Results 1 to 3 of 3

Math Help - Convex Functions

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    147

    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:

    Proof by contradiction

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member bondesan's Avatar
    Joined
    Jul 2010
    From
    Brazil
    Posts
    58
    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.
    Last edited by bondesan; October 8th 2010 at 10:23 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    147
    Quote Originally Posted by bondesan View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: December 1st 2010, 10:48 PM
  2. Convex functions and increasing derivatives
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 10th 2010, 07:54 PM
  3. Sum of convex functions has a minimum
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: May 17th 2010, 12:39 PM
  4. convex functions
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 12th 2009, 01:19 PM
  5. Convex Functions
    Posted in the Calculus Forum
    Replies: 0
    Last Post: January 28th 2009, 01:25 PM

Search Tags


/mathhelpforum @mathhelpforum