Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

If $\{f_{n}(x)\}$ is a sequence of non-zero real valued functions, which are continuous and bounded by some constant $A>0$. Let $a_{n}=\sup_{x\in \mathbb R}|f_{n}(x)|$ and the sup is attained at the points $p_{n}$, also $\lim_{n\to\infty}a_{n}=0$. Define a new function $F_{n}(x)=\frac{f_{n}(x)}{a_{n}}$, for all $n\geq 1$, and $x\in \mathbb R$. What can we say about $\lim_{n\to\infty}F_{n}(x)$? Is it zero? If not when it could be zero?

My guess is we cannot know! Unless $f_{n}$ converges to $0$ faster that $a_{n}$, but the given information doesn't tell if this is the case or not!

Looking to this from another point of view: Note that $1=\sup|F_{n}|$, so if $F_{n}\to 0$, then we must have $\sup |F_{n}|\to 0$, a contradiction!!

share|cite|improve this question
If I understood your setup correctly, then the sequence $f_n(x)= 1/n$ would satisfy your assumptions. So $a_n= 1/n$ and $F_n(x) =1$. – user20266 Jun 9 '12 at 18:33
Yes, this was an example for my second idea above. – Mathvisitor Jun 9 '12 at 18:36
No that's not the same. $f_n$ can be be a sequence of compactly supported bumps which escape to $\infty$ – user20266 Jun 9 '12 at 18:37
So, this means we only have to have the first guess above, like $f_{n}(x)=1/n^{2}$, and $a_{n}=1/n$, in this case $F_{n}\to 0$. – Mathvisitor Jun 9 '12 at 18:38

For all $n$ $\sup_{x\in\mathbb{R}}|F_n(x)|=1$, so there is absolutely no reason why $F_n(x)$ should converge to $0$ as $n\to\infty$ for any $x$. For instance, you could have $f_n(x)=1/n$ for all $x\in\mathbb{R}$, $a_n=1/n$ and $F_n(x)=1$ for all $x\in\mathbb{R}$.

share|cite|improve this answer

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.