Proving convexity from the non-negativity of the 2nd derivative

Asked May 03 '17 at 10:36

Active May 03 '17 at 11:15

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By definition, saying that $f$ is a convex function means

$$f(t x_1 + (1-t) x_2) \leq t f(x_1) + (1-t) f(x_2)$$

where $t \in [0,1]$. How can I prove that

$$f(tx_1+(1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$$

if and only if $f''(x) \geq 0$, assuming that the second derivative of $f$ exists?

edited May 03 '17 at 11:15

asked May 03 '17 at 10:36

Ray Cheng

1

One part: https://math.stackexchange.com/questions/1224955/proving-that-the-second-derivative-of-a-convex-function-is-nonnegative – larry01 May 03 '17 at 10:48
I am sure you also require linearity of $f$. – Bilbottom May 03 '17 at 10:49
Is the relation if and only if , or only A implies B? If yes, what implies what? – Ray Cheng May 03 '17 at 10:53
Other part: https://math.stackexchange.com/questions/513887/second-derivative-positive-implies-convex – A.Γ. May 03 '17 at 10:59
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I think linearity is not required in this question. – Ray Cheng May 03 '17 at 11:03

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