# Convergence of sequence of uniformly bounded continuous functions

Let $$f_n:[0,2]\rightarrow\mathbb{R}$$ be a sequence of continuous functions, which is uniformly bounded, i.e., $$|f_n(x)|\leq c$$ for all $$n\geq 1$$ and $$x\in[0,2]$$. Let $$a_k$$ be a sequence of nonnegative real number such that $$a_k\to 0$$ as $$k\to\infty$$. Is it always true that $$f_n(x+a_k)-f_n(x)$$ is converging to $$0$$?

I am trying to find counterexample, but I get stuck when considering the assumption uniformly bounded. If not uniformly bounded, then it is easy to find, such as $$f_n(x)=nx$$, and $$a_n=1/n$$. In this case $$f_n(x+a_k)-f_n(x)=1$$. However, I don't know how to construct a counterexample when $$f_n$$ is uniformly bounded. Can anyone help?

You can take a sequence of continuous functions which converges to a non-continuous function.

Take $$f_n: [0,2] \to [0,1] \begin{cases} x \mapsto x^n && x \in [0,1]\\ x\mapsto (2-x)^n && x\in[1,2] \end{cases}$$, those are continuous and uniformly bounded, as it is $$x\mapsto x^n$$ mirrored at the vertical at 1.
Now take $$a_n = 1/n$$. and $$x = 1$$ then $$f_n(x+a_n)-f_n(x)$$ does not converge to zero:
As $$f_n(x+a_n)-f_n(x) = (2-(1+1/n))^n-1 = (1-1/n))^n-1 \rightarrow e^{-1} -1 \not = 0$$ for $$n \to \infty$$

• +1. Nice answer! Dec 12, 2022 at 23:35
• Thanks for your help. Good example. Dec 13, 2022 at 3:53