Convexity of continuous real function, midpoint convex

[boombaby]

Homework Statement

Assume f is a continuous real function defined in (a,b) such that f(\frac{x+y}{2})<=\frac{f(x)+f(y)}{2} for all x,y in(a,b) then f is convex.

Homework Equations

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\cup{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!

 

[boombaby]

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\capE = 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...
Thanks!

 

[Dick]

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.

 

[boombaby]

The set of all x such that f(x)>g(x) is open (since both functions are continuous)

This is cool! I didn't realize it... Thanks a lot