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

By [L3] and the holomorphic algebra laws, $h(z):={\sin }^{2}z+{\cos }^{2}z-1$ is entire.

For every real $x$, [L1] gives $h(x)=0$.

The real axis has accumulation point $0$ in the complex domain $\mathbb{C}$, so [L2] applied to $h$ and the zero function makes $h$ identically zero. Hence ${\sin }^{2}z+{\cos }^{2}z=1$ for every complex $z$.

At $0$, one has $f(z)-f(0)=z^2$, so [L1] gives ${\mathrm{deg}}_{0}f=2$. By [L2], every sufficiently small nonzero $w$ has exactly two local preimages; explicitly they are the distinct roots $z$ and $-z$, while $w=0$ has only the preimage $0$ with multiplicity $2$.

At $1$, $f(z)-f(1)=(z-1)(z+1)$ and the factor $z+1$ is nonzero at $1$, so [L1] gives ${\mathrm{deg}}_{1}f=1$. Hence [L3] makes $f$ biholomorphic between neighbourhoods of $1$ and $1$.

The injectivity can also be seen explicitly on $D(1,1\mathrm{/}2)$. If $z,w$ lie in that disc and $z^2=w^2$, then $z=w$ or $z=-w$; the second alternative would give $2=\mathrm{\mid }(z-1)+(w-1)\mathrm{\mid }<1$, which is impossible.

If $\mathrm{\mid }z\mathrm{\mid }\le 1$, then $\mathrm{\mid }p(z)\mathrm{\mid }\le \mathrm{\mid }z{\mathrm{\mid }}^2+2\mathrm{\mid }z\mathrm{\mid }+2\le 1+2+2=5$.

By [L2], $p$ is entire, so [L1] applies to the open unit disc and confirms that its maximum on the closed disc occurs on the unit circle.

The point $-1$ lies on that circle and $p(-1)=1+2+2=5$. Together with step 1.1, this proves that the exact maximum is $5$.

The set $\mathrm{\Omega }$ is open and is connected by [L3] under the plane dictionary, so it is a complex domain (A complex domain is a nonempty connected open subset of $\mathbb{C}$). The chain and quotient rules make $f$ holomorphic there, and $g$ is holomorphic as a constant.

For every natural $k\ge 1$, put $z_k=1\mathrm{/}(k\pi )$. Then $z_k\in \mathrm{\Omega }$, [L1] gives $f(z_k)=\sin (k\pi )=0=g(z_k)$, the points are distinct, and $z_k\to 0$.

The accumulation point $0$ is not in $\mathrm{\Omega }$, so [L2] does not apply. Moreover, $2\mathrm{/}\pi \in \mathrm{\Omega }$ and $f(2\mathrm{/}\pi )=\sin (\pi \mathrm{/}2)=1\mathrm{\ne }0=g(2\mathrm{/}\pi )$. Thus the functions agree on a set accumulating only at the boundary but are not identical.

For $x\mathrm{\ne }0$, induction using [L2], [L3], [L4], and [L5] gives ${\psi }^{(n)}(x)=P_n(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ for a real polynomial $P_n$: $P_0=1$, and differentiating one such expression produces another polynomial in $1\mathrm{/}x$ times the same exponential.

For every $x\mathrm{\ne }0$, [L8] gives $\psi (x)=\exp (-1\mathrm{/}{x}^2)>0$.

Extend each expression in step 1.1 by the value $0$ at $x=0$. By [L1], both $P_n(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ and its difference quotient divided by $x$ tend to $0$ as $x\to 0$ from either side. Inductively, every derivative exists at $0$, equals $0$, and is continuous there by [L6]; hence $\psi$ is smooth by [L7].

Suppose, for contradiction, that a holomorphic function $F$ on a complex neighbourhood of $0$ agrees with $\psi$ on a real interval about $0$. Derivatives along the real axis then give $F^{(n)}(0)={\psi }^{(n)}(0)=0$ for every natural $n$, and [L10] makes every Taylor coefficient of $F$ at $0$ equal to $0$.

By [L9], $F$ equals that zero Taylor series on a complex disc about $0$, so $F$ vanishes there.

