I was assigned a few problems in my Honors Calculus II class, and one of them was kind of interesting to do:

Suppose that $f_{n}$ are nonnegative bounded functions on $A$ and let $M_{n} = \sup f_{n}$. If $\displaystyle\sum\limits_{n=1}^\infty f_{n}$ converges uniformly on $A$, does it follow that $\displaystyle\sum\limits_{n=1}^\infty M_{n}$ converges (a converse to the Weirstrass $M$-test)?

I know that this question has been asked before, but I'm trying not to just copy an answer off the internet and instead to come up with an example of my own to see if I can actually understand the theorems that I'm learning.

To provide a counterexample, I tried to create a function which has a diverging $\sup$, but I'm not too confident that my proof is valid. Here it goes:

$$ \text Let \ f_{n}(x) = \begin{cases} \begin{cases} \frac{1 + x}{2}, & \text{if }x \in (-1,0]\\ \frac{1 - x}{2}, & \text{if }x \in (0,1)\\ 0, & \text{elsewhere} \end{cases}, & \text{if }n \text{ even}\\ \begin{cases} x, & \text{if }x \in (0,1]\\ 2 - x, & \text{if }x \in (1,2)\\ 0, & \text{elsewhere} \end{cases}, & \text{if }n \text{ odd} \end{cases} $$

Now, $\text Let f(x) = 0$.

From this definition, I can conclude that $$ \sup\{f_{n}\} = \begin{cases} \frac{1}{2}, & \text{if }n \text{ even}\\ 1, & \text{if }n \text{ odd} \end{cases} $$

Now, to show that ${f_{n}(x)}$ is uniformly convergent, the definition of uniform convergence is used:

$$ \forall_{\epsilon > 0}\ \exists_{N}\ \text s.t.\ \forall_{x}\ \text if\ n > N, |f(x) - f_{n}(x)| < \epsilon $$

Since $f_{n}(x)$ is strictly nonnegative and $f(x) = 0$, $|f(x) - f_{n}(x)| = f_{n}(x)$. By definition, $\epsilon > 0$, and since $f_{n}(x) = 0$ for $x \geq 2, f(x) - f_{n}(x) = 0 < \epsilon$ for $x \geq 2$. Therefore there exists a $N$ (namely, $N = 1$) which proves that the sequence is uniformly convergent.

Since $\lim_{n\to\infty} f_{n} \neq 0$, by the Limit Test, the infinite sum $\displaystyle\sum\limits_{n=1}^\infty \sup{f_{n}}$ diverges, which disproves the converse of the Weierstrass $M$-Test. $\blacksquare$

This is the first time I've actually used LaTeX, so I'm sorry for the way it looks. Is there anything that I can do to make this proof better (or even valid, if it's wrong), or is it fine the way it is?

This might be a bit of a long question...

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  • Well, it's about time I took a break from SO ;) – Blender Mar 10 '11 at 22:54
  • Hmm, why would you do it "the hard way" when it can be done much easier? – JT_NL Mar 10 '11 at 22:55
  • $f_n$ is strictly non-negative seems to be false. – Aryabhata Mar 10 '11 at 22:57
  • @Jonas, would you care to elaborate? I would love to know an easier way. @Moron, I think changing $1-x$ to $2-x$ fixes it, right? – Blender Mar 10 '11 at 22:58
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    @Blender: I didn't go through the whole thing, but you seem to be going about the wrong way. To prove uniform convergence, you will have to use some test _other_ than the M-Test. After all, you are trying to disprove the converse... – Aryabhata Mar 10 '11 at 23:04
  • Thanks @Moron. I looked over the proof after an 1 hour, and I see what you mean. I have the Weierstrass $M$-Test and the definition of uniform convergence completely mixed up, so I think the changes that I made would remove the dependency upon the Weierstrass $M$-Test. Thanks for your help! – Blender Mar 11 '11 at 01:06
  • Why do you say that $f_n(x)=0$ for $x\ge 1$? Isn't $f_n(x)=2-x\neq0$ if $n$ is odd and $x\in(1,2)$? – joriki Mar 11 '11 at 02:25
  • There are two things that I do not get about your counterexample. (1) The original problem is about convergence of a series of functions, but your example is a sequence, for which the M-test is meaningless. (2) You say that the sequence $f_n$ converges uniformly to the constant $0$. This is not so: if $-1 – Julián Aguirre Mar 11 '11 at 15:14
  • Hmm, that's not very good then. Would defining $f(x) = f_{n}(x)$ fix the convergence problem, as $f(x) - f_{n}(x) = 0$, for all $x$ and $n$? – Blender Mar 11 '11 at 16:50
  • @Blender: NO. The limit cannot depend on n! I get the impression that you do not grasp the concept of convergence of a series or sequence of functions. – Julián Aguirre Mar 11 '11 at 17:37
  • Ignore that reference to $n$ (I'm not sure why I even wrote that). I'm beginning to think that my function is unusable for proving the converse false... I'll try and work with what @yoyo posted, but I have no idea what the *chi* with the subscript means. – Blender Mar 11 '11 at 17:50

1 Answers1


let $f_n(x):\mathbb{R}\to\mathbb{R}$ be $\frac{1}{n}\chi_{(n-1,n)}$. then $\sum f_n$ converges uniformly but $\sum M_n=\sum 1/n$ diverges. something like this would work (not trying to give an answer away, if this is valid, but trying to show along what lines you should be thinking)

(for a set $A\subseteq\mathbb{R}$, $\chi_A(x)=1$ if $x\in A$, $\chi_A(x)=0$ if $x\not\in A$. sometimes called the indicator function of the set $A$, also denoted by $1_A$)
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