Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(f_n)$ be a sequence of functions defined on $[0,1]$. Show that if $(f_n)$ converges to zero uniformly on $[0,1]$, then for any sequence of points $(x_n)$ with $x_n \in [0,1]$ for every $n$, the sequence $f_n(x_n)$ has limit zero.

share|cite|improve this question
What have you tried so far? Where are you stuck? Kindly show your work, even if only partial. – cardinal Jan 25 '12 at 2:54
I think I have the main idea, but I'm not quite sure of the wording. Since (fn) converges uniformly to zero for all n, the limit clearly exists, but I am not sure what reasoning can be used to show the limit is zero. – phillips0023 Jan 25 '12 at 2:59
Assume not. Argue by contradiction. – mixedmath Jan 25 '12 at 3:00
For every $n$, $|f_n(x_n)| \le \sup_{x \in [0,1]} |f_n(x)|$. But the right side goes to 0 as $n \to \infty$... – Nate Eldredge Jan 25 '12 at 3:46
You can prove, as a previous exercise, that $f_n\to f$ uniformly in $A$ if and only if the sequence of numbers $\sup_{x\in A} \vert f_n(x)-f(x)\vert$ converges to $0$. Then, look at the Nate's comment. – leo Jan 25 '12 at 5:42

Since $\{f_n\}$ converges uniformly to $0$, given $\epsilon>0$, there is an $N$ so that $|f_m(x)|<\epsilon$ for all $m>N$ and all $x\in[0,1]$.

To show that the sequence $\{f_n(x_n)\}$ has limit $0$, you need to show that for every $\epsilon>0$, there is an $N$ so that $|f_m(x_m)|<\epsilon$ for all $m>N$.

Can you see how to put these together to get what you want?

share|cite|improve this answer
But how can the |fn(xm)|< epsilon if the function isn't explicitly defined? – phillips0023 Jan 25 '12 at 3:04
@phillips0023: Because you use the structure available to you. You've been told that you have a sequence of functions $(f_n)$ which converge uniformly to zero on $[0,1]$. That is a rather strong statement to make about the $(f_n)$. Indeed, it is enough to allow you to arrive at the conclusion without having to say anything more explicit about the nature of $(f_n)$. – cardinal Jan 25 '12 at 3:09
@phillips0023 I'm sorry; what function? The limit function is the zero function: $|f_m(x_m)-0|=|f_m(x_m)|$. From the first paragraph, you can make $|f_n(x)|$ as small as you like $for\ all$ numbers $x\in[0,1]$ as long as $n$ is sufficiently large, and that includes the $x$'s in the given sequence. – David Mitra Jan 25 '12 at 3:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.