Goursat's Theorem and Cauchy's Theorem in a Convex Domain

In the order fixed by the triangle definition, the four boundary edge lists are $(a\to p,p\to r,r\to a)$, $(p\to b,b\to q,q\to p)$, $(r\to q,q\to c,c\to r)$, and $(p\to q,q\to r,r\to p)$.

In their sum, the interior pairs $p\to r,r\to p$, $q\to p,p\to q$, and $r\to q,q\to r$ cancel by reversal, while concatenation joins the surviving pairs $a\to p\to b$, $b\to q\to c$, and $c\to r\to a$ into the three outer edges.

Additivity under concatenation now gives exactly the displayed integral identity; reversal and concatenation also hold for constant and collinear segments, so no nondegeneracy was used.

For any parent triangle, [L1] and [L7] imply that at least one of its four indexed midpoint children has integral modulus at least one quarter of the parent's: otherwise the modulus of their sum would be strictly smaller than the parent modulus. Choose the least qualifying index, which also works when the parent integral is zero.

If $T_n=\mathrm{\Delta }[u,v,w]$, the continuous map $\mathrm{\Phi }(s,t)=u+s((1-t)(v-u)+t(w-u))$ takes the compact square $[0,1{]}^2$ onto $T_n$: its coefficients are nonnegative and sum to one, and conversely a barycentric point with coefficients $(\alpha ,\beta ,\gamma )$ is obtained by $s=\beta +\gamma$ and, when $s>0$, $t=\gamma \mathrm{/}s$, while $s=0$ gives $u$. Thus [L4] makes $T_n$ compact, closed, and bounded; it is nonempty because it contains $u$.

The least-index rule is a function of the ordered parent triangle, so recursion gives $(T_n)$; direct midpoint geometry shows every child is contained in its parent and is the image of it under a similarity of ratio $1\mathrm{/}2$, hence its perimeter and diameter are half those of the parent.

Induction applied to step 2.1 and the retained one-quarter estimate gives, including at $n=0$, $\mathrm{\mid }{I}_f(T_n)\mathrm{\mid }\ge 4^{-n}\mathrm{\mid }{I}_f(T_0)\mathrm{\mid }$, $P(T_n)=2^{-n}P(T_0)$, and $\mathrm{diam}(T_n)=2^{-n}\mathrm{diam}(T_0)$.

By [L6] and step 3.1, the diameters tend to zero, even when the initial diameter is zero. Steps 2.1 and 1.2 give a nested sequence of nonempty closed bounded subsets of the complete complex plane, so [L2] and [L3] give a unique common point $z_{\ast }$.

By [L5], $f$ is continuous, so [L1] gives selected triangles $T_n$ with common point $z_{\ast }$ and all the stated retention and scaling formulas.

By [L7], differentiability at $z_{\ast }$ gives $f(w)=f(z_{\ast })+f^{\mathrm{\prime }}(z_{\ast })(w-z_{\ast })+(w-z_{\ast })\eta (w)$, where $\eta (w)\to 0$ as $w\to z_{\ast }$; set $\eta (z_{\ast })=0$.

Put $P(w)=f(z_{\ast })w+\frac{1}{2}{f}^{\mathrm{\prime }}(z_{\ast })(w-z_{\ast }{)}^2$. By [L3], $P$ is entire and $P^{\mathrm{\prime }}(w)=f(z_{\ast })+f^{\mathrm{\prime }}(z_{\ast })(w-z_{\ast })$ is a polynomial, hence continuous. Every $\mathrm{\partial }{T}_n$ is a closed rectifiable contour, so all hypotheses of [L2] hold and ${\int }_{\mathrm{\partial }{T}_n}{P}^{\mathrm{\prime }}(w)dw=0$.

Given $\epsilon >0$, take $n$ so large that $\mathrm{\mid }\eta (w)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }w-z_{\ast }\mathrm{\mid }\le \mathrm{diam}(T_n)$; this is possible by step 1.2 and the diameter limit in [L1]. Since $z_{\ast }\in T_n$, [L4], [L6], and step 1.3 give $\mathrm{\mid }{I}_f(T_n)\mathrm{\mid }\le \epsilon \mathrm{diam}(T_n)P(T_n)$.

