Direct Proofs....deriviate non decreasing

fourwindschill

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Suppose that f: R->R is at least once differentiable. Prove, using direct proofs, that f is convex if and only if its first deriviative is non-decresing.
 
fourwindschill said:
Suppose that f: R->R is at least once differentiable. Prove, using direct proofs, that f is convex if and only if its first derivative is non-decreasing.

Please show us your work, indicating exactly where you are stuck, so that we know where to begin to help you

Hint:

Use definition of convex function;

f[tx[sub:2d4mstzd]1[/sub:2d4mstzd] + (1-t)x[sub:2d4mstzd]2[/sub:2d4mstzd]] ? tf[x[sub:2d4mstzd]1[/sub:2d4mstzd]] + (1-t)f[tx[sub:2d4mstzd]2[/sub:2d4mstzd]]

and definition of derivative.
 
Direct proofs, convexity, once differentiable.

We cannot assume that the function is twice differentiable, can we? Thats a greater restriction.....
Convexity is a type of concavity (the difference is the direction)... which is directly testable with the second derivative test. I dont see how, without second derivatives, we can possibly test concavity, directly.
 
fourwindschill said:
Direct proofs, convexity, once differentiable.

We cannot assume that the function is twice differentiable, can we? Thats a greater restriction.....
Convexity is a type of concavity (the difference is the direction)... which is directly testable with the second derivative test. I dont see how, without second derivatives, we can possibly test concavity, directly.

What does "non-decreasing" imply? Does it imply continuity?
 
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