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Goursat's triangle theorem remains valid for a continuous function holomorphic away from one point

Statement

Let $U \subseteq C$ be open, let $p \in U$ , and let $f : U \to C$ be continuous and holomorphic on $U ∖ {p}$ . If the filled triangle $T = Δ [a, b, c]$ of Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter is contained in $U$ , then

$\int_{\partial T} f (z) d z = 0.$

The exceptional point may lie outside, inside, or on the boundary of $T$ .

Facts & Assumptions

Given: An open set $U$ , a point $p \in U$ , a continuous function $f : U \to C$ holomorphic away from $p$ , and a filled triangle $T \subseteq U$ .

[L1]

Goursat's triangle theorem gives zero boundary integral when the function is holomorphic on an open set containing the filled triangle (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L2]

Reversing a contour negates its integral and concatenating compatible contours adds their integrals (Complex line integrals change sign under reversal and add under concatenation).

[L3]

The ML estimate bounds an integral by a uniform bound on the integrand times the contour length (ML estimate: a contour integral is bounded by a supremum bound times path length).

Proof

technique · direct

1.1

If $p \notin T$ , then $T \subseteq U ∖ {p}$ ; this punctured set is open because every $z \neq p$ has the ball $B (z, ∣ z - p ∣ / 2)$ 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.

givenL1L2

1.2

It remains to treat a nondegenerate triangle $Δ [p, u, v]$ 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 $Δ [u_{t}, u, v]$ and $Δ [u_{t}, v, v_{t}]$ lie in $Δ [p, u, v]$ and avoid $p$ , so [L1] makes their boundary integrals zero; adding their boundaries and cancelling opposite internal edges by [L2] gives $I_{f} [p, u, v] = I_{f} [p, u_{t}, v_{t}]$ .

L1L2

2.1

Continuity at $p$ gives a neighborhood on which $∣ f ∣ \leq ∣ f (p) ∣ + 1$ . For all sufficiently small $t > 0$ , the small triangle lies in that neighborhood, has perimeter $t P (Δ [p, u, v])$ , and [L3] gives $∣ I_{f} [p, u_{t}, v_{t}] ∣ \leq (∣ f (p) ∣ + 1) t P (Δ [p, u, v])$ .

step 1.2L3

3.1

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.

step 1.2step 2.1algebra

4.1

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

step 1.1step 3.1L2∎

Depends on

Used by

A continuous function holomorphic away from one point on a star-shaped domain has a primitive and zero closed-contour integrals Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 90 results over 17 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 1.4 (standard reference, not scraped)