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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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Write in Tex please –  Riccardo.Alestra Mar 7 '12 at 9:03
3  
What can you say about $\| \sum_{k=n+1}^m f_k \|_{C([0,1])}$ for large $m$, $n$? –  martini Mar 7 '12 at 9:25
    
I see that I am not sure why I mentioned the harmonic series since these "peaks" aren't actually being added to each other--they actually occur on different neighborhoods each time. Back to the drawing board. –  cap Mar 7 '12 at 9:32
    
Misread the question. My comment's been removed. Btw, is there a way to delete comments? –  Patrick Mar 7 '12 at 15:36

2 Answers 2

up vote 1 down vote accepted

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. –  cap Mar 8 '12 at 8:08

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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What was the point of introducing the function a? Maybe that's a form of Cauchy I have yet to see. That was clever, is that some sort of common analysis trick? –  cap Mar 7 '12 at 20:49
    
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). –  Davide Giraudo Mar 7 '12 at 21:41

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