This is a problem from Abbott's Analysis:
Let $f$ be uniformly continuous on all of $\mathbb{R}$, and define a sequence of functions $f_n(x)=f(x+\frac{1}{n})$. Show that $f_n\rightarrow f$ uniformly. Give an example to show that this proposition fails if $f$ is only assumed to be continuous and not uniformly continuous on $\mathbb{R}$.
I know I want $|f(x+1/n)-f(x)|<\epsilon$. I am struggling with finding an $N$ so I can manipulate the quantity on the left to get eventually to epsilon.
On
It goes like this:
By uniform continuity of $f$, for every $\varepsilon > 0$ there is some $\delta > 0$ such that $$ |x-y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon, $$ namely such that $$ x + \frac{1}{n} - x = \frac{1}{n} < \delta \Rightarrow \bigg| f\bigg( x+\frac{1}{n} \bigg) - f(x) \bigg| < \varepsilon, $$ namely such that $$ n > 1/\delta \Rightarrow \bigg| f\bigg( x + \frac{1}{n} \bigg) - f(x) \bigg| < \varepsilon. $$
For the proof, given $\epsilon > 0$, since $f$ is uniformly continuous on $\mathbb{R}$, there exists $\delta > 0$ such that $$|f(x) - f(y)| < \epsilon, \forall x, y \text{ such that } |x - y| < \delta. \tag{$*$}$$ Now let $N$ be chosen such that $\frac{1}{N} < \delta$ (notice that the choice of $N$ doesn't depend on $x$). Then for all $n > N$, by $(*)$ it follows that for every $x \in \mathbb{R}$ $$|f_n(x) - f(x)| = |f(x + 1/n) - f(x)| < \epsilon,$$ since $|x + 1/n - x| = 1/n < 1/N < \delta$. That is, $f_n$ converges to $f$ uniformly.
One counterexample: $f(x) = x^2, x \in \mathbb{R}$. It can be seen that \begin{align*} & \sup_{x \in \mathbb{R}}\left[\left(x + \frac{1}{n}\right)^2 - x^2\right] \\ = & \sup_{x \in \mathbb{R}}\left[\frac{2}{n}x + \frac{1}{n^2}\right] \\ \geq & \frac{2}{n}\times n + \frac{1}{n^2} \\ = & 2 + \frac{1}{n^2} \end{align*} which doesn't converge to $0$ as $n \to \infty$.