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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?

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  • $\begingroup$ Just pick the constant functions $1,-1,\frac{1}{2},-\frac{1}{2},\ldots$ for the purpose. $\endgroup$ – AdditIdent Nov 29 '18 at 0:48
  • $\begingroup$ Ah!! Thanks, I see what you mean. $\endgroup$ – Dong Le Nov 29 '18 at 0:50
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    $\begingroup$ 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. $\endgroup$ – RRL Nov 29 '18 at 0:52
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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.

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

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