# Continuity and bounded sequence

Let $$f: \mathbb{R}^N\rightarrow \mathbb{R}$$ be a continuous function, let $$x_n$$ be a bounded sequence.

Show that for every $$\epsilon >0$$, there is a $$\delta>0$$ such that when $$|d|<\delta$$ we have $$\forall n$$ $$|f(x_n+d)-f(x_n)|<\epsilon$$

Suppose $$x_n$$ was made of finitely many points. Then one could apply the definition of continuity for each $$x_n$$ and obtain $$\delta_n$$, such that when $$|d|<\delta_n$$ we have $$|f(x_n+d)-f(x_n)|<\epsilon$$. Taking $$\delta$$ to be the smallest of those $$\delta_n$$ solves the problem.

However we have infinitely many points in the sequence. The sequence is bounded, so there must be only finitely many isolated in the sequence, then the idea would be to apply the reasoning in the finite case but I'm not sure if that would work.

Do you have any hint?

One way to solve this would be to take a large closed ball containing all of your point, call it $$B$$. Since $$B$$ is closed and bounded it is compact. This tells us $$f$$ is uniformly continuous over $$B$$. Then by definition of uniform continuity for all $$\epsilon>0$$ you can choose $$d>0$$ such that $$|f(x+d)-f(x)|<\epsilon$$ for all $$x\in B$$. In particular, for all $$x_n.$$

• This is by far the simplest IMO. – Henno Brandsma Aug 16 '20 at 9:12

I think the inequality is supposed to be $$|f(x_n+d)-f(x_n)| <\epsilon$$.

Prove this by contradiction. If the result is false there exists $$\epsilon >0$$ and $$(n_i)$$ and $$(d_i)$$ such that $$|f(x_{n_i}+d_i)-f(x_{n_i})| \geq \epsilon$$ for all $$i$$ and $$|d_i|<\frac 1 i$$. Since $$(x_{n_i})$$ is bounded it has convergent subsequence. If the limit is $$x$$ then letting $$i \to \infty$$ in above inequality an using continuity of $$f$$ at $$x$$ gives the contradiction $$|f(x)-f(x)| \geq \epsilon$$.