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According to my notes, the function

$$u(x,t)=\sum_{n=0}^{\infty} e^{(1-n^2)t}\cos{nx}$$ should be continuous on $(0,\pi) \times (0,\infty)$, but I'm unable to prove it.

We have

$$|u(x+h_1,t+h_2)-u(x,t)|\le \sum_{n=0}^{\infty} e^{(1-n^2)t}|e^{(1-n^2)h_2}\cos{n(x+h_1)}-\cos{nx}|,$$

but I can't find a way to bound this as $(h_1,h_2)\to (0,0)$. Maybe I'm missing some useful inequality ?

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Why is this tagged differential equations :x) –  user9413 Jun 17 '11 at 17:52
3  
You should be able to show that the sequence converges uniformly on $(0, \pi) \times (\epsilon, \infty)$ for all $\epsilon > 0$ using the fact that $e^{(1-n^2)t}$ eventually becomes very small. See en.wikipedia.org/wiki/Uniform_convergence –  Qiaochu Yuan Jun 17 '11 at 18:08
    
Just a quick hint: split the series of the differences into $\sum_{n=0}^{N} + \sum_{n=N+1}^{\infty}$, where $N = N(t,x,\epsilon)$, so that the second sum is guaranteed to be $< \epsilon / 2$ independent of $h_1,h_2$. Then you can choose $\delta = \delta(\epsilon,t,x,N)$ such that the first term is also $< \epsilon / 2$ whenever $|h_1| + |h_2| < \delta$. –  Willie Wong Jun 17 '11 at 18:34

1 Answer 1

up vote 2 down vote accepted

It is enough to consider differences on $x$ and $t$ separately: $|u(x+h_1,t+h_2)-u(x,t)|\le |u(x+h_1,t+h_2)-u(x+h_1,t)|+|u(x+h_1,t)-u(x,t)|$. Оr it is possible to prove uniform convergence of partial derivatives on $[0,2\pi]\times[\varepsilon,\infty]$ for any fixed $\varepsilon>0$ and use corresponding theorem to obtain that $u\in C^\infty([0,2\pi]\times]0,\infty))\,$.

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Thank you, Andrew, Qiaochu Yuan, and Robert Israel. Using uniform convergence is a very good idea. –  Klaus Jun 17 '11 at 18:34

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