Best estimate using Cauchy integral formula: why is a circle the optimal path?

I once encountered this question from Ahlfors' Complex Analysis.

An analytic function $f$ has the property that for $|z|<1$, $|f(z)|\leq \frac{1}{1-|z|}$. Find the best estimate of $|f^{(n)}(0)|$ that Cauchy's formula will yield.

I solved it by finding an estimate via the Cauchy integral formula over a circle of general radius and then optimized the result by differentiating it with respect to the radius. I later wondered, why should we expect the optimal path to be a circle? I applied the Euler-Lagrange Equations to this problem and got a completely untractable O.D.E. I expect that a circle is the optimal path, but is there a concrete reason why?

Edit

Here is my application of the Euler-Lagrange equations:

Suppose we want to integrate on a path $\gamma(t)=r(t)e^{i\theta(t)}$ for $0\leq t\leq 1$ with the stipulation the $\gamma(0)=\gamma(1)$ and $0<r(t)<1$. This leaves us with $|f(z)|<\frac{1}{1-r(t)}$, and $|dz|=|r'(t)+ir(t)\theta'(t)|dt.$ Then the problem is to minimize $$|f^{(n)}(0)|\leq\frac{n!}{2\pi}\int_0^1\frac{1}{r^{n+1}(t)(1-r(t))}|r'(t)+ir\theta'(t)|dt.$$

It is sufficient that we optimize the integrand. Let the integrand be written

$$F(r,\theta,r',\theta')=\frac{\sqrt{r'(t)^2+r(t)^2\theta'(t)^2}}{r^{n+1}(t)(1-r(t))}.$$

I simply apply the variational derivative: $$\frac{\partial F}{\partial r}-\frac{d}{dt}\frac{\partial F}{\partial r'}=0,$$ $$\frac{d}{dt}\frac{\partial F}{\partial \theta'}=0.$$

As much as I hate to spoil the joy of working out the differential equations on your own, I have computed them form you:

$$\frac{r(t)^{-n} \theta '(t) \left(r(t) \theta '(t) \left(r(t) \left(r''(t)-n \theta '(t)^2\right)+(n+1) r(t)^2 \theta '(t)^2-r''(t)\right)+(n r(t)-n+1) r'(t)^2 \theta '(t)-(r(t)-1) r(t) r'(t) \theta ''(t)\right)}{(r(t)-1)^2 \left(r'(t)^2+r(t)^2 \theta '(t)^2\right)^{3/2}}=0,$$ $$-\frac{r(t)^{-n} r'(t) \left(r(t) \theta '(t) \left(r(t) \left(r''(t)-n \theta '(t)^2\right)+(n+1) r(t)^2 \theta '(t)^2-r''(t)\right)+(n r(t)-n+1) r'(t)^2 \theta '(t)-(r(t)-1) r(t) r'(t) \theta ''(t)\right)}{(r(t)-1)^2 \left(r'(t)^2+r(t)^2 \theta '(t)^2\right)^{3/2}}=0$$

It is not too hard to see that these reduce to equations solved by the result for the circle assuming that $r'(t)=0$ and $\theta'(t)=\text{const}$. Thus, the circle I initially found is a stationary point, but it may not be unique. I am stymied at the prospect of finding other solutions.

Solution 1:

The idea here is that we're trying to use an estimate like $$ |f^{(n)}(z)| \;\leq\; \frac{n!}{2\pi}\oint_\gamma \frac{|dz|}{|z|^{n+1}(1-|z|)} $$ for a closed curve $\gamma$ in the unit disk that has winding number one around the origin.

Let $m$ be the minimum value of $\dfrac{1}{r^n(1-r)}$ for $0<r<1$. Then $$ \oint_\gamma\frac{|dz|}{|z|^{n+1}(1-|z|)} \;\geq\; \oint_\gamma \frac{m\,|dz|}{|z|} \;\geq\; m \left|\oint_\gamma \frac{dz}{z}\right| \;=\; 2\pi m $$ for any such curve $\gamma$, and this value is achieved if $\gamma$ is a circle whose radius is the minimizing value of $r$. (Indeed, the circle is the only curve that achieves this value, since the second inequality is only an equality when $\gamma$ is a circle centered at the origin.)