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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.

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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.

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  • $\begingroup$ Thank you. Clear counterexample $\endgroup$ – A. PI Feb 22 '18 at 12:47
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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.

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  • $\begingroup$ Thank you for this detailed answer $\endgroup$ – A. PI Feb 22 '18 at 12:48
  • $\begingroup$ you can upvote if you liked it! $\endgroup$ – Tommaso Seneci Feb 22 '18 at 14:29

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