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Let $f:\mathbb{R}\to \mathbb{R}$ be a function uniformly continuous on $\mathbb{R}$; let $f_n(x)=f(x+1/n)$ be a sequence. Prove that $f_n$ converges uniformly to $f$.

Proof:

Since $f$ is uniformly continuous, then, $\forall\varepsilon>0, \exists \delta >0$ such that $\lvert x-y \rvert < \delta \implies \lvert f(x)-f(y) \rvert<\varepsilon, \forall x\in\mathbb{R}$. Let $N>0, N\in\mathbb{N}$, such that $1/N<\delta$. Then $\lvert x+1/N-x \rvert = 1/N < \delta$. Then $1/n < 1/N < \delta$ whenever $n>N$. Thus $\lvert f(x+1/n)-f(x) \rvert < \varepsilon$. This implies that $f_n$ converges to $f$. Since $1/n<\delta$ for all $n>N$, $f_n(x)$ converges to $f(x)$ for all $x$. Hence, $f_n$ converges to $f$ uniformly.

Please let me know if this proof isn't lacking some details and/or has some flaws.

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    $\begingroup$ What is that for $x\in{\bf{R}}$ in the second line? I guess you want to state $x,y\in{\bf{R}}$ instead of $x$ only. And you should demonstrate $|x+1/n-x|=1/n\leq 1/N<\delta$ rather than $|x+1/N-x|=...$. $\endgroup$ – user284331 Nov 30 '16 at 5:37
  • $\begingroup$ Right, errors of omission. Thanks for pointing out. $\endgroup$ – sequence Nov 30 '16 at 6:47

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