# Thread:

1. ## Uniform convergence of a function sequence

Hey everyone
Given function f(x) which is differentiable on R , and f'(x) is continuous on R, and a sequence (f_n(x)) such as : $f_n(x)=n(f(x+\frac{1}{n}))-f(x))$
I need to prove that f_n(x) is uniformly converges on each interval [a,b].

I have managed to show that (f_n(x)) approaches f'(x), but I couldn't get any further.
I know nothing about the values of f(x), so I can't know if f_n(x) is monotonic or not , and therefore I can't use Dini's theorem. Plus , I can't say much about |f_n(x)-f'(x)| , so I can't figure out how do I show that for every epsilon > 0 , there is N such as for every n>N , |f_n(x)-f'(x)| < epsilon

Any ideas how to get any further? Thanks people!

2. Originally Posted by Gok2
Hey everyone
Given function f(x) which is differentiable on R , and f'(x) is continuous on R, and a sequence (f_n(x)) such as : $f_n(x)=n(f(x+\frac{1}{n}))-f(x))$
I need to prove that f_n(x) is uniformly converges on each interval [a,b].

I have managed to show that (f_n(x)) approaches f'(x), but I couldn't get any further.
I know nothing about the values of f(x), so I can't know if f_n(x) is monotonic or not , and therefore I can't use Dini's theorem. Plus , I can't say much about |f_n(x)-f'(x)| , so I can't figure out how do I show that for every epsilon > 0 , there is N such as for every n>N , |f_n(x)-f'(x)| < epsilon

Any ideas how to get any further? Thanks people!
http://www.math.uchicago.edu/~ershov/16300/uniform2.pdf

3. Thank you , but I knew that.
That's what I explained,
I don't know nothing about |f_n(x)-f'(x)| , therefore I can't find a sequence c_n that |f_n(x)-f'(x)| <= c_n , and from the seam reason I can't find sup |f_n(x)-f(x)|, therefore I am kinda stuck...
Any ideas people?

4. Anyone?
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