Analyticity of Holomorphic Functions; Liouville and Morera — Examples

Subtracting $z$ from [L1] gives $h(z)=-z^3\mathrm{/}3!+z^5\mathrm{/}5!-z^7\mathrm{/}7!+\cdots$.

By [L5], the convergent representation in step 1.1 is the Taylor series of $h$ at zero. Its coefficients in degrees $0$, $1$, and $2$ vanish, while the coefficient in degree $3$ is $-1\mathrm{/}3!\mathrm{\ne }0$, so [L2] gives ${\mathrm{ord}}_0(h)=3$.

The factorization in [L3] therefore has $h(z)=z^{3}g(z)$ with $g(0)=-1\mathrm{/}3!\mathrm{\ne }0$; independently, [L4] and [L1] give $h^{\mathrm{\prime }}(0)=h^{\mathrm{\prime }\mathrm{\prime }}(0)=0$ and $h^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)=-\cos 0=-1$, confirming the same order and sign.

For $t>0$ define $\phi (t,z)=\exp (z\log t)$ by [L1], and define $\phi (0,z)=0$ as stated in the example.

For fixed $t>0$, the map $z\mapsto z\log t$ is complex linear and [L2] with [L8] makes $\phi (t,\cdot )$ entire; for $t=0$ the slice is the constant zero function and is entire.

On $(0,1]\times \mathrm{\Omega }$, [L3] and [L4] give $\mathrm{\mid }\phi (t,z)\mathrm{\mid }=\exp ((\mathrm{Re}z)\log t)$; [L6] gives $\log t\le 0$, so $\mathrm{Re}z>1$ and [L5] yield $\mathrm{\mid }\phi (t,z)\mathrm{\mid }\le \exp (\log t)=t$. Thus $\phi (t,z)\to 0$ uniformly for $z$ near any fixed point of $\mathrm{\Omega }$ as $t\to 0^{+}$; away from $t=0$, continuity follows from [L6], [L7], and the multiplication estimate from [L9], so $\phi$ is jointly continuous on $[0,1]\times \mathrm{\Omega }$.

Steps 2.1 and 2.2 satisfy [L10] on the finite interval $[0,1]$, so $F(z)={\int }_0^1\phi (t,z)dt$ is holomorphic on $\mathrm{\Omega }$.

On $\mathrm{\mid }z\mathrm{\mid }\le 1$, one has $\mathrm{\mid }{z}^n\mathrm{/}{n}^2\mathrm{\mid }\le 1\mathrm{/}{n}^2$, so [L1] and the convergent series in [L2] give absolute pointwise and uniform convergence of the displayed series.

For real $0<r<1$, [L3] gives $F^{\mathrm{\prime }}(r)={\sum }_{n\ge 1}{r}^{n-1}\mathrm{/}n$.

Suppose, for contradiction, that a function $G$ holomorphic on $D(0,S)$ for some $S>1$ agrees with $F$ on $\mathrm{\mid }z\mathrm{\mid }<1$. By [L9], $G^{\mathrm{\prime }}$ is holomorphic near $[0,1]$, hence continuous by [L5]; [L8] makes $\mathrm{\mid }{G}^{\mathrm{\prime }}\mathrm{\mid }$ continuous, [L7] makes $[0,1]$ compact, and [L4] bounds $\mathrm{\mid }{G}^{\mathrm{\prime }}\mathrm{\mid }$ there.

Given $B>0$, divergence in [L2] supplies $N\ge 2$ with ${\sum }_{n=1}^{N}1\mathrm{/}n>2B$; by [L6] and [L5], the finitely many monomials $r^{n-1}$ are continuous at $r=1$, so choose $0<r<1$ with $r^{n-1}>1\mathrm{/}2$ for every $1\le n\le N$, and step 1.2 then gives $F^{\mathrm{\prime }}(r)>\frac{1}{2}{\sum }_{n=1}^{N}1\mathrm{/}n>B$.

Step 2.1 makes $\mathrm{\mid }{F}^{\mathrm{\prime }}(r)\mathrm{\mid }=\mathrm{\mid }{G}^{\mathrm{\prime }}(r)\mathrm{\mid }$ exceed every proposed bound for points $0<r<1$, contradicting step 1.3; no such extension exists, and the refuted claim is false.

