# Showing uniform continuity implies uniform convergnce

Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be uniformly continuous on $\mathbb{R}$ and let $f_n(x) = f(x+1/n)$ for $x\in \mathbb{R}$. Im trying to show that $f_n$ converges uniformly on $\mathbb{R}$ to $f$.

I know that for $f_n(x)$ to converge uniformly to $f$ we must have that $|f(x+1/n)-f(x)|<\epsilon$ for $\epsilon$ independent of $x$. The condition that $f$ is uniformly continuous provides me that $|x-c|< \delta \rightarrow |f(x)-f(c)| < \epsilon$, but this doesnt seem to be a good method for proving uniform continuity directly.

Im trying to think of some criteria from uniform continuity of $f$ that I could use to help me show uniform convergence, but im drawing a blank..perhaps it be better to prove the contrapositive here?

Any hints would be appreciated!

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The $\epsilon$-$\delta$ method is the right approach here. How far apart are $x$ and $c = x + 1/n$ when $n$ is large? How does this distance compare to $\delta$? –  froggie Dec 2 '12 at 21:31
Is is really so basic? Let $c = x+1/n$, then $|x-c| = 1/n < \delta$ if we choose an $N \in \mathbb{N}$ sufficiently large. Then we know that $|f(x)-f(x+1/n)| < \epsilon$ implies that $|f(x+1/n)-f(x)| < \epsilon$ which satisfies our definition of uniform continuity for $f_n$, and thus our result holds. –  MSEoris Dec 2 '12 at 21:53
Yes, it really is that basic! –  froggie Dec 2 '12 at 23:37