# Reciprocal of the Weierstrass $M$ test.

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?

• I´m not sure but I think its key consider A open or compact – JoseSquare Feb 12 '19 at 22:32
• I don't know what that means. – davidllerenav Feb 12 '19 at 22:35
• 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 – JoseSquare Feb 12 '19 at 22:38
• See the comment on top answer of math.stackexchange.com/questions/26273/… for a counterexample. – cofnmarol Feb 12 '19 at 23:09
• But what's the meaning of compact set and open set? – 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.