For and EDIT: and f(0)=0 and f(1)=1
Is it possible for f to be convex but not increasing?
And,
Is it possible for f to be increasing but not convex?
I'm pretty sure the first question is a "no", but how about the second one? If the answer is yes could you please provide an example?
Thanks
ln(x) is negative for all x in (0,1) !!
But I see the gist of what you're saying... a function with that shape could be increasing but not convex and satisfy the requirements I said...
OK, I'm edditing the original post with extra conditions for f.
Now is convexity the same as "increasingness"?
Hey, I don't set at the computer all day!
Okay, you want an example of an increasing function that stays within the triangle with vertices (0,0), (1, 0), (1,1), that is increasing but not always convex. Take, as a "middle" point, (1/2, 1/4). The function f(x)= has graph that goes from (0,0) to (1/2, 1/4) and is always increasing (and convex). Further it has slope 2(1/2)= 1 at (1/2, 1/4). Now smoothly attach another function to that:
Take a cubic, that
1) passes through (1/2, 1/4)
2) has slope 1 at (1/2, 1/4)
3) passes through (1, 1)
4) has slope 1 at (1, 1)
Those are 4 conditions that will allow us to determine the four coefficients, a, b, c, and d.
The graph passes through (1/2, 1/4): .
The slope at (1/2, 1/4) is 1: \frac{3a}{4}+ b+ c= 1[/tex].
The graph passes through (1,1): a+ b+ c+ c= 1.
The slope at (1,1) is 1: 3a+ 2b+ c= 1.
Solving form a, b, c, and d and attaching that cubic to should give a function which is always increasing but is becomes "non-convex" somewhere between x= 1/2 and x= 1.