The retention and scaling formulas yield $\mathrm{\mid }{I}_f(T_0)\mathrm{\mid }\le 4^n\mathrm{\mid }{I}_f(T_n)\mathrm{\mid }\le \epsilon \mathrm{diam}(T_0)P(T_0)$. Because this holds for every $\epsilon >0$, $I_f(T_0)=0$; the argument never divides by the initial integral, diameter, or perimeter, so it also covers zero integrals and degenerate triangles.

The diagonal from $a$ to $c$ splits $R$ into the filled triangles $\mathrm{\Delta }[a,b,c]$ and $\mathrm{\Delta }[a,c,d]$, both contained in $U$.

By [L1], the integrals over the positively oriented boundaries $a\to b\to c\to a$ and $a\to c\to d\to a$ are both zero.

Adding those identities, the diagonal $c\to a$ in the first boundary cancels the diagonal $a\to c$ in the second by [L2].

The surviving directed sides are $a\to b\to c\to d\to a$, exactly the displayed positive boundary $\mathrm{\partial }R$, so its integral is zero.

If $p\mathrm{\notin }T$, then $T\subseteq U\setminus \{p\}$; this punctured set is open because every $z\mathrm{\ne }p$ has the ball $B(z,\mathrm{\mid }z-p\mathrm{\mid }\mathrm{/}2)$ inside it, so [L1] applies. If $T$ is degenerate, order its distinct collinear vertices along their common line and split each directed edge at the intervening vertices; every resulting subsegment occurs equally often in both orientations, so [L2] gives zero.

It remains to treat a nondegenerate triangle $\mathrm{\Delta }[p,u,v]$ having $p$ as a vertex. For $0<t<1$, put $u_t=p+t(u-p)$ and $v_t=p+t(v-p)$. The triangles $\mathrm{\Delta }[u_t,u,v]$ and $\mathrm{\Delta }[u_t,v,v_t]$ lie in $\mathrm{\Delta }[p,u,v]$ and avoid $p$, so [L1] makes their boundary integrals zero; adding their boundaries and cancelling opposite internal edges by [L2] gives $I_f[p,u,v]=I_f[p,u_t,v_t]$.

Continuity at $p$ gives a neighborhood on which $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }f(p)\mathrm{\mid }+1$. For all sufficiently small $t>0$, the small triangle lies in that neighborhood, has perimeter $tP(\mathrm{\Delta }[p,u,v])$, and [L3] gives $\mathrm{\mid }{I}_f[p,u_t,v_t]\mathrm{\mid }\le (\mathrm{\mid }f(p)\mathrm{\mid }+1)tP(\mathrm{\Delta }[p,u,v])$.

Letting $t$ be arbitrarily small in steps 1.2 and 2.1 forces $I_f[p,u,v]=0$, so every triangle having $p$ as a vertex has zero boundary integral.

If $p\in T$, the three filled triangles $\mathrm{\Delta }[p,a,b]$, $\mathrm{\Delta }[p,b,c]$, and $\mathrm{\Delta }[p,c,a]$ cover $T$ with compatible orientations; their internal edges cancel by [L2], leaving $\mathrm{\partial }T$. Their integrals vanish by steps 1.1 and 3.1, so $I_f(T)=0$. Together with the case $p\mathrm{\notin }T$ from step 1.1, this proves the claim for every position of $p$.

Fix $z\in U$. The segment ${\mathrm{\ell }}_{az}$ is rectifiable and $f$ is continuous on it, so $F(z)$ exists by [L5]. Since $U$ is open, choose $\rho >0$ with $B(z,\rho )\subseteq U$. If $0<\mathrm{\mid }h\mathrm{\mid }<\rho$, the short segment from $z$ to $z+h$ lies in that ball, and every segment from $a$ to a point of the short segment lies in $U$ by star-shapedness; hence $\mathrm{\Delta }[a,z,z+h]\subseteq U$.

Parametrizing the short edge by $\zeta =z+th$ and using [L2] gives ${\int }_{{\mathrm{\ell }}_{z,z+h}}f(z)d\zeta =f(z)h$.