Every real interval about $0$ contains a nonzero $x$, where step 1.2 gives $F(x)=\psi (x)>0$, contradicting step 4.1. Thus the smooth function $\psi$ has no holomorphic extension to any complex neighbourhood of $0$.

If $a=0$, the denominator is $1$. If $a\mathrm{\ne }0$, choose $R$ with $1<R<1\mathrm{/}\mathrm{\mid }a\mathrm{\mid }$; for $\mathrm{\mid }z\mathrm{\mid }<R$ one has $\mathrm{\mid }\bar{a}z\mathrm{\mid }<1$, so $1-\bar{a}z\mathrm{\ne }0$. Thus [L2] makes $B_a$ holomorphic on a neighbourhood of the closed unit disc, and hence continuous there.

When $\mathrm{\mid }z\mathrm{\mid }=1$, direct expansion gives $\mathrm{\mid }z-a{\mathrm{\mid }}^2=1-z\bar{a}-a\bar{z}+\mathrm{\mid }a{\mathrm{\mid }}^2=\mathrm{\mid }1-\bar{a}z{\mathrm{\mid }}^2\mathrm{.}$ The denominator is nonzero by step 1.1, so $\mathrm{\mid }{B}_a(z)\mathrm{\mid }=1$ on the entire unit circle.

Since $B_a(a)=0$ and its boundary modulus is $1$, the function is nonconstant. It therefore realizes the zero alternative in [L1] and refutes the proposed implication, including the case $a=0$, where $B_0(z)=z$.

The two discs are nonempty, open, and disjoint, so $U=D(-2,1)\cup D(2,1)$ is open but disconnected and therefore is not a complex domain (A complex domain is a nonempty connected open subset of $\mathbb{C}$).

Define $f=0$ on $D(-2,1)$ and $f=1$ on $D(2,1)$. Every point has a neighbourhood on which $f$ is constant, so $f$ is holomorphic on $U$.

At every point of the second disc, $\mathrm{\mid }f\mathrm{\mid }=1$ is a local maximum, but $f$ is not constant on $U$ because it is $0$ on the first disc. Thus the statement is false; the connected-domain hypothesis in [L1] is exactly what prevents this componentwise witness.

The identity function $f(z)=z$ is holomorphic on the nonempty complex domain $\mathbb{D}$.

Its modulus satisfies $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\ge 0=\mathrm{\mid }f(0)\mathrm{\mid }$, so it has a global, and hence local, minimum at the interior point $0$.

The map is not constant, since $f(0)=0$ and $f(1\mathrm{/}2)=1\mathrm{/}2$. The valid theorem [L1] does not apply because $f$ vanishes at the minimizer, so nonvanishing is essential.

If $F(x,y)=F(u,v)$, then $y=v$ and $0=x^3-u^3=(x-u)((x+u\mathrm{/}2{)}^2+3u^2\mathrm{/}4)$. The second factor is nonnegative and vanishes only when $x=u=0$, so in every case $x=u$. Thus $F$ is injective.

By [L1], the derivative matrix is $DF(x,y)=\left(\begin{array}{cc}3x^2& 0\\ 0& 1\end{array}\right),$ so $F$ is differentiable and $\det DF(x,y)=3x^2$.

On the entire vertical axis $x=0$, the determinant in step 1.2 is zero even though step 1.1 shows that $F$ is injective. Hence the real-differentiable statement is false; [L2] shows the contrasting conclusion that does hold for injective holomorphic maps.

The function $f(z)=\exp (-iz)$ is entire and hence is holomorphic on $\mathbb{H}$ and continuous on its closed half-plane.

For $z=x+iy$, one has $-iz=y-ix$, so [L2] gives $\mathrm{\mid }f(x+iy)\mathrm{\mid }=e^y$. Thus $\mathrm{\mid }f(x)\mathrm{\mid }=1$ on the real boundary $y=0$, while $\mathrm{\mid }f(iy)\mathrm{\mid }=e^y$ is unbounded as $y\to +\mathrm{\infty }$.

Step 2.1 violates the proposed conclusion. The valid unbounded-domain result [L1] requires control at infinity as well as finite-boundary control, and this example fails exactly that additional hypothesis.