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Suppose that $f:\Bbb{R}\to\Bbb{R}$ is uniformly continuous. Let $f_n(x)=f(x+1/n)$.

a) Prove that $f_n$ converges uniformly to $f$ on $\Bbb{R}$

b) Does this remain true if $f$ id just continuous? Prove it or provide a counterexample.

My attempt:-

For a, since $f$ is uniformly continuous, $f(x+1/n)$ exists and $\limsup\left|f_n-f\right|=\limsup\left|f(x+1/n)-f(x)\right|=0$. Therefore $f_n\to f$ uniformly on $\Bbb{R}$.

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The example of a function which is continuous but not uniformly continuous on $\mathbb{R}$ that you should keep in mind is $f(x)=x^2$. Here you find

$$\left |f_n(x)-f(x) \right |=\left |x^2+2x/n+1/n^2-x^2 \right |= \left |2x/n+1/n^2 \right |.$$

Note the $x$ term can be made large for each fixed $n$. Can you finish from here?

Also I would say that your proof of part (a) is not really complete. You need to argue somewhere that the $N$ you choose for convergence can be chosen to not depend on $x$. This much is not so hard, because uniform continuity tells you that $|f(x)-f(y)|$ can be made small by only controlling $|x-y|$, which for your case is just $1/n$.

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  • $\begingroup$ So uniform continuity implies independence of x, and merely continuous implies dependence on x and hence only pointwise convergence? $\endgroup$ – dd19 Mar 13 '15 at 3:16
  • $\begingroup$ Just continuity doesn't necessarily imply dependence on $x$, but it makes it possible, as in the example I gave. $\endgroup$ – Ian Mar 13 '15 at 3:16
  • $\begingroup$ I see, poor choice of words on my part. It makes sense now, thank you $\endgroup$ – dd19 Mar 13 '15 at 3:17
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Since $f$ is uniformly continuous, for any $\epsilon > 0, \exists \delta > 0$ such that $$ |x-y| < \delta \Rightarrow |f(x)-f(y)| < \epsilon $$ So for $N \in \mathbb{N}$ such that $1/N < \delta$, it happens that $$ |f_n(x) - f(x)| < \epsilon \quad \forall n\geq N \quad \forall x\in \mathbb{R} $$ This gives uniform convergence.

For part (b), take $f(x) = e^x$, then suppose $f_n \to f$ uniformly, then for $\epsilon = 1>0, \exists N \in \mathbb{N}$ such that $|f_n(x) - f(x)| < 1$ for all $x\in \mathbb{R}$ and $n\geq N$. Choose $M \in \mathbb{R}$ such that $$ e^x > N \quad\forall x>M $$ Then for any $x > M$, the mean-value theorem implies that there is some $\zeta \in [x,x+1/N]$ such that $$ |f_N(x) - f(x)| = e^{\zeta}/N > e^x/N > 1 $$

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