Unif. conv. of fx=Σc_n*x^n on |x|<h implies unif. conv. on |x|<=h?

[omyojj]

[SOLVED] unif. conv. of fx=Σc_n*x^n on |x|<h implies unif. conv. on |x|<=h?

my whole question is on the title..uniform convergence of power series..

The answer is yes..how can I prove that? help me lol

and I`d like to know..the existence of limit..

[1 3 5 ''' (2n-1)]/[2 4 6 ''' (2n)] -> ? as n goes infinity..

(sorry for my bad Eng..)

 

[EnumaElish]

Suppose there is no unif. conv. on the closure (|x| < h). What does that imply about the behavior of the function as |x| ---> h (versus at |x| = h)?

 

[rudinreader]

I liked this problem because I hadn't thought about it before and my initial thought was to apply Abel's theorem, but it turned out not to be the thing to do here.


Fact 1) Suppose f_n -> f uniformly on [a,b), and f_n(b) -> f(b).
prove: f_n -> f uniformly on [a,b].


Fact 2) If f_n is a sequence of continuous functions on [a,b] that converges
uniformly to f(x) on [a,b), then lim f_n(b) exists.

proof:

Fix e > 0.
By uniform convergence, choose N such that such that |f_n(x)-f_m(x)| < e for n,m > N, x in [a,b).
By continuity, |f_n(b)-f_m(b)| = lim |f_n(x)-f_m(x)| <= e.
Hence f_n(b) is Cauchy, thus converges.


Applying 1),2) above, you get:

If f_n is a sequence of continuous functions on [a,b] such that f_n -> f(x) uniformly on [a,b),
then f_n -> f(x) uniformly on [a,b]. (Where in the hypothesis, it is not assumed lim f_n(b) exists.)

(*note: f(b) is defined to be lim f_n(b), in the conclusion of the proof.)

 

[omyojj]

thx much..:)

but after showing that lim f_n(b) exists(f_n Cauchy seq. at x=b)..

isn`t it necessary to show that the limit tends to f(b)=Σc_n b^n??

so |f_n(b)-f(b)| -> 0 as n-> infinity..?

 

[rudinreader]

You have $$f_n(b) = \sum_{i=0}^n c_i b^i$$ is a Cauchy sequence. By definition, $$\sum_{i=0}^\infty c_i b^i$$ is the limit of this sequence. So no, there's nothing to show.

In the above proof, I didn't assume lim f_n(b) existed. Once I showed it existed, of course we have to define f(b) = lim f_n(b). But in the case of power series, that is precisely what the definition of f(x) is.

 

[omyojj]

Got it~! thanks very much for saving me!