Could any one tell me in which interval the sequence of functions converges uniformly?
$f_n(x)= e^{-n\cos^2 x}$
at $x=0$ we have limit function $f(x)=0$ at $x={\pi\over 2}, f(x)=1$ again at $x=\pi$ $f(x)=0$, but I am not able to find out about the interval of uniform convergence, point wise it converges that I agree but the limit function is not continuous but our sequence consists all continuous function so for uniform convergence our limit function must be continuos
It seems the following.
Put $A=\{\pi/2+n\pi:n\in\mathbb Z\}$. The sequence $\{f_n\}$ converges pointwise to the function $f$ which is a characterictic function $\chi_A$ of the set $A$, that is $f(x)=1$ if $x\in A$ and $f(x)=0$ if $x\in\mathbb R\setminus A$. We claim that the sequence $\{f_n\}$ converges (to the zero function) uniformly on an interval $[a,b]$ iff $[a,b]\cap A=\emptyset$. Indeed, if the sequence $\{f_n\}$ converges uniformly on the interval $[a,b]$ then the function $f|[a,b]$ is continuous as a uniform limit of a sequence $\{f_n|[a,b]\}$ of continuous functions. So $[a,b]\cap A=\emptyset$. From the other side, if $[a,b]\cap A=\emptyset$ then $0\le q=\sup\{\cos^2 (x):x\in [a,b]\}<1$. Then for each $x\in [a,b]$ and each $n$ we have $|f_n(x)|\le e^{-nq}$. Since the sequence $\{ e^{-nq}\}$ converges to the zero, the sequence $\{f_n\}$ converges to the zero function uniformly on the interval $[a,b]$.