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Let $\phi$ be a rapidly decreasing test function on $\mathbf R$ and define $\tau_n$ so that $\tau_n\phi(x)=\phi(x+n)$. I would like to show that the series

$\begin{align}\sum_{n \in \mathbf Z} \tau_n \phi\end{align}$

and all its derivatives converge uniformly. I know that since $\phi$ is rapidly decreasing, there exists a $c>0$ such that $|\phi(x+n)| < c(1+|x+n|^2)^{-1}$ for all $x$ and $n$. However, I need a bound that is independent on $x$ so that I may apply the Weierstrass M-test. I have no idea where this bound should come from. Perhaps someone can shed some light on this for me.

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Where do you want to show the uniform convergence? We can try to show it on $\left[0,1\right]$, and I guess it will be enough for the uses of this result, since the function defined by the series is $1$-periodic. –  Davide Giraudo Oct 29 '11 at 22:17
    
Can it be shown that the convergence is uniform on $\mathbf R$? Otherwise, I will settle for uniform convergence on an nonempty compact $K$. –  echoone Oct 29 '11 at 22:22

1 Answer 1

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We don't have the uniform convergence on $\mathbb R$. Indeed, let $\phi(x)=e^{-x^2}$. We have that for $N\in\mathbb N$ fixed $$a_N:=\sup_{x\in\mathbb R}\left|\sum_{|n|\geq N}e^{-(x+n)^2}\right|=\sup_x\sum_{n\geq N}e^{-(x+n)^2}+\sum_{n\leq -N}e^{-(x+n)^2}\geq\sup_x\sum_{n\geq N}e^{-(x+n)^2}$$ hence $$a_N\geq\sup_x\sum_{n\geq 0}e^{-(x+N+n)^2}\geq \sup_x\sum_{n\geq 0}e^{-(x+n)^2}\geq \sum_{n\geq 0}e^{-n^2}\geq 1.$$ If the convergence were uniform, we would have $\lim_{N\to\infty}A_N=0$, which cannot be the case. Anyway, we have normal convergence on nonempty compact $K$. Indeed, we can find $A\in\mathbb N^*$ such that $K\subset\left[-A,A\right]$, and for all $n>A$, $x\in\left[-A,A\right]$, $(x+n)^2\geq (n-A)^2$ hence $\frac 1{(x+n)^2}\leq \frac 1{(n-A)^2}$. Since $\phi\in\mathcal S(\mathbb R)$, we can find $C>0$ such that for all $x\in\mathbb R$, $|\phi(x+n)|(x+n)^2\leq C$. We get $\sup_{x\in K}|\phi(n+x)|\leq \frac C{(n-A)^2}$. For $n<-A$, we have $x+n\leq A+n<0$ hence $(x+n)^2\geq (A+n)^2$ and $|\phi(x+n)|\leq \frac C{(n+A)^2}$ and we get the normal convergence on $K$. Since $(\tau_n \phi)^{(d)}\tau_n =\phi^{(d)}$ for all $d$ and $\phi^{(d)}\in\mathcal S(\mathbb R)$, a similar argument shows the normal convergence on the compacts of the derivatives.

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Very nice. I am having a bit of trouble though: when $n < -A$, then it is not true that $|\phi(x+n)|\le C/(n-A)^2$. –  echoone Oct 30 '11 at 1:18
    
Yes, I only looked at $n>A$, but for $n<-A$ we have $A-n\leq x+n\leq A+n<0$ hence $(x+n)^2\geq (A+n)^2$ and you can conclude. –  Davide Giraudo Oct 30 '11 at 9:22

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