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Goursat bisection selects nested triangles retaining one quarter of the boundary-integral magnitude, with halving diameters and a one-point intersection

Statement

Let $U \subseteq C$ be open, let $f : U \to C$ be continuous, and let $T_{0} = Δ [a, b, c] \subseteq U$ . There is a sequence $(T_{n})_{n \in N}$ of filled triangles such that $T_{n + 1}$ is one of the four midpoint subtriangles of $T_{n}$ and, for every $n \in N$ ,

$T_{n + 1} \subseteq T_{n}, ∣ I_{f} (T_{n}) ∣ \geq 4^{- n} ∣ I_{f} (T_{0}) ∣,$

$P (T_{n}) = 2^{- n} P (T_{0}), diam (T_{n}) = 2^{- n} diam (T_{0}) .$

Every $T_{n}$ is nonempty, compact, closed, and bounded, and there is a unique $z_{*} \in C$ with

$⋂_{n \in N} T_{n} = {z_{*}} .$

Here $I_{f} (T)$ denotes the integral over the oriented boundary prescribed by Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter.

Facts & Assumptions

Given: An open set $U$ , a continuous $f : U \to C$ , and a filled triangle $T_{0} \subseteq U$ .

[L1]

The four midpoint subtriangle boundary integrals sum to the parent boundary integral (Midpoint subdivision of a triangle cancels every interior edge and preserves its outer boundary integral).

[L2]

A nested sequence of nonempty closed bounded subsets of a complete metric space whose diameters tend to zero has intersection consisting of exactly one point (In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness).

[L3]

The complex plane with its usual metric is complete (The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts).

[L4]

The unit square in $R^{2}$ is compact, a continuous image of a compact metric space is compact, and a compact subset of a metric space is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, A compact subset of a metric space is closed and bounded).

[L5]

Recursion constructs a sequence from an initial value and a self-map; every nonempty set of natural numbers has a least element; induction proves a statement for all natural indices (The recursion theorem, The well-ordering principle, The principle of mathematical induction).

[L6]

If $0 < r < 1$ , then the real sequence $(r^{n})$ tends to zero (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L7]

The complex modulus satisfies the triangle inequality $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

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.

L1L5L7

1.2

If $T_{n} = Δ [u, v, w]$ , the continuous map $Φ (s, t) = u + s ((1 - t) (v - u) + t (w - u))$ 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 $(α, β, γ)$ is obtained by $s = β + γ$ and, when $s > 0$ , $t = γ / s$ , while $s = 0$ gives $u$ . Thus [L4] makes $T_{n}$ compact, closed, and bounded; it is nonempty because it contains $u$ .

L4algebra

2.1

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 $1 / 2$ , hence its perimeter and diameter are half those of the parent.

step 1.1L5algebra

3.1

Induction applied to step 2.1 and the retained one-quarter estimate gives, including at $n = 0$ , $∣ I_{f} (T_{n}) ∣ \geq 4^{- n} ∣ I_{f} (T_{0}) ∣$ , $P (T_{n}) = 2^{- n} P (T_{0})$ , and $diam (T_{n}) = 2^{- n} diam (T_{0})$ .

step 1.1step 2.1L5

4.1

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 $z_{*}$ .

step 2.1step 3.1step 1.2L2L3L6∎

Depends on

Used by

Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 172 results over 34 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Theorem 1.1 (standard reference, not scraped)