Is there Cauchy-type estimate for real analytic functions? A real analytic function $f: I\subset \mathbb{R}\to \mathbb{C}$ can be expressed as a power series on a neighbourhood of each point in $I.$ (See here). I wonder whether there is a Cauchy estimate for such functions. Namely, 
$$f^{(n)}(a) \leq \frac{M n!}{r^n}$$
for $[a-r, a+r]\subset I$ and $\max_{t\in [a-r, a+r]}|f(t)|\leq M.$
Such estimate exists for "complex" analytic functions and can be derived using the Cauchy integral formula (see here).
Thank you in advanced. 
 A: You can't bound the derivatives of a real-analytic function just using
the values of the function on the real line. For example take $f(x)=\sin Ax$
where $A$ is a positive constant. Then on $\Bbb R$, $|f(x)|\le 1$, but $f'(0)=A$, which can be arbitrarily large.
A: Suppose that $f$ is analytic at $(x_0 \pm \delta)$, i.e.
$$ f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n. $$
You can extend it to a holomorphic function $\tilde{f}$ on $B_\delta(x_0)\subset \mathbb{C}$, namely
$$ \tilde{f}(z) = \sum_{n=0}^\infty a_n (z-x_0)^n. $$
Note that such function convereges because of the triangular inequality, and it is also analytic. This process is called analytic continuation.
Now you get the estimate you wanted, but $M$ is the sup over the extended $\tilde{f}$
$$f^{n}(x) \leq \frac{n!}{\delta^n} \sup_{z\in B_\delta(x_0)} |\tilde{f}(z)|. $$
Triangular inequality does not give you further information. Actually, such inequality, where $M$ is the sup over the real line, is impossible. Take for example $f(x)=\sin(x)$ and suppose that such estimate is true
$$ \sin^{2[n]}(x) = \sin(x) \leq \frac{n!}{\delta^n} \sup_{|x|\leq \delta} |\sin(x)|. $$
For $\delta$ big enough, the sup on the rhs is $1$. Altough
$$\sin(x) \leq \frac{n!}{\delta^n} \xrightarrow{\delta\to\infty} 0. $$
This contradiction and the anlyticity of $\sin$ show that you cannot prove such estimate. Note that analytic functions on $\mathbb{C}$ are proven to be unbounded, so you never get such contradiction.
