# Series of continuous functions

• Mar 16th 2014, 10:24 PM
phys251
Series of continuous functions
"Prove that if a series of continuous functions converges uniformly, then the sum function is also continuous."

I'm trying to interpret what this is saying so that I can know what to prove. Does the given portion mean that the individual f_n's uniformly converge to f, or that $\sum_{n=1}^{\infty} f_n \to \sum_{n=1}^{\infty} f$?
And what does "the sum function is continuous" mean? Simply that $\sum_{n=1}^{\infty} f$ is continuous?
• Mar 16th 2014, 10:54 PM
romsek
Re: Series of continuous functions
Quote:

Originally Posted by phys251
"Prove that if a series of continuous functions converges uniformly, then the sum function is also continuous."

I'm trying to interpret what this is saying so that I can know what to prove. Does the given portion mean that the individual f_n's uniformly converge to f, or that $\sum_{n=1}^{\infty} f_n \to \sum_{n=1}^{\infty} f$?
And what does "the sum function is continuous" mean? Simply that $\sum_{n=1}^{\infty} f$ is continuous?

This has the definition you need. Right at the top.
• Mar 16th 2014, 11:06 PM
phys251
Re: Series of continuous functions
Quote:

Originally Posted by romsek

The part about "A series...converges uniformly"? OK, thanks.

Again, though, what does "the sum function is continuous" mean? S_n(x) --> ?
• Mar 16th 2014, 11:49 PM
romsek
Re: Series of continuous functions
Quote:

Originally Posted by phys251
The part about "A series...converges uniformly"? OK, thanks.

Again, though, what does "the sum function is continuous" mean? S_n(x) --> ?

if the $f_k(x)$ are continuous then

let $f(x)=\displaystyle{\lim_{n \to \infty}}S_n(x)\text{ where }S_n=\displaystyle{\sum_{k=0}^n} f_k(x)$

if $S_n(x)$ converges uniformly then $f(x)$ is continuous.
• Mar 17th 2014, 12:39 PM
phys251
Re: Series of continuous functions
Quote:

Originally Posted by romsek
if the $f_k(x)$ are continuous then

let $f(x)=\displaystyle{\lim_{n \to \infty}}S_n(x)\text{ where }S_n=\displaystyle{\sum_{k=0}^n} f_k(x)$

if $S_n(x)$ converges uniformly then $f(x)$ is continuous.

OK. Can we just use the fact that if (1) f_n is continuous for all n's, and (2) f_n -> f uniformly, then (3) f is continuous? I.e., does the result I'm trying to prove immediately follow, since the sum of continuous functions on an interval is also continuous on that interval?
• Mar 17th 2014, 01:11 PM
Plato
Re: Series of continuous functions
Quote:

Originally Posted by phys251
OK. Can we just use the fact that if (1) f_n is continuous for all n's, and (2) f_n -> f uniformly, then (3) f is continuous? I.e., does the result I'm trying to prove immediately follow, since the sum of continuous functions on an interval is also continuous on that interval?

Yes you can.
Use what Ken Ross calls the famous $\dfrac{\varepsilon }{3}>0$ argument.

$|f(x)-f(x_0)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x)|$

You select an $n$ large enough to insure that on RHS the first and third term is less that the magic number from the continuity of $f_n$.

The middle comes from the uniform convergence of the sequence.
• Mar 17th 2014, 02:41 PM
phys251
Re: Series of continuous functions
Thanks, Plato, but I think you have a few errors in there. Isn't that last term supposed to be f_n(x_0) - f(x_0)? And uniform convergence produces the first and third terms, not the second one, right?
• Mar 17th 2014, 02:55 PM
Plato
Re: Series of continuous functions
Quote:

Originally Posted by phys251
Thanks, Plato, but I think you have a few errors in there. Isn't that last term supposed to be f_n(x_0) - f(x_0)? And uniform convergence produces the first and third terms, not the second one, right?

You are correct. But the same idea holds.