# Limit of a sequence of functions

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!!

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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 $x\in\mathbb{R}$.
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