2

In a proof of Cauchy's theorem where $\oint_C f(z)\,dz = 0$ for analytic $f$ they first showed this holds for a triangular contour. The argument used the distance between any two interior points is no greater than the longest side of the triangle.

It's very obvious. I set out proving rigorously but got stuck.

My approach:

Let $\bf v_1,v_2, v_3 \in \mathbb{R}^2$ be the vertices of a triangle and let $\bf p,q$ be interior points. Since a triangle is convex there are numbers $a_1,a_2,a_3,b_1,b_2,b_3$ where $0 \leq a_i,b_i \leq 1$, $a_1 + a_2 + a_3 =1 $ and $b_1 + b_2 + b_3 = 1$ so that $\mathbf{p} = a_1\mathbf{v_1} + a_2 \mathbf{v_2} + a_3 \mathbf{v_3}$ and $\mathbf{q} = b_1\mathbf{v_1} + b_2 \mathbf{v_2} + b_3 \mathbf{v_3}$ and

\begin{align} \mathbf{p} - \mathbf{q} = {} & (a_1 - b_1)\mathbf{v_1} + (a_2 - b_2)\mathbf{v_2} +(a_3 - b_3)\mathbf{v_3} \\[10pt] |\mathbf{p} - \mathbf{q}|^2 = {} & (a_1 - b_1)^2| \mathbf{v_1}|^2 + (a_2 - b_2)| \mathbf{v_2}|^2 +(a_3 - b_3)^2| \mathbf{v_3}|^2 + 2(a_1-b_1)(a_2 - b_2) \mathbf{v_1} \cdot \mathbf{v_2} \\ & {} + 2(a_1-b_1)(a_3 - b_3)\mathbf{v_1} \cdot \mathbf{v_2} + 2(a_2-b_2)(a_3 - b_3) \mathbf{v_2} \cdot \mathbf{v_3} \end{align}

Now I can't see how to show $|\mathbf{p} - \mathbf{q}| < \max |\mathbf{v_i}|$

Michael Hardy
  • 1
scobaco
  • 597

1 Answers1

1

Any point in the triangle can be written as $p = \sum_k \mu_k x_k = x_1 + \sum_{k>1} \mu_k (x_k-x_1)$ where $\sum_k \mu_k = 1$ and $\mu_k \ge 0$.

Then $\|p_1-p_2\| \le \sum_{k>1} \mu_k \|x_k - x_1\| \le \max_{i,j} \|x_i -x_j\|$.

copper.hat
  • 172,524