# Thread: Uniform convergence of series

1. ## Uniform convergence of series

Hi,
If a uniformly convergent series on a finite open interval (a,b) is multiplied by a continuous function f(x) on (a,b),is the resulting series also uniformly convergent?

2. ## Re: Uniform convergence of series

You have $\displaystyle \sum_{n\ge 0} a_n = k, k \in (a,b)$ and $f(x) \to (a,b)$ and you are looking to see if $\displaystyle f(x)\sum_{n\ge 0} a_n$ is uniformly convergent? Well, I can tell you right now that sum may not be in $(a,b)$, but otherwise, you are not generating a new series. The series is still uniformly convergent. Its result is just being multiplied by a function. I am not sure I understand your question.

3. ## Re: Uniform convergence of series

Originally Posted by hedi
If a uniformly convergent series on a finite open interval (a,b) is multiplied by a continuous function f(x) on (a,b),is the resulting series also uniformly convergent?
Please define what is meant by a uniformly convergence of series. In my experience the idea of uniform convergence is applied to sequences of functions. Are these sequences partial sums of functions?

5. ## Re: Uniform convergence of series

Is the function $f(x)$ just continuous or uniformly continuous? If $f(x)$ were defined on a closed interval or if it were uniformly continuous, then the answer would be yes, the product is uniformly convergent. Otherwise, no, it is not uniformly convergent (but is still convergent).

6. ## Re: Uniform convergence of series

@Hedi, that is exactly what I said. The question is about a sequence of functions not about series.

Originally Posted by SlipEternal
Is the function $f(x)$ just continuous or uniformly continuous?
It is given that $f(x)$ is continuous on $(a,b)$ The fact that $(a,b)$ is an open set makes me suspect that this problem is looking for a counter-example. If $f(x)$ were continuous on $[a,b]$ then it is uniformly continuous. I think the statement follows easily.

7. ## Re: Uniform convergence of series

Here's a quick counterexample:

$(a,b) = \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
$S_n(x) = 1-\dfrac{1}{n}$
$f(x) = \tan x$

Let $\varepsilon = 1$. Given any $N$, we want to find $x\in (a,b)$ such that
$\left|\tan(x)\left(1-\dfrac{1}{N} - 1 \right) \right| > 1$

That is $|\tan x| > N$. We can just find $\tan x > N$. That will occur at $x = \arctan(N+1)$.