limit function / uniform convergence

bmixon · #1 $Old$ June 14th, 2009, 08:22 PM

Examine the functions f_n(x) = x^(2n) / (1+x^(2n)) on R for n = 1,2,...
a) calculate the limit function f of the sequence [f_n]
b) Complete the following sentence: If [f_n] converges uniformly on the interval I=[a,b], then I cannot contain the point(s) x=____. Justify your answer, and then show the convergence is uniform on any closed intervals not containing those points.

Chandru1 · #2 $Old$ June 14th, 2009, 09:24 PM

The function $f_{n}(x)$ can also be written as $\displaystyle f_{n}(x)=1- \frac{1}{1+x^{2n}}$ . From here we see that $\lim_{n \to \infty} f_{n}(x)=1$ if $x \neq 0$ and $\lim_{n \to \infty} f_{n}(x)=0 \ \textrm{if x=0}$ . As for the next part i think that $I$ cannot contain $x<0$

Enrique2 · #3 $Old$ June 28th, 2009, 03:39 PM

Quote:

Originally Posted by bmixon $View Post$

Examine the functions f_n(x) = x^(2n) / (1+x^(2n)) on R for n = 1,2,...
a) calculate the limit function f of the sequence [f_n]
b) Complete the following sentence: If [f_n] converges uniformly on the interval I=[a,b], then I cannot contain the point(s) x=____. Justify your answer, and then show the convergence is uniform on any closed intervals not containing those points.

If |x|<1 the function goes to 0. If |x|>1 the function goes to 1. f(1) =
f(-1)=1/2. Thus, the convergence cant be uniform in intervals contaiining 1 or -1.