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Looking for some input on my approach to this problem:

Suppose that $f: \mathbb{R} \to \mathbb{R}$ is uniformly continuous, and let $f_n(x)=f(x+\frac{1}{n})$. Prove that $(f_n)^{\infty}_{n=1}$ converges uniformly to $f$ on $\mathbb{R}$. Does this hold true if $f$ is just continuous?

If $f$ is uniformly continuous, then for every $\epsilon > 0$ there is a positive real number $r>0$ such that

$$|f(x)-f(a)|< \epsilon$$

whenever

$$|x-a|<r$$

for $x,a \in \mathbb{R}$. Let $a=x+\frac{1}{n}$. Then we have

$$\left | f\left (x+\frac{1}{n}\right )-f(x) \right | < \epsilon$$

when

$$\left | \frac{-1}{n} \right |<r.$$

Now, $(f_n)$ converges uniformly if for every $\epsilon > 0$ there is an integer $N$ such that

$$|f_n(x)-f(x)|< \epsilon$$

for all $x \in \mathbb{R}$ and $n \geq N.$ Choose an $N$ such that $1/N < r$ (do I need this condition?). Then the conditions are met for uniform convergence. It seems to me that this would hold if $f$ is just continuous as well. As $n$ increases $1/n$ can be made arbitrarily small. There does not need to be continuity with an $r$ that is independent of the point chosen as $|x-a| \to 0$ as $n \to \infty$.

Thanks for any help.

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Yes, you need this condition $1/N<r$, good job. The fact that $r$ is independent of $x$ is precisely uniform continuity. –  1015 Mar 15 '13 at 12:26

2 Answers 2

up vote 1 down vote accepted

Your proof is fine and does use heavily the uniform continuity of $f$, as one would expect.

Now take the first non uniformly continuous yet continuous function you can think of. Say $f(x)=x^2$. Then $$ f_n(x)=\left(x+\frac{1}{n}\right)^2=x^2+\frac{2x}{n}+\frac{1}{n^2}. $$ What is the pointwise limit $f$? Does $\sup_\mathbb{R}|f_n-f|$ tend to $0$? You might want to consider points such as $x=n$.

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The first part of the proof seem ok to me.

For the second part: If $r$ could depend on $x$, then $r$ can become arbitrarily small for certain $x$ and $N>1/r$ arbitrarily large. So, $f_n$ would not converge uniformly. Just try it with some function that does not converge uniformly on the whole real numbers. Something easy like $f(x)=x^2$ should do the trick, I think.

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