# Is $\lim\sup=\sup\lim$?

Assume $(a_n(x))_{n=1}^{\infty}$ is a bounded sequence in $\mathbb R$, when $x$ is $\in\mathbb R$ and is relevant to the sequence in some way that doesn't really interest us in my question.

Assume $\lim_{n\to\infty} |a_n(x)|=0$ for all $x\in A$ for some $A\subset \mathbb R$.

Is it true that $\lim_{n\to\infty} \sup\{|a_n(x)|:x\in A\}=0$? Why?

I believe it is (i want it to be, for my proof to be good...), but I can see how it can turn out false.

thanks

• In situations like that (reversing infinite processes), the answer is usually no unless extra assumptions are made. – Matthew Leingang Jan 20 '15 at 13:59

The answer is no. Consider $A = \mathbb N$ for simplicity and let $a_n(x) = \delta_{nx}$ the Kronecker delta. Since for fixed $x$, $a_n(x) = 0 \ \forall n > x$, the limit of each $(a_n(x))_n$ is $0$, but $$\lim_{n\to\infty} \sup_{x\in A}|a_n(x)| = \lim_{n\to\infty} 1 = 1 \ne 0$$
No: take $a_n$ a piecewise linear function on $[0,1]$ whose graph joins the points $(0,0)$, $(1/n,0)$, $(2/n,1)$, $(3/n,0)$ and $(1,0)$ for $n\geqslant 4$.
Hence even when $a_n$ is a continuous bounded function, the result has no reason to be true.