# Proving a Set is Convex

1. Oct 11, 2012

### WilcoRogers

1. The problem statement, all variables and given/known data
Using the definition of a convex set, show that the set in R^2
$\{(x,y) \in R^2 \colon y \ge 1/x, x\ge 0\}$

2. Relevant equations
An object is convex if for every pair of points within the object, every point on the straight line segment that joins them is also within the object.

3. The attempt at a solution
I know the solution involves playing with inequalities, I just don't know how to set this one up. Any help is appreciated.

2. Oct 11, 2012

### Ray Vickson

Call the given set S. Suppose $(x_1,y_1) \in S \text{ and } (x_2,y_2) \in S.$ First write out in detail what these statements mean, then go on from there.

RGV

3. Oct 11, 2012

### WilcoRogers

I know that we can say $y_1\ge 1/x_1$ and $y_2\ge 1/x_2$, and that we can add those constraints, but they don't get me anywhere. I know that I have to show, somehow that $(1-\lambda)P_1 + \lambda P_2$ is also in the set. I just don't know how to set up the proof and inequalities, or how that whole thing works. I very much understand the idea of convexity, i'm just very shaky on the process of actually taking a function and showing it's convex, algebraically.

4. Oct 11, 2012

### Ray Vickson

Do you know some simple characterizations of a convex function? For example, is the function f(x) = 1/x convex in the region {x > 0}? How do you know? If you were able to say one way or another that f(x) is or is not convex, would that help you solve the original problem?

RGV

5. Oct 11, 2012

### Zondrina

EDIT : Even better try using what RGV said.

If you SHOW that both (x1, y1) and (x2, y2) are BOTH in your set S and the line you draw between them is still also inside your set, then the set is convex.

Last edited: Oct 11, 2012
6. Oct 11, 2012

### WilcoRogers

Well i know the function is convex on that region, intuitively, because of how it looks. I know the definition of a convex line, but I can't show it. I feel, given the nature of this problem, the answer is staring me in the face, but I can't seem to get an algebraic statement that makes sense.

7. Oct 11, 2012

### Zondrina

Edited my last post.

8. Oct 11, 2012

### WilcoRogers

Uhh not if the set has a hole in the middle... then it's not convex.

9. Oct 11, 2012

### Zondrina

Sorry i had a hiccup there, i also meant to say your line is also contained within the set.

So start by showing you have two points in the set, and then assume that some straight line y = mx also fits the constraints of your set and will join them together.

10. Oct 11, 2012

### Ray Vickson

Does your textbook or course notes really not have the necessary material on checking for convexity of a function? If this is the case, a simple Google search will give you more than you need.

RGV