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Hadamard's three circle theorem is given as follows:

$$A(a,b)=\{z:a<|z|< b\},$$ and $f$ is a holomorphic function in this annulus $A$.

Let $$M(r) = \max_{|z|=r}|f(z)| $$for $a < r < b$. Then

$$\log\left(\frac ba\right)\log(M(r)) \le\ \log\left(\frac br\right)\log(M(a)) + \log\left(\frac ra\right)\log(M(b))$$

For which non-constant $f$ does equality hold in this inequality?

  • 1
    According to Wikipedia (http://en.wikipedia.org/wiki/Hadamard_three-circle_theorem), $f=cz^\lambda$ for some $c\in\mathbb C$, $\lambda\in\mathbb Z$.2012-12-12

2 Answers 2

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If $f=cz^\lambda$ for some $c\in\mathbb C$ and $\lambda\in\mathbb Z$, then $f$ is holomorphic on $\mathbb C-\{0\}$, and $M(r)=|c|r^\lambda$. Thus $\log M(r)=\log c+\lambda \log r$, and

$$\log\frac br\log M(a)+\log\frac ra\log M(b)=\log\frac br\left(\log c+\lambda \log a\right)+\log\frac ra\left(\log c+\lambda \log b\right)$$ $$=(\log b\log c-\log r\log c+\lambda\log b\log a-\lambda\log r\log a)$$ $$\qquad+\ (\log r\log c-\log a\log c+\lambda\log r\log b-\lambda\log a\log b)$$ $$=\log b\log c-\lambda\log r\log a-\log a\log c+\lambda\log r\log b$$ $$=\log\frac ba\left(\log c+\lambda \log r\right)=\log\frac ba\log M(r).$$

Thus the inequality is saturated. Conversely, if $\log\frac br\log M(a)+\log\frac ra\log M(b)=\log\frac ba\log M(r)$ for all $a

$$\log r\log\frac{M(b)}{M(a)}+(\log b\log M(a)-\log a\log M(b))=\log\frac ba\log M(r)$$

so if we let $\lambda\log\frac ba=\log\frac{M(b)}{M(a)}$ and $c'\log\frac ba=\log b\log M(a)-\log a\log M(b)$, then

$$\lambda\log r+c'=\log M(r)\Rightarrow M(r)=cr^\lambda$$

where $c=e^{c'}$. Now since $M(r)=\max_{|z|=r}|f(z)|$, for each point $r_0$ there is a corresponding point $z_0$ such that $|f(z_0)|=M(r_0)$. Since $f$ is continuous on the circle of radius $r_0$, $\frac d{d\theta}|f(z)|=0$ at $z_0$. Also, $\frac d{dr}|f(z)|=M'(r)$. Thus, $\frac d{dz}f(z)=\psi z/|z|M'(|z|)=c\psi\lambda z^{\lambda-1}$ at $z_0$ (where $|\psi|=1$) by the Cauchy-Riemann relations (which is to say, the "full" derivative of $f$ at $z_0$ is consistent with the derivative in the radial direction being $M'(|z_0|)$.

The $|f(z_0)|\leq M(|z_0|)$ constraint is strong enough to constrain the higher derivatives in the same manner, so that in fact $f(z)$ looks like $g(z)=\alpha z^\lambda$ (where $\alpha=c\psi$) to all orders. To show this, note that we have here a tight inequality on "one side" of $f$ (the outer radial direction) against another analytic function $g$ in some neighborhood of $z_0$ - we are going to show that any perturbation of $f$ will push up against this wall somewhere.

If the derivatives of $f$ and $g$ do not coincide, we can zoom in close enough that the lowest different derivative is the dominant factor in the difference $f-g$. Then $f(z+z_0)\approx g(z+z_0)+dz^n+O(z^{n+1})$ for some $n\in\Bbb N$ and $d\in\Bbb C$. We choose a direction $\phi$ such that $d\phi^n$ has the same phase as $g(z_0)$, and then taking $z=\epsilon\phi$ for sufficiently small $\epsilon$ yields $|f(z+z_0)|\approx|g(z+z_0)|+|d||z|^n+O(|z|^{n+1})$, which violates the bound $|f(z+z_0)|\le |g(z+z_0)|$.

