Proving Convexity of Functions Using the Mean Value Theorem

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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.
 
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There's two directions to prove, so which one are you asking about? And what is the definition of a convex function?
 
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:

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.
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:
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)
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?

That's certainly not true in general. If we stick to the direction you mentioned, you can rearrange and get:

f(x) \geq f(x_0) + f'(x_0)(x-x_0)

Or in other words, f lies above every line tangent to f. From here it's easy to show the function is convex, it's just a matter of plugging into the defintion (which I'm not going to copy for you). The other direction will be a little harder.
 
Halls of Ivy,

I meant for theorems which state if and only if. Are there if and only if statements where the logic cannot backwards?
 
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