The Identity Theorem, the Maximum Principle and the Open Mapping Theorem

If $a\in L(h)$, one of the neighbourhoods appearing in the definition of $L(h)$ is contained in $L(h)$, so $L(h)$ is open in $U$; this also covers $L(h)=\mathrm{\varnothing }$.

Let $b\in U\setminus L(h)$. If $h(b)\mathrm{\ne }0$, [L2] gives a neighbourhood on which $h$ is nonzero. If $h(b)=0$, then [L1] and $b\mathrm{\notin }L(h)$ make the order finite, so $h(z)=(z-b{)}^{m}g(z)$ near $b$ with $g(b)\mathrm{\ne }0$; after shrinking by [L2], $g$ is nowhere zero there, and $b$ is the only zero. In either case a neighbourhood of $b$ contains no point of $L(h)$, so $U\setminus L(h)$ is open.

Thus $L(h)$ is open and its complement in $U$ is open, so $L(h)$ is both open and closed in $U$.

The function $h$ vanishes at points arbitrarily close to $a$. If it had finite order there, [L2] would give $h(z)=(z-a{)}^{m}q(z)$ with $q(a)\mathrm{\ne }0$, and [L3] would make $q$ nonzero on a smaller neighbourhood; then $h$ would have no zeros there other than possibly $a$, contrary to accumulation. Hence $h$ has infinite order at $a$, so [L2] makes it vanish on a neighbourhood of $a$.

By [L1], the locally zero locus of $h$ is open and closed in $\mathrm{\Omega }$; it is nonempty by step 1.1. Since $\mathrm{\Omega }$ is connected, that locus is all of $\mathrm{\Omega }$.

Therefore $h(z)=0$ for every $z\in \mathrm{\Omega }$, which means $f(z)=g(z)$ throughout $\mathrm{\Omega }$.

The function $f$ cannot vanish on any neighbourhood of $a$, for otherwise its zero set would have the interior point $a$ as an accumulation point and [L1], applied to $f$ and the zero function, would make $f$ identically zero on $\mathrm{\Omega }$.

By step 1.1 and [L2], the order of $f$ at $a$ is finite, so $f(z)=(z-a{)}^{m}g(z)$ near $a$ with $g(a)\mathrm{\ne }0$. By [L3], after shrinking the neighbourhood, $g$ is nowhere zero there; since $f(a)=0$, the finite order $m$ is positive, and $a$ is the only zero of $f$ in that neighbourhood.

The zero $a$ was arbitrary, so every zero of $f$ is isolated; if $f$ has no zeros, the conclusion is vacuous.

By the holomorphic algebra laws in the given facts, $H(\mathrm{\Omega })$ contains the constant functions and is closed under pointwise addition, subtraction, and multiplication; [L2] and the field laws therefore make it a commutative subring of ${\mathbb{C}}^{\mathrm{\Omega }}$.

Since $\mathrm{\Omega }$ is nonempty and $0\mathrm{\ne }1$ in $\mathbb{C}$, the constant zero and one functions take different values at any point of $\mathrm{\Omega }$ and are distinct.

Suppose $fg$ is the zero function and $f$ is not the zero function. Choose $a\in \mathrm{\Omega }$ with $f(a)\mathrm{\ne }0$. The given holomorphic algebra fact makes $f$ nonzero on a neighbourhood of $a$, so $g$ vanishes there; [L1] then makes $g$ the zero function on $\mathrm{\Omega }$. Thus a zero product has a zero factor.

Steps 1.1, 1.2, and 1.3 verify all clauses of [L3], so $H(\mathrm{\Omega })$ is an integral domain.

By the star-shapedness in the Given and [L1], $h^{\mathrm{\prime }}\mathrm{/}h$ has a holomorphic primitive $K$ on the disc. Put $H(z):=K(z)-K(a)$; then $H(a)=0$ and $H^{\mathrm{\prime }}=h^{\mathrm{\prime }}\mathrm{/}h$.

The complex product and chain rules give $(h\exp (-H){)}^{\mathrm{\prime }}=h^{\mathrm{\prime }}\exp (-H)-h{H}^{\mathrm{\prime }}\exp (-H)=0$, so, the disc being a complex domain by [L4], [L2] makes $h\exp (-H)$ constant; its value at $a$ is $h(a)$.