Thus, $f(z)=\alpha z^\lambda$, but in analytically continuing around the pole at zero, one may find that $f$ does not match up, i.e. a branch cut is necessary. This contradicts the assumption that $f$ is holomorphic on $A$, so this constrains the set of possible $f$'s satisfying the criteria from $\{cz^\lambda:c,\lambda\in\mathbb C\}$ to $\{cz^\lambda:c\in\mathbb C,\lambda\in\mathbb Z\}$.

  • 0
    How can you get $\frac{d}{dr}|f(z)| = M'(r)?$2015-10-10
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    @user124697 Because $|f(z)|\le M(|z|)$ for all $z\in A$, and this inequality is saturated at $z=z_0$. Basically it is an application of [Fermat's theorem](https://en.wikipedia.org/wiki/Fermat's_theorem_%28stationary_points%29) to the local minimum of the function $M(|z|)-|f(z)|$ at $z_0$.2015-10-10
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    Could you please explain how do you get $f'(z)=\psi\frac{z}{|z|}M'(|z|)$?2016-01-21
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    @pozio Because we have the directional derivatives in two directions for one component of an analytic function, the C-R relations allow us to extrapolate to the full derivative at $z_0$. The listed derivative is the unique complex number that satisfies the known values of $\frac d{d\theta}|f(z)|$ and $\frac d{dr}|f(z)|$.2016-01-22
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    Could you explain more in detail for higher derivatives? And it seems that your result about f(z) being the form αz^λ only holds for z_0. Does it hold for other z in the punctured plane as well?2017-08-20
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    @Keith If $f(z)$ has all the same derivatives as $\alpha z^\lambda$ at $z_0$, then they are equal everywhere in the domain, because $f$ is analytic (this follows from the definition of analyticity).2017-08-20
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    But z_0 is only a point. It is not some open set or a sequence having a limit point in the domain.2017-08-20
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    @Keith It doesn't matter. To say a function is analytic at $z_0$ means exactly that it equals its Taylor series about $z_0$ (at a finite distance away from $z_0$). Meaning, if all derivatives of $f$ match derivatives of $\alpha z^\lambda$ at a single point, then they are equal as functions. (Compare with the real case, where this does not hold for smooth functions but still holds for real analytic functions, again by definition.)2017-08-22
  • 0
    Oh I understand. Thank you for your explanation. Could you explain more or give a hint about higher derivatives? I think I can understand completely if the higher derivative part is handled.2017-08-22
  • 0
    @Keith I added a slightly sketchy argument of the higher derivatives stuff in the answer.2017-08-22
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Let $\lambda=\frac{\log(b/r)}{\log(b/a)}$. Then $1-\lambda=\frac{\log(r/a)}{\log(b/a)}$. Dividing both sides of your equation by $\log(b/a)$ gives:

$$\log(M(r))\leq \lambda \log(M(a))+(1-\lambda)\log(M(b)).$$

Notice that $a^{\lambda}b^{1-\lambda}=\exp(\lambda\log(a)+(1-\lambda)\log(b))=r$ (verify this!). Thus,

$$\log(M(a^\lambda b^{1-\lambda})\leq \lambda\log(M(a))+(1-\lambda)\log(M(b))$$

which is saying that $M$ is log-convex, in that $\log(M(\exp(z))$ is convex in $z$. It is not hard to show that the only times you get strict equality for convex functions is iff the function is affine (linear): $\log(M(\exp(z))=Az+B$. Thus, $M(\exp(z))=Ce^{Az}$, or $M(z)=Cz^A$

Addendum: to show that all strictly equal convex functions are affine, write:

$f''(z)=\lim_{h\rightarrow 0}\frac{f(z+h)+f(z-h)-2f(z)}{h^2}=\lim_{h\rightarrow 0}\frac{2f(z)-2f(z)}{h^2}=0$

where we used $z=\frac{1}{2}(z+h)+\frac{1}{2}(z-h)$ and equal convexity.