# Math Help - Convex Functions

1. ## Convex Functions

Let f(x) be a convex function st f(0)=0 and f(1)=1

Prove that if A > B, then we cannot have that for all c in (0,1);

A*f(c) + B*f(1-c) > B

I was actually given an answer to this, but can't seem to understand it, I was hoping for some help:

Here is the solution:

The above statement is equivalent to:

$\frac{f(c)}{1-f(1-c)} > \frac{A}{B}$ for all c in (0,1)

But since f is convex, then (omitting a few steps):

$\frac{f(c)}{1-f(1-c)} < 1$ for all c in (0,1)

Up to here I understand perfectly, the prof then seemed to make a giant leap by saying that since this implies that

$\frac{f(c)}{1-f(1-c)}$ can take on all values between in (0,1) for all c in (0,1), we can find a c* such that

$\frac{f(c*)}{1-f(1-c*)} < \frac{A}{B}$ and we have a contradiction

That's what I was given: no limits, continuity or derivatives.

Can someone see how that last step follows smoothly from what came before?

2. There's a typo when you say f(c)/1-f(1-c) > A/B, the correct would be f(c)/1-f(1-c) > B/A by the enunciate.

First you assumed you had A > B, which tell us for any values of A and B that the fraction B/A is always less than 1. So, f(c)/1-f(1-c) > B/A implies f(c)/1-f(1-c) < 1.

You showed also that f(c)/1-f(1-c) < 1 for any c in (0,1) by the convex propriety. Since you have f(0)=0 and f(1)=1 and all the other proprieties of this kind of function, you can say that f(c)/1-f(1-c) can be any number between the interval (0,1). As B/A is also a number between (0,1), by saying that exists a c* which makes f(c*)/1-f(1-c*) < B/A you are proving that there exists at least one c in (0,1) which makes the first inequality false (because you started with the premise f(c)/1-f(1-c) > B/A) and this completes the proof, because it is the contrapositive of the first affirmative (not for all c's implies that exists a c which does not hold the inequality).

I hope it helps.

3. Originally Posted by bondesan
You showed also that f(c)/1-f(1-c) < 1 for any c in (0,1) by the convex propriety. Since you have f(0)=0 and f(1)=1 and all the other proprieties of this kind of function, you can say that f(c)/1-f(1-c) can be any number between the interval (0,1).
Can you please expand on this part?