Solving Convexity of Open Disc Problem in Complex Plane

[e(ho0n3]

[SOLVED] Convexity of Open Disc

I'm trying to prove that the open disc in the complex plane given by D = {z : |z - w| < r} is convex.

Let p and q be two points in D. The line segment from p to q is L = {(1 - t)p + tq : 0 <= t <= 1}. Let u be a point an arbitrary point on this segment. If I can show that |u - w| < r, I'm done.

This is essentially a geometry problem. There's probably a proposition in Euclid's Elements that the line segment from any two points on the circle is contained in the circle. How do you show this analytically though?

 

[Dick]

Use the triangle inequality on w(t)=|(1-t)p+tq|. The result is a linear function in t. |w(0)|=|p|<=1 and |w(1)|=|q|<=1. Hence?

 

[e(ho0n3]

Using the triangle inequality is a great idea:

|p| - |w| < |p - w| < r so |p| < r + |w|. Similarly |q| < r + |w|.

|(1 - t)p + tq - w| <= (1 - t)|p| + t|q| + |w| < (1 - t)(r + |w|) + t(r + |w|) + |w| = r + 2|w| so (1 - t)|p| + t|q| - |w| < r.

However, since (1 - t)|p| + t|q| - |w| <= |(1 - t)p + tq - w|, I don't know if the latter is greater than r.

 

[Dick]

I can how my first reply could be a little difficult to digest. I was taking w=0 and r=1 and not telling you that. Furthermore I meant to define u(t)=p(1-t)+qt to fit in with your notation better. Let's try that again. u(t)-w=(p-w)(1-t)+(q-w)t, right? So |u(t)-w|<=|p-w|(1-t)+|q-w|t. Can you finish it from there?

 

[e(ho0n3]

Ah, I see. Writing u(t)-w as (p-w)(1-t)+(q-w)t is the right idea. It never occurred to me. From then on, it trivially follows that |u - w| < r. Thanks a lot.