Theoretical problem on uniform convergence of sequence of functions

Question

I'm having difficulties trying to solve this problem :

If $ f $ is differentiable and $ \forall x, y \in \Bbb R :\lvert f'(x) - f'(y) \rvert \le 3 \lvert x - y \rvert $, show that $ f_n $ converge uniformly to $ f' $, where $$ f_n(x) = \frac{f(x + \frac{1}{n}) - f(x)}{\frac{1}{n}}$$

This problem is very different from the usual investigation of uniform convergence for sequences of functions and I don't know how to proceed.

Answer 1

By the MVT, $f_n(x)= f'(c(n,x))$ for some $c(n,x) \in (x,x+1/n).$ Thus $$f_n(x) - f'(x) = f'(c(n,x))-f'(x),$$ which in absolute value is no more than $3/n.$ Therefore $f_n \to f'$ uniformly on $\mathbb R.$

Answer 2

Write $$ f_n(x)=n·(f(x+\tfrac1n)-f(x))=\int_0^1 f'(x+\tfrac sn)\,ds $$

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This is a specialization of the more general formula $$ F(x+v)-F(x)=\int_0^1F'(x+sv)v\,ds, $$ used for instance in the vector version of the mean-value theorem, which in turn is a consequence of the fundamental theorem of calculus for $G(s)=F(x+sv)$ with $G'(s)=F'(x+sv)v$.

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In consequence, $$ |f_n(x)-f'(x)|\le\int_0^1|f'(x+\tfrac sn)-f'(x)|\,ds\\ \le\int_0^13\left|\frac sn\right|\,ds=\frac3{2n} $$ which implies the postulated uniform convergence.