Let $K\subseteq \mathbb{D}$ be compact. If $K=\mathrm{\varnothing }$, uniform convergence on $K$ is vacuous. Otherwise [L4] and [L1] give $q={\max }_{z\in K}\mathrm{\mid }z\mathrm{\mid }$, and $q<1$ because the maximum is attained at a point of $K\subseteq \mathbb{D}$; then $\mathrm{\mid }{f}_n(z)\mathrm{\mid }=\mathrm{\mid }z{\mathrm{\mid }}^n\le q^n\to 0$ by [L2], uniformly for $z\in K$.

At the boundary point $z=1$, one has $f_n(1)=1$ for every natural $n$, including $n=0$, so the sequence does not converge pointwise to zero there and fails the uniform condition [L3] on $\bar{\mathbb{D}}$.

Step 1.1 proves local uniform convergence on the open disc, while step 1.2 proves failure on its closure, so the claimed implication is false.

Fix an arbitrary triangle and read its boundary integral in the edgewise sense of [L1] for the point-supported function $f$.

The function is discontinuous at $0$, since every punctured neighbourhood contains points where its value is $0$ while $f(0)=1$; by [L3], it is not complex differentiable, and hence not holomorphic, at $0$.

On a nonconstant affine edge, the edge map is injective and therefore meets $0$ at most once, so each real component of its velocity-weighted pullback differs from zero at at most one parameter and [L2] makes its integral zero; on a constant edge the velocity is zero, so its pullback integral is also zero. Hence [L1] gives zero around every triangle, including repeated or collinear vertices.

Steps 2.1 and 1.2 exhibit vanishing edgewise triangle integrals without holomorphy, so removing the continuity hypothesis from [L4] makes the implication false.

For every natural $m$, repeated use of [L1], [L2], and [L3] gives a real polynomial $P_m$ such that ${\varphi }^{(m)}(x)=P_m(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ for $x\mathrm{\ne }0$: take $P_0=1$, and differentiation replaces $P_m(y)$ by the polynomial $-y^2{P}_m^{\mathrm{\prime }}(y)+2y^3{P}_m(y)$.

As $x\to 0$, every expression $P(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ and its quotient by $x$ tends to $0$: with $y=1\mathrm{/}\mathrm{\mid }x\mathrm{\mid }\to +\mathrm{\infty }$, polynomial growth is bounded by a natural power of $y$, which for $y\ge 1$ is bounded by a natural power of $y^2$, and [L4] applied to $u=y^2$ makes that power times $\exp (-u)$ tend to zero.

Inductively set every derivative value ${\varphi }^{(m)}(0)=0$: step 2.1 makes ${\varphi }^{(m)}$ continuous at $0$ and makes its difference quotient there tend to $0$, so the next derivative exists and has value $0$. Thus $\varphi$ is smooth, and [L5] makes $F(x,y)=(\varphi (x),0)$ smooth with every mixed derivative at $(0,0)$ equal to $0$.

By step 3.1, the total-degree Taylor series of $F$ at $(0,0)$ is the zero map, but [L7] gives $F(x,0)=(\exp (-1\mathrm{/}{x}^2),0)\mathrm{\ne }(0,0)$ for every $x\mathrm{\ne }0$; such points occur in every neighbourhood of the origin, so the equality required by [L6] fails there.

The map $F$ is smooth by step 3.1 and not real analytic by step 4.1, so it refutes the false claim.

By [L1] and [L4], complex sine is entire; by [L2] and [L3], its restriction to the real axis satisfies $\mathrm{\mid }\sin x\mathrm{\mid }\le 1$ for every real $x$.

By [L4], the derivative of sine is cosine, and the series in [L1] gives $\cos 0=1$, so sine has a nonzero derivative and is not constant.

Steps 1.1 and 1.2 give an entire function bounded on the real axis but not constant, refuting the claim; this does not contradict Liouville's theorem: every bounded entire function is constant, whose hypothesis is boundedness on the whole complex plane.

By [L1], the complex exponential is entire and is its own entire antiderivative.

By [L2] and [L4], $\exp z\exp (-z)=\exp 0=1$, so the exponential never vanishes; [L1] then makes its derivative nonzero everywhere, and it is nonconstant.

If the complex exponential were a nonconstant polynomial, [L3] would give it a complex zero, contradicting step 1.2.

It is not a constant polynomial because its derivative is nonzero by step 1.2, and step 2.1 excludes every nonconstant polynomial; together with step 1.1, the exponential is an entire nonpolynomial function with an entire antiderivative, refuting the claim.