# Converse of Weierstrass M-Test: counterexample to the statement

The statement is If $$\sum_{n=1}^{\infty} f_n$$ converges uniformly on a domain say $$A$$, then there exist constants $$M_n$$ such that $$|f_n(x)| \leq M_n$$ for all $$n$$ and $$x \in A$$ and $$\sum M_n$$ converges. My intuition is that the statement is false because if it is true then the converse statement of Weierstrass M-Test is true. I know that $$\sum 1/n$$ diverges so my attempt is to produce $$\sum f_n$$ converges where $$|f_n(x)| < 1/n$$. How can I produce such a series for that?

• Just pick the constant functions $1,-1,\frac{1}{2},-\frac{1}{2},\ldots$ for the purpose. – AdditIdent Nov 29 '18 at 0:48
• Ah!! Thanks, I see what you mean. – Dong Le Nov 29 '18 at 0:50
• The more interesting (harder) question is if the series converges uniformly and absolutely, does such $M_n$ exist -- because the Weierstrass test always gives absolute convergence. Otherwise we have trivial examples. – RRL Nov 29 '18 at 0:52

## 1 Answer

For an absolutely convergent example, take $$f_n(x) = \frac{1}{n}\mathbf{1}_{(n-1,n)}$$. The sums converge to the function: $$f(x) = \frac{1}{\lceil x \rceil}$$ because summation in this case is joining piecewise constant functions, each of height $$\frac{1}{n}$$ on the interval $$(n-1,n)$$. Clearly, convergence is absolute and also uniform, but $$\sup f_n(x) = \frac{1}{n}$$ for any $$f_n$$, so the right hand side of the sum doesn't converge.

Moral of the story: you can create annoying counterexamples by partially defining a function on some domain and then setting it equal to $$0$$ everywhere else.

• Good answer to the "correct" question. A Weierstrass converse would require the series to be absolutely convergent. (+1) – RRL Nov 29 '18 at 1:05