Holomorphic surjection theorem (Picard's little theorem)

post by dkl9 · 2024-07-21T13:24:18.300Z · LW · GW · 0 comments

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Consider an entire function (complex-differentiable everywhere) $f\left(z\right)$. I will intuitively prove that certain lemmas hold on any such $f$. If $f$ is a polynomial, I can combine those lemmas with another one, showing that the holomorphic surjection theorem (more commonly, Picard's little theorem) applies. Any entire function is the limit of a sequence of polynomials, so a theorem holding for all polynomials is a compelling hint that it may also hold for all entire functions.

Pick an input point $a$. Suppose $\epsilon$ is a small positive real number, and $b$ is a unit-modulus complex number. Incrementing $a$ to $a+\epsilon b$ takes $f\left(a\right)$ to $f(a+\epsilon b)\approx f\left(a\right)+\epsilon b{f}^{\prime }\left(a\right)$. That approximation holds sith $f$ is entire, and is closer to exact for smaller $\epsilon$.

From any complex $f\left(a\right)$ — except 0 — exactly two directions of increment keep its modulus constant, which are those directions along the circle in output space centred at 0. $b$ is arbitrary, so the direction of the increment $\epsilon b{f}^{\prime }\left(a\right)$ is arbitrary. Hence, for any input $a$, there are two directions to increment it, along which to develop a contour of constant modulus. We can repeat this operation at the ends of the contour, so every such contour is either closed or infinite. The exceptions are $f\left(a\right)=0$, which form degenerate "contours" of a single point.

Every point in the input space (complex numbers) lies on some contour, either degenerate, closed, or infinite. The contours we care about are defined by the modulus of the outputs along them, which precludes intersections. So any closed contour encloses some area, within which all points are on contours either closed — fitting entirely within the enclosing contour — or degenerate.

Pick a closed contour $C$, as described above. Suppose the smallest-area contour within it is $D$, enclosing area exceeding zero. But $D$ is a closed contour, holding points within it belonging to other contours, which must enclose smaller areas. Contradiction. The smallest-area contour within $C$ has area zero.

A zero-area contour is either open or degenerate. An open contour within a closed contour must be finite. Our earlier contour construction forbade finite open contours. The smallest-area contour within $C$ is degenerate, i.e. a single point, which only arises when $0=\left|f\right(C\left)\right|=f\left(C\right)$. Any closed contour of constant modulus of $f$ contains a zero.

$f$ is entire, so it contains zero poles. By the argument principle, any closed contour of constant modulus, mapped to output space, winds around zero at least once. Thus, if $f$ has a closed contour for output modulus $k$, f maps at least once to every complex number with modulus $k$.

Any path from a closed contour to a zero within must, by the IVT, cross (in output space) every modulus less than the modulus along the contour. That $\left|f\right(z\left)\right|=k$ and $z$ is in the interior of a constant-modulus contour implies that $z$ lies on a closed contour of constant modulus $k$. So, if $f$ has a closed contour for output modulus $K$, $f$ maps at least once to every complex number with modulus $k\le K$.

Let's take as a lemma that entire functions, unless constant, output numbers with arbitrarily large moduli (Liouville's theorem). A stronger similar theorem holds for polynomials $P\left(z\right)$: that for any modulus $K$, there exists a radius $R$ past which ($\left|z\right|>R$) all outputs are bigger than $K$ ($\left|P\right(z\left)\right|>K$). Intuitive proof: visualise each ring (points $z$ with $\left|z\right|=R$) of input mapping to a sum of rings, one for each term in the polynomial. For large enough $R$, higher-degree terms dominate, such that any possible sum of the ring-vectors stays outside a circle of radius $K$.

Say $f$ is a polynomial. Pick a starting point $a$. Plug in $\left|f\right(a\left)\right|$ as $K$ for the preceding lemma. Outside some ring (radius $R$), all outputs are bigger than $f\left(a\right)$. The contour for the starting point is bound to that ring, and so must be closed. So $f$ maps at least once to every complex number with modulus $k\le \left|f\right(a\left)\right|$.

By Liouville's theorem, $\left|f\right(a\left)\right|$ can be made arbitrarily large. So polynomial $f$ maps at least once to every complex number. QED.

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