle) = \langle u,-\phi' \rangle.$$ By definition, we have that $\langle u',\phi \rangle = -\langle u,\phi' \rangle$. The problem for me is to show the convergence $\frac{\phi_{-r} - \phi}{r} \rightarrow -\phi'$ in $\mathcal{C}_0^\infty$, as $r \rightarrow 0$.

Answer 1

Take a sequence $r_n \to 0.$

Given $\phi \in C_c^\infty(\mathbb{R})$ let $\psi_n(x) = r_n^{-1}(\phi(x+r_n) - \phi(x)).$ It's obvious that $\psi_n \in C_c^\infty(\mathbb{R}).$

We want to show that $\psi_n \to \phi'$ in $C_c^\infty(\mathbb{R}).$ For this we must show two things:

1. All $\psi_n$ have their support contained in one fixed compact set.
2. $\| \partial^k\psi_n - \partial^k\phi' \| \to 0$ for all $k=0, 1, 2, \ldots$

Point 1 is easy. If $\operatorname{supp} \phi \subseteq [a, b]$ then $\operatorname{supp} \phi_n \subseteq [a-r_n, b-r_n] \subseteq [a-\sup_n r_n, b+\inf_n r_n]$ so all $\phi_n$ have their supports contained in one compact set.

For point 2 we use the mean value theorem. This says that $\phi(x+r_n) - \phi(x) = r_n \phi'(\xi_n)$ for some $\xi_n$ between $x$ and $x+r_n$. Therefore, $$ | \psi_n(x) - \phi'(x) | = | r_n^{-1} (\phi(x+r_n) - \phi(x)) - \phi'(x) | = | \phi'(\xi_n) - \phi'(x) | . $$

Now, the mean value theorem again says that $\phi'(\xi_n) - \phi'(x) = (\xi_n-x) \phi''(\eta_n)$ for some $\eta_n$ between $\xi_n$ and $x$. Thus, $$ | \psi_n(x) - \phi'(x) | \leq | \xi_n-x | |\phi''(\eta_n)| \leq | r_n | \sup_{y \in \mathbb{R}} |\phi''(y)| \to 0 \cdot \sup_{y \in \mathbb{R}} |\phi''(y)| = 0 \text{ as $n\to \infty$}, $$ since $\sup_{y \in \mathbb{R}} |\phi''(y)|$ is finite. This implies that $\| \psi_n - \phi' \| \to 0$ as $n \to \infty.$

This shows point 2 for $k=0.$ For $k=1,2,\ldots$ we just have to apply the same reasoning, replacing $\psi_n$ with $\partial^k \psi_n(x) = r_n^{-1}(\phi^{(k)}(x+r_n) - \phi^{(k)}(x))$ and $\phi'$ with $\partial^k \phi' = \phi^{(k+1)}$.