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Trying to show that the infinite sum of $f_n(x)$ converges uniformly on $[0,1]$.

For $n=1,2,\ldots$ define continuous functions on $[0,1]$ by

  • $f_n(x)=0$ on $\left[0,\frac{1}{2n+1}\right]$ or $\left[\frac{1}{2n-1},1\right]$,
  • $f_n(x)=\frac{1}{n}$ if $x=\frac{1}{2n}$,
  • $f_n(x)$ is linear on $[\frac{1}{2n+1},\frac{1}{2n}]$ and $\left[\frac{1}{2n},\frac{1}{2n-1}\right]$.

I sketched some graphs. I can see that when I plot a few of them the linear portions never overlap and that the peaks continue to decrease, however isn't the $1/n$ divergent by the Harmonic Series Theorem?

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    Misre$a$d the question. My comment's been removed. Btw, is there a way to delete comments?2012-03-07

2 Answers 2

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The point is that every $x$ is below exactly one "peak", so you are not adding the $1/n$ as in the harmonic series.

Take any $x\in[0,1]$. Then there exists a unique $n\in\mathbb{N}$ such that $ \frac1{2n+1}\leq x<\frac1{2n-1}. $ So, as Davide mentioned, the functions $\{f_n\}$ satisfy $f_k(x)f_j(x)=0$ for all $k\ne j$ and all $x\in[0,1]$. So, for any $x\in[0,1]$, for any $m\in\mathbb{N}$, $ 0\leq\sum_{k=n}^mf_k(x)\leq\frac1n. $ This shows that the convergence is uniform, as the bound for the tail of the series does not depend on $x$.

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    Yes! I finally came to that conclusion which wasn't obvious to me right away. Nice use of the Cauchy Criterion. Davide got me off to the right start, but I haven't learned of a thm about using the lim sup norm thing yet so this will do for now. Thanks for the input guys.2012-03-08
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Let $a_k(x):=\begin{cases}1&\mbox{ if }x\in \left[\frac 1{2j+1},\frac 1{2j-1}\right]\\\ 0&\mbox{ otherwise}.\end{cases}$. Fix $N\in\mathbb N$. For $m\geq 1$ we have $\sup_{x\in [0,1]}\left|\sum_{k=N}^{N+m}f_k(x)a_k(x)\right|=\max_{N\leq j\leq N+m}\sup_{x\in \left[\frac 1{2j+1},\frac 1{2j-1}\right]}f_j(x)=\max_{N\leq j\leq N+m}\frac 1j=\frac 1N,$ so the sequence $\{\sum_{k=0}^Nf_k(x)\}$ is Cauchy for the uniform norm on $[0,1]$.

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    In fact the $a_k$ are not very usefull, since I just use the fact that $f_k(x)f_j(x)=0$ if $k\neq j$ (that's the trick here).2012-03-07