If $f:\mathbb{R} \to \mathbb{R}$ is uniformly continuous on $\mathbb{R}$, show that $f(x-1/n)$ converges uniformly on $\mathbb{R}$

Question

This is question 5 of Section 8.2, of Introduction to Real Analysis by Robert G. Bartle, Donald R. Sherbert.

Question statement: Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be uniformly continuous on $\mathbb{R}$ and let $f_n(x) := f(x - 1/n)$ for $x \in \mathbb{R}$. Show that $f_n$ converges uniformly on $\mathbb{R}$ to $f$.

My attempt: Let $\epsilon>0$ be given. Since $f$ is uniformly continuous on $\mathbb{R}$, there exists $\delta>0$ such that for any $x,y$ in $\mathbb{R}$ with $|x-y|<\delta$, we have that $|f(x)-f(y)|<\epsilon$. I'm not really sure how to proceed.

I also tried taking limits. For any $x \in \mathbb{R}$, we have, since $f$ is continuous, that$$ \displaystyle \lim_{n \rightarrow \infty}f_n(x) = \lim_{n \rightarrow \infty}f(x - 1/n)=f(\lim_{n \rightarrow \infty}(x - 1/n))=f(x).$$

But I think this proves pointwise convergence, rather than the uniform convergence that I need.

Can someone please give me a hint?

Thank you.

Answer 1

Choose $ N > \frac 1 \delta$, then for all $n>N$ we have

$$|f_n(x) -f(x)|= |f(x -\frac 1 n) -f(x)| <\epsilon.$$

This is because we have from the uniform convergence and that

$$|x- x+\frac 1 n| = \frac 1 n < \frac 1 N < \delta$$