Since $h(a)\mathrm{\ne }0$, choose by [L3] a complex number $c$ with $\exp c=h(a)$, and set $L:=H+c$.

The exponential addition law and step 2.1 give $\exp L=\exp H\exp c=\exp Hh(a)=h$ throughout the disc, so $L$ is the required holomorphic logarithm.

Take from [L1] a holomorphic function $L$ with $\exp L=h$.

Because $m\ge 1$, division by $m$ is defined. Put $q:=\exp (L\mathrm{/}m)$. Repeated use of [L2] gives $q^m=\exp L=h$, and $q$ is holomorphic; for $m=1$ this construction gives $q=h$.

For the prescribed value, both $q(a)$ and $\xi$ are nonzero and have $m$th power $h(a)$. Thus $c:=\xi \mathrm{/}q(a)$ satisfies $c^m=1$, and $\tilde{q}:=cq$ is holomorphic with ${\tilde{q}}^m=h$ and $\tilde{q}(a)=\xi$.

By [L1], $\det Df(a)=\mathrm{\mid }{f}^{\mathrm{\prime }}(a){\mathrm{\mid }}^2>0$, so the real derivative $Df(a)$ is invertible.

Apply [L2] to the underlying smooth real map: there are open neighbourhoods $V_0$ of $a$ and $W_0$ of $f(a)$ such that $f{\mathrm{\mid }}_{V_0}:V_0\to W_0$ is bijective with a $C^1$ inverse $g$.

At every $w\in W_0$, the derivative $Df(g(w))$ is multiplication by $f^{\mathrm{\prime }}(g(w))$ by [L3], and it is invertible by the inverse-function construction. Hence $Dg(w)=Df(g(w){)}^{-1}$ is multiplication by $1\mathrm{/}{f}^{\mathrm{\prime }}(g(w))$; [L3] makes $g$ complex differentiable there with the displayed derivative.

Choose a disc $V$ centred at $a$ with closure contained in $V_0$, and put $W:=f[V]$. The disc $V$ is nonempty and open, and the triangle inequality keeps the segment between any two of its points inside it, so [L5] makes $V$ connected and hence a complex domain. The homeomorphism $f{\mathrm{\mid }}_{V_0}$ makes $W$ open, while [L4] makes it connected as the continuous image of $V$. Thus $V$ and $W$ are complex domains, and steps 2.1 and 3.1 show that $f{\mathrm{\mid }}_V$ and its inverse are holomorphic. Hence $f{\mathrm{\mid }}_V:V\to W$ is biholomorphic.

Apply [L1] to $f-f(a)$: on a neighbourhood of $a$ one has $f(z)-f(a)=(z-a{)}^{m}h(z)$, where $h$ is holomorphic and $h(a)\mathrm{\ne }0$.

Since $h(a)\mathrm{\ne }0$, [L4] permits shrinking to a disc on which $h$ is nowhere zero. By [L2] there is a holomorphic $q$ on that disc with $q^m=h$.

Define $\varphi (z):=(z-a)q(z)$. Then $\varphi (a)=0$ and the product rule gives ${\varphi }^{\mathrm{\prime }}(a)=q(a)\mathrm{\ne }0$, so [L3] makes $\varphi$ biholomorphic after one further shrinking around $a$.

On that final neighbourhood, steps 1.1 and 2.1 give $f(z)-f(a)=(z-a{)}^{m}q(z{)}^m=\varphi (z{)}^m$, which is the required normal form.

Take a complex domain $V_0$ and biholomorphic coordinate $\varphi$ from [L1]. Since $N$ is a neighbourhood of $a$, choose an open set $O$ with $a\in O\subseteq N$. The set $\varphi [V_0\cap O]$ is open and contains $0$, so choose $\rho >0$ with $D(0,\rho )\subseteq \varphi [V_0\cap O]$ and put $V:={\varphi }^{-1}[D(0,\rho )]\subseteq N$.

If $0<\mathrm{\mid }w-f(a)\mathrm{\mid }<{\rho }^m$, then [L2] gives exactly $m$ distinct roots $u$ of $u^m=w-f(a)$, and each satisfies $\mathrm{\mid }u\mathrm{\mid }=\mathrm{\mid }w-f(a){\mathrm{\mid }}^{1\mathrm{/}m}<\rho$.

