3
$\begingroup$

I need help with this problem:

Suppose that $f_n$ are continous non-negative fucntions bounded on A and let $M_n=\sup f_n$. If $\sum_{n=1}^\infty f_n$ converges uniformly on A, it follows that $\sum_{n=1}^\infty M_n$ converges? (a reciprocal of the Weierstrass $M$ test

I don't know how to solve this problem. Can you give me some help please?

$\endgroup$
  • $\begingroup$ I´m not sure but I think its key consider A open or compact $\endgroup$ – JoseSquare Feb 12 '19 at 22:32
  • $\begingroup$ I don't know what that means. $\endgroup$ – davidllerenav Feb 12 '19 at 22:35
  • $\begingroup$ My intuition tells me that if A is a compact set then the theorem might be true, but if A is an open set then I think a counterexample might be found. This is only intuition, nothing serious $\endgroup$ – JoseSquare Feb 12 '19 at 22:38
  • 1
    $\begingroup$ See the comment on top answer of math.stackexchange.com/questions/26273/… for a counterexample. $\endgroup$ – cofnmarol Feb 12 '19 at 23:09
  • $\begingroup$ But what's the meaning of compact set and open set? $\endgroup$ – davidllerenav Feb 12 '19 at 23:38
0
$\begingroup$

The comments point to a counterexample where $A$ is unbounded, and there is some speculation that the theorem is true if $A$ is compact.

For a counterexample where $A$ is compact (closed and bounded) take $A = [0,1]$ and the sequence of continuous, nonnegative and bounded functions

$$f_n(x) = \begin{cases}0, & 0 \leqslant x \leqslant \frac{1}{2^{n+1}}\\ \frac{\sin^2(2^{n+1}\pi x)}{n},& \frac{1}{2^{n+1}} < x < \frac{1}{2^{n}} \\ 0, & \frac{1}{2^{n}} \leqslant x \leqslant 1 \end{cases}$$

We have $|f_n(x)| \leqslant \frac{1}{n}$ and the supremum is attained at $x^* = \frac{3}{2}\frac{1}{2^{n+1}} \in \left(\frac{1}{2^{n+1}}, \frac{1}{2^n}\right)$ where

$$M_n = \sup_{x \in [0,1]}f_n(x) = f_n(x^*) = \frac{\sin^2(\frac{3}{2}\pi)}{n} = \frac{1}{n},$$

and $\sum M_n = \sum \frac{1}{n}$ diverges.

Nevertheless the series $\sum f_n(x)$ is uniformly convergent.

For any fixed $x$, we have $f_n(x) = 0$ if $x \in \left[\frac{1}{2},1\right]$ or if $x = \frac{1}{2^n}$ for any $n \in \mathbb{N}$. Otherwise for any fixed $x \in \left(0,\frac{1}{2}\right)$ such that $x \neq \frac{1}{2^n}$ for all $n$, we have

$$x \in \bigcup_{n=1}^\infty \left(\frac{1}{2^{n+1}}, \frac{1}{2^n} \right),$$

Since this is a union of disjoint intervals there exists one and only one integer $m(x)$ such that

$$x \in \left(\frac{1}{2^{m(x)+1}}, \frac{1}{2^{m(x)}} \right), $$

and $f_n(x) = 0 $ if $n \neq m(x)$ but $f_n(x) = \frac{1}{m(x)} \sin^2 \left(2^{m(x) +1}\pi x \right)$ if $n = m(x)$.

Thus, for any $x$ we have $f_n(x) \leqslant \frac{1}{m(x)}$.

For any $p > n$, either $m(x) \notin \{n+1,\ldots, p\}$ and

$$\sum_{k=n+1}^p f_k(x) = 0,$$

or $m(x) \in \{n+1,\ldots, p\}$ and,

$$\sum_{k=n+1}^p f_k(x) \leqslant \frac{1}{m(x)} < \frac{1}{n}$$

From here we can conclude that the series $\sum f_n(x)$ is uniformly convergent on $[0,1]$ by the uniform Cauchy criterion.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.