The zero boundary integral of that triangle reads $F(z)+{\int }_{{\mathrm{\ell }}_{z,z+h}}f-F(z+h)=0$ by [L1], and therefore $F(z+h)-F(z)={\int }_{{\mathrm{\ell }}_{z,z+h}}f(\zeta )d\zeta$.

By [L3], steps 2.1 and 1.2 imply $\mid (F(z+h)-F(z))\mathrm{/}h-f(z)\mid \le {\sup }_{0\le t\le 1}\mathrm{\mid }f(z+th)-f(z)\mathrm{\mid }$.

Continuity of $f$ at $z$ makes the right side of step 3.1 tend to zero as $h\to 0$. Thus $F^{\mathrm{\prime }}(z)=f(z)$ for arbitrary $z\in U$, including $z=a$; [L4] says exactly that $F$ is a primitive, and $h=0$ was only the excluded difference-quotient value.

By [L1], $f$ is continuous, and by [L2] its integral around every filled triangle contained in $U$ is zero.

The hypotheses of [L3] now hold, so the displayed function is holomorphic on $U$ and has derivative $f$.

By [L1] and [L2], there is a holomorphic $F$ on $U$ with $F^{\mathrm{\prime }}=f$; by [L3], this derivative is continuous.

The given $\gamma$ is closed and rectifiable, so every hypothesis of [L4] holds and ${\int }_{\gamma }f={\int }_{\gamma }{F}^{\mathrm{\prime }}=0$.

By [L1], choose any $a\in U$ and regard $U$ as star-shaped with respect to $a$.

Now [L2] applied to the given $f$ and $\gamma$ gives the displayed zero integral.

By [L1], every contained triangle integral vanishes.

Since $f$ is continuous and $U$ is star-shaped, [L2] gives a primitive $F$, and [L3] makes explicit that $F$ is holomorphic and $F^{\mathrm{\prime }}=f$.

The derivative $F^{\mathrm{\prime }}=f$ is continuous by hypothesis; for any given closed rectifiable contour $\gamma$ in $U$, all hypotheses of [L4] are therefore satisfied, and ${\int }_{\gamma }f=0$.

By [L1], the off-point quotient tends to $f^{\mathrm{\prime }}(z)=g(z)$ as $\zeta \to z$, which is exactly continuity of $g$ at $z$; this also covers constant $f$.

On $U\setminus \{z\}$, the numerator and denominator are holomorphic and the denominator is nonzero, so [L2] makes $g$ holomorphic there.

By [L3], step 1.2 also makes $g$ continuous away from $z$; together with step 1.1 this proves continuity on all of $U$, without claiming differentiability at the filled point.

Define $g(\zeta )=(f(\zeta )-f(z))\mathrm{/}(\zeta -z)$ for $\zeta \mathrm{\ne }z$ and $g(z)=f^{\mathrm{\prime }}(z)$. By [L1], $g$ is continuous on $D(a,R)$ and holomorphic away from $z$.

If $z=a$, [L4] gives ${\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta =2\pi i$. If $z\mathrm{\ne }a$, the finite identity $1\mathrm{/}(\zeta -z)={\sum }_{k=0}^N(z-a{)}^k\mathrm{/}(\zeta -a{)}^{k+1}+(z-a{)}^{N+1}\mathrm{/}((\zeta -a{)}^{N+1}(\zeta -z))$ holds on $\gamma$, and its remainder has modulus at most $(\mathrm{\mid }z-a\mathrm{\mid }\mathrm{/}r{)}^{N+1}\mathrm{/}(r-\mathrm{\mid }z-a\mathrm{\mid })$ there.

The disc is star-shaped with respect to $a$, since $\mathrm{\mid }(1-t)a+t\zeta -a\mathrm{\mid }=t\mathrm{\mid }\zeta -a\mathrm{\mid }<R$. The circle lies in the disc and is closed and rectifiable, so [L2] gives ${\int }_{\gamma }g(\zeta )d\zeta =0$.

In the second case, [L6] makes the remainder tend uniformly to zero; [L5] and [L4] then give ${\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta =2\pi i$, because only the $k=0$ term has exponent $-1$. Thus the same kernel integral value holds also in the first case.

