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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!

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2 Answers 2

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

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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.

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