Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like your help with proving that for a function uniformly differentiable $f$, and a series of functions, $f_n(x)=n[f(x+\frac{1}{n})-f(x)]$ is uniformly converges in closed interval $[a,b]$, for a fixed x. I proved that the function pointwise converges to $f'(x)$ and for the uniformly convergence I tried to use Dini's theorem but I don't see why $f_n(x)$ is monotonic for a fixed x. I tried to use the $\epsilon$ definition, but I didn't managed to show that $|f_n(x)-f'(x)|< \epsilon$.

Any suggestion?

Thanks a lot!

share|cite|improve this question
up vote 2 down vote accepted

Doesn't this follows directly from the uniformly differentiable property?

We need to show that, for any given $\epsilon >0$, there exists some natural number $N$ such that for any $n\ge N$ , $|f_n(x)-f'(x)|< \epsilon$

By uniform differentiabily, we know that, any given $\epsilon >0$ there exists some $\delta >0$ such that

$$\left| \frac{f(x+h)-f(x)}{h}-f'(x) \right|<\epsilon$$ for any $h$ with $|h|<\delta$. Pick $N=ceil(1/\delta)$. Then, $n\ge N$ iff $\frac{1}{n}<\delta$, and you are done

share|cite|improve this answer

If $f$ is uniformly differentiable we know that $$ \forall \epsilon \ \exists \delta \ \forall x,y \ \ : \ |x-y|< \delta \ \Rightarrow \ |(f(x)-f(y))/(x-y) - f'(x)| < \epsilon$$ But simply choosing $y= x+1/n$ and $n>1/\delta$ in this definition will give you a proof of uniform convergence.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.