interval the sequence of functions converges uniformly

Question

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

What is the limit function on intervals like $ [\alpha_n,, \beta_n],$ where $\alpha_n$ and $\beta_n$ are such that $-\frac{\pi}{2}+n\pi < \alpha_n < \beta_n < \frac{\pi}{2}+n\pi,?$ — M. Strochyk, Sep 11 '13 at 07:18
okay taking $n=1$ say, $\pi/2 <a<b<3\pi/2$ I think in this case the limit function is $0$ only and hence continuos. — Myshkin, Sep 11 '13 at 07:31

score 1 · Answer 1 · answered Sep 17 '13 at 15:29

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 $[a,b]$.

interval the sequence of functions converges uniformly

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