The discussion revolves around proving the convexity of a differentiable function using the Mean Value Theorem. The original poster presents a statement that a function is convex if and only if a specific inequality involving its derivative holds for all points in a given interval.
Participants are actively engaging with the problem, raising questions about the proof structure and the nature of "if and only if" statements. Some guidance is offered regarding the relationship between the function and its tangent lines, but there is no consensus on the approach to take.
There is an emphasis on understanding the definitions and implications of convexity, as well as the limitations of certain logical statements in mathematical proofs. Participants express uncertainty about the completeness of their reasoning and the necessity of exploring both directions of the proof.
I'm not sure what you mean here. You want to prove that the straight line between (a,f(a)) and (b,f(b)), which is y= (f(b)-f(a))/(b-a) (x- a)+ f(a) lies above the curve y= f(x). That is, that (f(b)-f(a))/(b-a) (x- a)+ f(a)> f(x) for all x between a and b.barksdalemc said:Homework Statement
Let f be differentiable on (a,b). Show that f is convex if and only if for every x,y in (a,b), f(y)-f(x)>= (y-x)f'(x)
The Attempt at a Solution
The mean value theorem says that there exists an x' in (a,b) such that f'(x') is the average rate of change of the functions. So I have the equation for that tangent line. I am stuck there.
Good heaven's no! There are plenty of theorems that are true in one direction but false in the other!barksdalemc said:If I prove one direction, is the proof in the other direction just the logic going the other way? In any case, let's say I want to show it is convex given for every x,y in (a,b), f(y)-f(x)>= (y-x)f'(x)
barksdalemc said:If I prove one direction, is the proof in the other direction just the logic going the other way?