$\frac{\varphi(x+h)-\varphi(x)}{h}\to\varphi'(x)$ uniformly as $h\to 0$?

Question

Let $\varphi$ be a bounded, differentiable function on $\mathbb{R}$ such that $\varphi'$ is bounded and uniformly continuous on $\mathbb{R}$.

We want to prove that $\displaystyle\frac{\varphi(x+h)-\varphi(x)}{h}\to\varphi'(x)$ uniformly as $h\to 0$

I can prove $\varphi$ is uniformly continuous, but I don't know what to do with it.

Any hints?

Answer 1

Hint: use the mean value theorem. $$ \phi(x+h) - \phi(x) - h\phi'(x) = h (\phi'(x + \theta_x h) - \phi'(x)) $$ for some $\theta_x\in(0,1)$.

Answer 2

You have $\phi(x+h)-\phi(x)-\phi'(x)h = \int_0^1 (\phi'(x+th)-\phi'(t))h dt$.

Let $\epsilon>0$ and choose $\delta$ such that if $\|\xi\| < \delta$ then $\|\phi'(x+\xi)-\phi'(x) \| < \epsilon$.

Then $\|\phi(x+h)-\phi(x)-\phi'(x)h \| \le \epsilon \|h\|$.

(There is no need to assume that $\phi,\phi'$ are bounded.)