Pointwise boundedness is uniform for a sequence of derivatives Let $f \in C^\infty ([a, b])$ be an infinitely differentiable function defined on a closed interval $[a, b]$ with the following property: for any $x \in [a, b]$ the sequence $|f^{(n)}(x)|$ is bounded, i.e.:
$$ \forall x \in [a, b] \;\; \exists C_x \in \mathbb R \;\; \forall n \geq 0 \;\; |f^{(n)}(x)| \leq C_x $$
Prove that there is a constant $C$ such that $|f^{(n)}(x)| \leq C$ for all $n \geq 0$ and $x \in [a, b]$.
Clarification: the derivatives are defined on closed intervals by taking one-side limits.

My ideas:


*

*$C_x$ is actually a function if we set:
$$ C_x := \sup_{n \geq 0} |f^{(n)}(x)| $$

*The task is to prove that this function is bounded on $[a, b]$. By contradiction?

*Suppose $C_x$ is unbounded. Then there is a sequence $(n_k, x_k)$ such that
$$ \lim_{k \to \infty} |f^{(n_k)}(x_k)| = \infty $$

*$n_k$ cannot be bounded since otherwise there would be infinitely many pairs $(n_0, x_k)$ for some $n_0$. The limit above would still be infinity for that subsequence, but for a fixed function $f^{(n_0)}$ that is impossible since it is a continuous function defined on a closed interval.

*The sequence $(n_k, x_k)$ can be chosen in such way that:

*

*$n_k$ are strictly increasing

*$x_k$ converge to some $x_0 \in [a, b]$



However, I have no idea how to use this sequence. Any advice? What direction should I look in?
 A: You can get your result by using a nice theorem sometimes called the "Pringsheim-Boas theorem". This theorem was stated by Pringsheim in 1934 but there was a gap in the proof, and Boas gave a correct proof in 1935.
To state this theorem, one needs to explain (or recall) what an analytic function of a real variable is. A function $f\in\mathcal C^\infty([a,b])$ is said to be analytic if for any $s\in [a,b]$, one can find an open interval $I_s$ with center $s$ such that $f(x)= \sum_0^\infty \frac{f^{(k)}(s)}{k!}\, (x-s)^k$ on $I_s\cap [a,b]$. 
Note that the definition of analyticity is local: the above power series may not converge to $f(x)$ for all $x\in [a,b]$, even if it is convergent for all $x\in [a,b]$.
The Pringsheim-Boas theorem reads as follows. Let $f$ be a $\mathcal C^\infty$ function on $[a,b]$. For each $s\in [a,b]$, denote by $\rho(s)$ the radius of convergence of the power series $\sum \frac{f^{(k)}(s)}{k!}\, (x-s)^k$. If there exists $\delta >0$ such that $\forall s\in [a,b]\;:\; \rho(s)\geq \delta$, then $f$ is analytic on $[a,b]$.
The prooof of this theorem is not at all trivial, and does use the Baire category theorem. Here is a link to Boas' paper: http://www.ams.org/journals/bull/1935-41-04/S0002-9904-1935-06049-5/S0002-9904-1935-06049-5.pdf
Let us see how the Pringsheim-Boas theorem can be used to get what you want.
Your assumption on $f$ clearly implies that $\rho(s)=\infty$ for every $x\in [a,b]$; so $f$ is analytic on $[a,b]$. Let $s_0$ be any point of $(a,b)$, for example $s_0=\frac{a+b}2\cdot$ let $f_0$ be the function defined on $\mathbb R$ by $$f_0(x)=\sum_{n=0}^\infty \frac{f^{(k)}(s_0)}{k!}\, (x-s_0)^k\, . $$
Being the sum of a convergent power series, this function is analytic on $\mathbb R$ (this is one of the basic results of the theory of analytic functions). Moreover, $f_0$ agrees with $f$ on a neighbourhood of $s_0$. Since $f_0$ and $f$ are both analytic, it follows (by the so-called identity principle for analytic functions) that $f_0=f$ on $[a,b]$. So we have
$$\forall x\in [a,b]\; :\; f(x)=\sum_{n=0}^\infty \frac{f^{(k)}(s_0)}{k!}\, (x-s_0)^{k}\, . $$
By the standard theory of power series, you can differentiate under the $\Sigma$. This gives for all $n\geq 0$:
\begin{eqnarray}f^{(n)}(x)&=&\sum_{k=n}^\infty \frac{k(k-1)\cdots (k-n+1)}{k!}f^{(k)}(s_0)\, (x-s_0)^{k-n}\\
&=&\sum_{k=n}^\infty\frac{f^{(k)}(s_0)}{(k-n)!}\,(x-s_0)^{k-n} .
\end{eqnarray}
Now, by assumption there is a constant $C_0=C_{s_0}$ such that $\vert f^{(k)}(s_0)\vert\leq C_0$ for all $k\geq 0$. So we have
\begin{eqnarray}\forall x\in [a,b]\;\forall n\;:\; \vert f^{(n)}(x)\vert&\leq& C_0\, \sum_{k=n}^\infty \frac{(x-s_0)^{k-n}}{(k-n)!}\\
&=& C_0\, e^{x-s_0}\\
&\leq& C_0e^{b-a}\, .
\end{eqnarray}
Since $C:=C_0e^{b-a}$ is independent of $x$ and $n$, this concludes the proof.
