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?

  • $\begingroup$ 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

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.

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