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Let $f_1, f_2, \ldots$ be a sequence of continuous positive functions of $[0,1]$ and let $a_n = \sup\{ f_n(x) : x \in [0,1]\}$.

In class, we showed that if $\sum f_n$ uniformly converges on $[0,1]$, then it is not always true that $\sum a_n < \infty$.

We did this by visually constructing a sequence of continuous functions of disjoint support on $[0,1]$ such that $a_n = \frac{1}{n}$.

My question is: Does the same conclusion hold if $f_n$ is a decreasing sequence of continuous positive functions on $[0,1]$, or not?

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up vote 3 down vote accepted

You could probably take a function $f$ which is constant $=0$ except around the $\frac{1}{n}$ where it has non overlapping peaks of height $\frac{1}{n\ln(n)}$ (say for $n\geq 3$), and define for every integer $m$ a function $f_m$ which coincides with $f$ except the first $m$ peaks of $f$ (coming from the right) have been erased and replaced by $0$. I haven't carried out the details but I think this might work.

The resuling sequence has $f_m$ is a decreasing sequence of non negative functions with $\sup f_m = \frac{1}{m \ln(m)}$ so $\sum \sup f_m =+\infty$ yet if I'm not mistaken $\sum f_m$ should be continuous and uniformly convergent since $\sup \sum_{m=N}^{\infty} f_m \leq \frac{1}{\ln(N)}$.

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Thank you Olivier! For the most part, I am able to follow everything except for the part where you say $\sum f_m$ converges uniformly. Can you explain this in more detail? Thanks. –  josh Nov 6 '11 at 19:20
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