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?


1 Answer 1


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 $n \to \infty$

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .