# Uniform convergence of a series

1. Jul 15, 2010

### WaterPoloGoat

1. The problem statement, all variables and given/known data

We know that f is uniformly continuous.

For each n in N, we define fn(x)=f(x+1/n) (for all x in R).

Show that fn converges uniformly to f.

2. Relevant equations

http://en.wikipedia.org/wiki/Uniform_convergence

3. The attempt at a solution

I know that as n approaches infinity, that fn(x)=f(x), which implies that fn converges to f.

I'm currently trying to apply the fact that if fn is uniformly convergent, then

limn->infinity Sup {fn(x): x in R}=0.

But I keep getting stuck on the fact that there's an function in the definition of fn i.e., fn(x)=f(x+1/n). Is there a way to work with it?

Last edited: Jul 15, 2010
2. Jul 15, 2010

### losiu99

It's actually pretty straightforward. Given $$\epsilon > 0$$, for each x there is a $$\delta>0$$ such that $$|f(x+h)-f(x)|<\epsilon$$ whenever $$|h|<\delta$$. Hence, for all $$n > \delta ^{-1}$$ we have $$|f_n(x)-f(x)|<\epsilon$$.