Since $\zeta \mathrm{\ne }z$ on $\gamma$, [L3] expands step 2.1 as ${\int }_{\gamma }f(\zeta )\mathrm{/}(\zeta -z)d\zeta =f(z){\int }_{\gamma }1\mathrm{/}(\zeta -z)d\zeta$.

Substitute step 2.2 into step 3.1 and divide by the nonzero number $2\pi i$ to obtain the formula.

For every $z\in V$, the function $\zeta \mapsto \phi (\zeta )\mathrm{/}(\zeta -z{)}^n$ is continuous on the trace, so $F_n(z)$ exists by [L1]. Moreover, [L3] applied to $\phi \circ \gamma$ gives a finite $M\ge 0$ with $\mathrm{\mid }\phi (\zeta )\mathrm{\mid }\le M$ on the trace.

Fix $z_0\in V$. Choose $\rho >0$ with $B(z_0,\rho )\subseteq V$; because the trace is disjoint from $V$, $w=\zeta -z_0$ satisfies $\mathrm{\mid }w\mathrm{\mid }\ge \rho$, and if $0<\mathrm{\mid }h\mathrm{\mid }<\rho \mathrm{/}2$ then $\mathrm{\mid }w-h\mathrm{\mid }\ge \rho \mathrm{/}2$.

The finite power identity gives $((w-h{)}^{-n}-w^{-n})\mathrm{/}h={\sum }_{j=0}^{n-1}1\mathrm{/}(w^{j+1}(w-h{)}^{n-j})$, and after subtracting $n\mathrm{/}{w}^{n+1}$ the remainder is $h{\sum }_{j=0}^{n-1}{\sum }_{k=0}^{n-j-1}1\mathrm{/}(w^{j+k+2}(w-h{)}^{n-j-k})$; by step 1.2 and [L4], its modulus is at most $\mathrm{\mid }h\mathrm{\mid }n(n+1)(2\mathrm{/}\rho {)}^{n+2}\mathrm{/}2$, uniformly on the trace.

By [L2], the difference between $(F_n(z_0+h)-F_n(z_0))\mathrm{/}h$ and $n{F}_{n+1}(z_0)$ has modulus at most a fixed finite constant times $ML(\gamma )\mathrm{\mid }h\mathrm{\mid }$, which tends to zero. Hence $F_n^{\mathrm{\prime }}(z_0)=n{F}_{n+1}(z_0)$; since $z_0$ was arbitrary, $F_n$ is holomorphic on $V$. The estimate also covers $n=1$ and a constant contour, while $h=0$ is only the excluded difference-quotient value.

For $n=0$, [L1] and $0!=1$ from [L3] give the formula and the existence of $f^{(0)}=f$ on $D(a,r)$.

Assume for a natural $n$ that $f^{(n)}$ exists on $D(a,r)$ and satisfies the displayed formula there.

By [L5], the boundary data $\phi =f{\mathrm{\mid }}_{\gamma }$ are continuous; the open disc $D(a,r)$ is disjoint from the radius-$r$ trace, so [L2] applies with $m=n+1$. Differentiating the induction formula gives $f^{(n+1)}(z)=n!(n+1)(2\pi i{)}^{-1}{\int }_{\gamma }f(\zeta )\mathrm{/}(\zeta -z{)}^{n+2}d\zeta$, which is the required formula because $n!(n+1)=(n+1)!$ by [L3].

Thus the property holds at $0$ and passes from $n$ to $n+1$; [L4] proves existence and the formula for every $n\in \mathbb{N}$, including $n=1$ and constant functions.

On the circle, $\mathrm{\mid }f(\zeta )\mathrm{/}(\zeta -a{)}^{n+1}\mathrm{\mid }\le M\mathrm{/}{r}^{n+1}$, so [L1], [L2], and [L3] give $\mathrm{\mid }{f}^{(n)}(a)\mathrm{\mid }\le n!(2\pi {)}^{-1}(M\mathrm{/}{r}^{n+1})(2\pi r)$.

Since $r>0$, simplifying step 1.1 gives $\mathrm{\mid }{f}^{(n)}(a)\mathrm{\mid }\le n!M\mathrm{/}{r}^n$. For $n=0$ this is $\mathrm{\mid }f(a)\mathrm{\mid }\le M$, and for $M=0$ or constant $f$ the same computation remains valid.