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I try to understand a demonstration from a book, but I have a problem with a line. We have the series $$u(x,t) = \sum_{k=0}^\infty \frac{g^{(k)}(t)}{(2k)!}x^{2k} \qquad (*)$$ where $$g(t) = \left\{ \begin{array}{ll} \exp\left(-\frac{1}{t^2}\right) & \mbox{if } t > 0 \\ 0 & \mbox{if } t \leq 0 \end{array} \right. $$ and we try to prove that it's well defined for $x$ and $t$ real, that we can take the derivatives, and so on. We have already proved that when $t > 0$, $$\left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \leq \frac{|x|^{2k}}{k!(\theta t)^k}\exp\left[-\frac{1}{2t^2}\right]$$ for some $\theta > 0$. We then have directly $$\sum_{k=0}^\infty \left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \leq \exp\left[\frac{|x|^2}{\theta t} - \frac{1}{2t^2}\right]$$ So they say that, by comparison, our series $(*)$ converges and that $\lim_{t\rightarrow 0}u(x,t) = 0$, I agree on that.

Here is the point I don't understand. It's said :

"The series $(*)$ as a power series in $x$ is majorised (they say that $f$ is majorised by $F$ if the derivatives of $f$ in $0$ are less than the corresponding derivatives of $F$ in $0$) by the power series for $$U(x,t) = \left\{ \begin{array}{ll} \exp\left(-\frac{x^2}{\theta t} - \frac{1}{2t^2}\right) & \mbox{if } t > 0 \\ 0 & \mbox{if } t \leq 0 \end{array} \right.$$ Since $U(x,t)$ is bounded uniformly for bounded complex $x$ and all real $t$, the series $(*)$ converges uniformly in $(x,t)$ for bounded $x$ and real $t$, and the same holds for the series obtained by the term by term $x$-differentiations."

I actually don't understand what they mean when they say that $U(x,t)$ is bounded uniformly. Do they mean that the power series of $U$ is uniformly bounded? But which series? The one with respect to $x$ ? And even with that, I don't understand how we can conclude that $(*)$ converges uniformly...

Don't hesitate to ask if something is unclear in what I said!

I hope you could help me,

Thank you in advance :)

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I believe a uniform bound on $U$ simply means there exists $C > 0$ such that for all $x \in (-1,1),~t \in \mathbb{R}$ we have $|U(x,t)| < C$ (in vector space terminology, its supremum-norm is finite).

Can you see why this implies the convergence of the power series?

EDIT: Okay, so let $B\subset\mathbb{C}$ be bounded, consider the domain $D:=B\times\mathbb{R}$. You want to show that for any $\varepsilon>0$ there is $N\in\mathbb{N}$ such that for all $n>N$: \begin{equation}\sup\limits_{(x,t)\in D}|u(x,t)-u_n(x,t)|<\varepsilon.\end{equation} We know that for any $(x,t)\in D$, \begin{align} |u(x,t) - u_n(x,t)| &= \left|\sum\limits_{k=n}^\infty \frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \\ &\leq \sum\limits_{k=n}^\infty \left|\frac{g^{(k)}(t)}{(2k)!}x^{2k}\right| \end{align} Can you take it from there?

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That's a possibility... I find it weird to say "uniform" since a bound is always uniform (saying that $U(x,t)$ is bounded for some $(x,t)$ is not very interesting). Actually, if we prove that the series of U converges uniformly, we are done. Is it true that a series which converges and whose limit is bounded necessarily converges uniformly? – Nicolas May 21 '13 at 14:03
No that's not true... :( – Nicolas May 21 '13 at 14:13
Yes I know what uniform and pointwise convergences are. Here the series $(*)$ converges (pointwise convergence), but I can't understand why the convergence is uniform. It is "majorised" by a sequence whose limit is bounded, I don't see why this implies that we have a uniform convergence... – Nicolas May 21 '13 at 14:26
I found the "uniform" weird because, for me, saying that the function was bounded for $x$ bounded and $t$ real wound have been enough (a bound on a certain space is always uniform, we never talk about a bound which is not uniform!) – Nicolas May 21 '13 at 14:32
Yes but how this $\sup\limits_{\substack{x \in [-1,1] \\ t \in \mathbb{R}}} | u(x,t) - u_n(x,t) | \rightarrow 0$ can be proved? It's supposed to be in the sentence "Since $U(x,t)$ is bounded uniformly for bounded complex $x$ and all real $t$, the series $(∗)$ converges uniformly in $(x,t)$ for bounded $x$ and real $t$" – Nicolas May 21 '13 at 14:37

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