Suppose $\phi \in C^{\infty}_0(\mathbb{R})$, ie $\phi$ in an infinitely differentiable function with compact support. Let $\{a_m\}_{m\in \mathbb{N}}\subseteq \mathbb{R}\setminus\{0\}$ a sequence of real numbers such that $a_m\to 0$ as $m\to \infty$.
For every $m$ we define $$\psi_m(x)=\frac{\phi(x-a_m)-\phi(x)}{a_m}\qquad x\in \mathbb{R}.$$ Then $\psi_m \in C^{\infty}_0(\mathbb{R})$ for every $m$ and also $$\psi_m\to -\phi^{'}$$punctually on $\mathbb{R}$ since for every $x \in \mathbb{R}$ $$\lim_{h\to 0}\frac{\phi(x+h)-\phi(x)}{h}=\phi^{'}(x).$$ I'm having trouble proving that also $$\psi_m\to -\phi^{'}$$ uniformly on $\mathbb{R}$.
Any hint would be really appreciated.
HINT
By M.V.T $\forall m \in \Bbb{N} ,\exists t_m$ such that $t_m \to 0$ and $x-a_m<x-t_m<x$ and
$|\psi_m(x)+\phi(x)|=|\phi'(x)-\phi'(x-t_m)|$
Thus since $\phi'$ has compact support $K$ you can use its uniform continuity on $K$ and prove the uniform convergence.
Continue from here...