Since $\varphi$ is bijective, its inverse transports those roots to exactly $m$ distinct points $z\in V$ satisfying $f(z)=w$. At $w=f(a)$, [L2] says the only coordinate root is $0$, so the only point is $a={\varphi }^{-1}(0)$, and the normal form records multiplicity $m$.

Thus every noncentral value in the stated target disc has exactly $m$ distinct preimages in $V$, while the central value has the one preimage of multiplicity $m$.

The function $f$ is not constant on any neighbourhood of any $a\in \mathrm{\Omega }$: if it were constant on one, [L1] would make it constant on the connected domain $\mathrm{\Omega }$.

Fix $a\in O$. Shrink the neighbourhood in [L2] so that it lies in $O$. The local multiplicity conclusion [L3], including its central value, gives a disc about $f(a)$ contained in the image of that neighbourhood and therefore in $f[O]$.

Every point of $f[O]$ is therefore interior. Hence $f[O]$ is open; when $O=\mathrm{\varnothing }$, its image is empty and the same conclusion holds.

Suppose $f$ is nonconstant. Choose an open disc $D$ about $a$ contained in $V$. By [L1], $f[D]$ is an open set containing $f(a)$, so it contains a disc $D(f(a),\rho )$ for some $\rho >0$.

If $f(a)\mathrm{\ne }0$, the point $w=(1+t)f(a)$ lies in $D(f(a),\rho )$ and has $\mathrm{\mid }w\mathrm{\mid }>\mathrm{\mid }f(a)\mathrm{\mid }$ for sufficiently small $t>0$. If $f(a)=0$, any nonzero $w$ with $\mathrm{\mid }w\mathrm{\mid }<\rho$ has larger modulus. Thus in either case $f[D]$ contains a value whose modulus is greater than $\mathrm{\mid }f(a)\mathrm{\mid }$.

Step 2.1 contradicts the local maximum on $V$. Hence $f$ cannot be nonconstant and must be constant on $\mathrm{\Omega }$.

Suppose $f$ is nonconstant. For a disc $D$ about $a$ contained in $V$, [L1] makes $f[D]$ an open neighbourhood of $f(a)$, so $D(f(a),\rho )\subseteq f[D]$ for some $\rho >0$.

The point $f(a)+\rho \mathrm{/}2$ belongs to that target disc and has real part $\mathrm{Re}f(a)+\rho \mathrm{/}2>\mathrm{Re}f(a)$, so some point of $D$ violates the assumed local maximum.

The contradiction in step 2.1 rules out nonconstancy, so $f$ is constant on $\mathrm{\Omega }$.

The closure $\bar{\mathrm{\Omega }}$ is nonempty, closed, and bounded in the Euclidean plane, hence compact by [L2]. The reverse triangle inequality and continuity of $f$ make $\mathrm{\mid }f\mathrm{\mid }$ continuous there, so [L3] gives a maximizer $z_0\in \bar{\mathrm{\Omega }}$.

If $z_0\in \mathrm{\Omega }$, then $\mathrm{\mid }f\mathrm{\mid }$ has an interior local maximum, and [L1] makes $f$ constant on $\mathrm{\Omega }$.

If $z_0\mathrm{\notin }\mathrm{\Omega }$, then $z_0\in \mathrm{\partial }\mathrm{\Omega }$. In the remaining branch $f$ is constant on $\mathrm{\Omega }$ by step 2.1 and hence on $\bar{\mathrm{\Omega }}$ by continuity. The boundary is nonempty: otherwise the nonempty open set $\mathrm{\Omega }$ would also be closed in the connected plane [L4] and therefore equal the unbounded plane. Thus any boundary point has the same modulus as $z_0$.

In either branch, a point of $\mathrm{\partial }\mathrm{\Omega }$ carries the global maximum of $\mathrm{\mid }f\mathrm{\mid }$ on $\bar{\mathrm{\Omega }}$.

Fix $\epsilon >0$ and form the superlevel set $K_{\epsilon }:=\{z\in \mathrm{\Omega }:\mathrm{\mid }f(z)\mathrm{\mid }\ge M+\epsilon \}$.

