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Convexity of continuous real function, midpoint convex

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume f is a continuous real function defined in (a,b) such that f([tex]\frac{x+y}{2}[/tex])<=[tex]\frac{f(x)+f(y)}{2}[/tex] for all x,y in(a,b) then f is convex.

    2. Relevant equations

    3. The attempt at a solution
    my attempt is to suppose there are 3 points p<r<q such that f(r)>g(r), where g(x) is the straight line connecting (p,f(p)) and (q,f(q)), and get a contradiction.
    1, There exists a neighborhood of r such that any x in N(r) implies that f(x)>g(x), followed by continuity.
    2, Let E_0={p,q}, define E_1 = E_0[tex]\cup[/tex]{x | x is a mid point of any two points in E_0}. we continue this procedure and let E= union of all E_n.
    3, For any points x in E, f(x)<=g(x), which can be proved straightforwardly
    4, then we can find a point t in E which is also in N(r) such that f(x)<=g(x), if E is proved to be dense in [p,q].
    5, we get a contradiction.

    Question 1, I got stuck when I came to prove that E is dense in [p,q], it is jammed...Could anyone give me some hint to prove E is dense?

    Question 2, Is there any other way(easier or harder are both welcome) to prove it?

    Question 3, by the way, how can I determine whether a point belongs to Cantor set or not? Say, 1/4?1/5?

    Thanks a lot!
  2. jcsd
  3. Sep 10, 2008 #2
    well, I think I know how to prove that E is dense in [p,q] by the following steps:
    suppose there exist a point in [p,q] and a neighborhood associated with it, such that N_0[tex]\cap[/tex]E = empty, and diam(N_1)=d, then I can always find another segment N_1, where diam(N_1)=2d... suppose we have find that N_n such that diam(N_n)=(2^n)*d. If diam(N_n)>(q-p)/2, the point (q+p)/2 is in N_n, contradicting that (q+p)/2 is also in E.
    Hence E is dense in [p,q].

    so, any other way to prove it? I think it could be done in an easier way...
  4. Sep 10, 2008 #3


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    I think you can prove it a lot more simply. You've picked r so f(r)>g(r). The set of all x such that f(x)>g(x) is open (since both functions are continuous). So there's an open interval around r in the set. Take the endpoints of the open interval to be x and y. Doesn't that work? I don't see why you have to worry about any set being dense.
  5. Sep 10, 2008 #4
    This is cool! I didn't realize it... Thanks a lot
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