Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Why is this tagged differential equations :x) – user9413 Jun 17 '11 at 17:52
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 – 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
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 $u\in C^\infty([0,2\pi]\times]0,\infty))\,$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.