# upper bound on derivatives of a function defined on an arc

This is a simple question I asked in math.SE last month but unfortunately no one gives any comment. So I decided to try some luck here. You can skip examples below and read from "General setting" at the bottom.

Given a smooth arc (part of an ellipse actually) on the complex plane by

$z=\cos t + 0.5 i \sin t,\; t\in[\pi/10,\pi/5]$ ,

and a non-analytic function $f(z) = \text{Re } z$ defined on the arc. Obviously, $f(z) = g(t) := \cos t.$
Suppose we compute the "derivatives" of $f$ on the arc recursively by
$f'(z) = g'(t)/z'(t),\quad$ $f''(z) = \dfrac{df'(z)}{dt}\dfrac{1}{z'(t)},\quad$ $f'''(z) = \dfrac{df''(z)}{dt}\dfrac{1}{z'(t)},\quad \dots$

Is there an estimate on the upper bound of magnitude of $n^\text{th}$ order derivative of $f$ ? For example, can we show something like

$|f^{(n)}(z)|\leq C n! r^n$, where $C$ and $r$ are positive constants independent of $n$ ?

Note that in the case above the form of $f$ is really simple. If $f$ is more complicated, for example, $f\circ z(t) := \frac{|z'(t)|}{z'(t)}$, what can we say about $|f^{(n)}(z)|$ ?

Update: According To Fedor's answer, function $f$ in first example actually coincides with an analytic function on the arc. I need to modify the curve so that it is not easy to find an analytic function that coincides with $f$ on the curve.

New curve: Suppose the smooth arc is given by

$z(t) = [1+0.5\cos(4t)]\cos t + i [1+0.5\sin(4t)]\sin t,\quad t\in[\frac{\pi}{8},\frac{3\pi}{8}],$

and the function $f$ defined on the arc is given by

$f\circ z(t) := \frac{|z'(t)|}{z'(t)}$.

With derivatives for $f$ defined recursively as before, can we derive an upper bound for $\lvert f^{(n)}(z) \rvert$ as above ?

General setting:
Given a smooth Jordan arc parametrized by $z(t)$ on complex plane with $z'(t)\neq 0,\; t\in [0,1]$, and a smooth function $f$ defined on the arc in the sense that $f\circ z(t) \in C^\infty$. Define derivatives of $f$ recursively as above, namely, let $g(t):=f\circ z(t)$,

$f'(z):= g'(t)/z'(t), \quad f''(z):= \dfrac{df'(z(t))}{dt}\dfrac{1}{z'(t)},\quad f'''(z) = \dfrac{df''(z(t))}{dt}\dfrac{1}{z'(t)},\quad \dots$

and we ask if there is an estimate

$||f^{(n)}(z)||_\infty \leq C n! r^n$, where $C$ and $r$ are positive constants independent of $n$ ?

In addition, suppose there exists an analytic function $F$ that equals $f$ on the arc as in Fedor's answer. Can we derive the upper bound on $||f^{(n)}(z)||_\infty$ only using recursive definition for $f^{(n)}(z)$ above instead of resorting to Cauchy's formula ? Why we want to do this is because if we use Cauchy's formula, then the constant $C$ in the estimate will depend on function values of $F$ outside the arc $\gamma$, which are unknown unless an explicit expression for $F$ is derived and also $r$ must depend on the region of analyticity of $F$, which is again not so traceable. It is to be hoped that the inequality can be proved in a manner such that the dependence of $C,r$ on $f,\gamma$ can be shown.

In general no, even if $z(t)=t$. Smooth function may have arbitrary sequence of values of derivatives at a point. It is a crucial difference with analytic case.
I think, we may think that $f(z)$ is analytic and derivatives are usual. Indeed, we have $4(z-f)^2=f^2-1$, thus $3f^2-8zf+(4z^2+1)=0$, $3f=4z+\sqrt{4z^2-3}$, where we should specify the branch of the root so that this equation holds. And if we look at $f$ from this point of view, we consider usual derivatives of $f(z)$, which admit usual bounds (radius of convergence of the Taylor series in $z_0$ is the distance from $z_0$ to the closest singular point.)
• Nice answer. I think I made a mistake in my post above because your argument shows that, in this particular setting, $f(z)=\text{Re} z$ coincides with an analytic function on the arc, so $f$ is actually analytic! It seems that I do need to modify the post later so that $f$ is indeed non-analytic. – booksee May 8 '16 at 22:48
• I've modified the example so it's not trivial to find an analytic function that equals $f$ on the curve. – booksee May 9 '16 at 14:56
• It is again algebraic equation for $\cos t$. Inverse functions for analytic functions are analytic again, right? – Fedor Petrov May 9 '16 at 15:12
• Yes, I understand. Alright, since I can't easily cook up an example to my needs, we shall forget about any concrete examples, and assume we have a case where the curve is smooth and $f$ cannot be extended analytically. – booksee May 9 '16 at 16:09