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?
Thank's in advance
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.
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).
@Hedi, that is exactly what I said. The question is about a sequence of functions not about series.
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.
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 $x = \arctan(N+1)$.