By [L5] and the reverse triangle inequality, $\mathrm{\mid }f\mathrm{\mid }$ is continuous, so $K_{\epsilon }$ is relatively closed in $\mathrm{\Omega }$: at a point where $\mathrm{\mid }f\mathrm{\mid }<M+\epsilon$, that strict inequality persists on a neighbourhood. Boundary control excludes every point of $\mathrm{\partial }\mathrm{\Omega }$ from the closure of $K_{\epsilon }$, so $K_{\epsilon }$ is closed in the plane. It is bounded because $\mathrm{\Omega }$ is bounded or, in the unbounded case, because infinity control excludes all points with sufficiently large modulus. Thus [L2] makes $K_{\epsilon }$ compact and it lies entirely inside $\mathrm{\Omega }$.

If $K_{\epsilon }$ were nonempty, [L3] would give a maximizer of $\mathrm{\mid }f\mathrm{\mid }$ on it. Outside $K_{\epsilon }$ the modulus is smaller than $M+\epsilon$, so this is also an interior global maximum on $\mathrm{\Omega }$; [L1] makes $f$ constant. When $\mathrm{\Omega }$ is bounded, its boundary is nonempty because otherwise it would be a nonempty clopen subset of the connected plane [L4] and hence the whole unbounded plane, so the constant contradicts finite-boundary control. When $\mathrm{\Omega }$ is unbounded, it contradicts infinity control. Hence $K_{\epsilon }$ is empty.

The set $K_{\epsilon }$ is empty for every $\epsilon >0$. If some $z$ had $\mathrm{\mid }f(z)\mathrm{\mid }>M$, taking $\epsilon =(\mathrm{\mid }f(z)\mathrm{\mid }-M)\mathrm{/}2$ would put $z$ in $K_{\epsilon }$, a contradiction; therefore $\mathrm{\mid }f\mathrm{\mid }\le M$ on $\mathrm{\Omega }$.

Since $f$ is nowhere zero, [L2] makes $1\mathrm{/}f$ holomorphic throughout $\mathrm{\Omega }$.

The local inequality $\mathrm{\mid }f(z)\mathrm{\mid }\ge \mathrm{\mid }f(a)\mathrm{\mid }>0$ is equivalent to $\mathrm{\mid }1\mathrm{/}f(z)\mathrm{\mid }\le \mathrm{\mid }1\mathrm{/}f(a)\mathrm{\mid }$, so $\mathrm{\mid }1\mathrm{/}f\mathrm{\mid }$ has an interior local maximum at $a$. By [L1], $1\mathrm{/}f$ is constant.

The reciprocal of that nonzero constant is $f$, so $f$ is constant on $\mathrm{\Omega }$.

If $M=0$, then [L1] gives $\mathrm{\mid }f\mathrm{\mid }\le 0$ on $\bar{\mathrm{\Omega }}$, so $f$ is the zero function and is constant.

Suppose $M>0$ and $f$ has no zero in $\mathrm{\Omega }$. On $\mathrm{\partial }\mathrm{\Omega }$ one has $\mathrm{\mid }f\mathrm{\mid }=M>0$, so $f$ has no zero on $\bar{\mathrm{\Omega }}$. At a point $c\in \bar{\mathrm{\Omega }}$ the identity $1\mathrm{/}f(z)-1\mathrm{/}f(c)=(f(c)-f(z))\mathrm{/}(f(z)f(c))$ together with $\mathrm{\mid }f(z)\mathrm{\mid }\ge \mathrm{\mid }f(c)\mathrm{\mid }\mathrm{/}2$ for $z$ near $c$ bounds $\mathrm{\mid }1\mathrm{/}f(z)-1\mathrm{/}f(c)\mathrm{\mid }$ by $2\mathrm{\mid }f(z)-f(c)\mathrm{\mid }\mathrm{/}\mathrm{\mid }f(c){\mathrm{\mid }}^2$, so $1\mathrm{/}f$ is continuous on $\bar{\mathrm{\Omega }}$; the given reciprocal law makes it holomorphic on $\mathrm{\Omega }$. Applying [L1] to $f$ and to $1\mathrm{/}f$ gives $\mathrm{\mid }f\mathrm{\mid }\le M$ and $1\mathrm{/}\mathrm{\mid }f\mathrm{\mid }\le 1\mathrm{/}M$ on $\bar{\mathrm{\Omega }}$. Hence $\mathrm{\mid }f\mathrm{\mid }=M$ throughout $\mathrm{\Omega }$.