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Let $f:\mathbb R\to\mathbb R$, maps a point $x \in \mathbb R$. $f$ is twice differentiable. Show that if second derivative is positive for all $x$ then $f$ is convex

Is there anyway to prove this without using MVT/definitions of derivatives?

Basically can this be proven with just the definition of convexity/concavity and maybe quasi-convexity and lower contours. The definition we learned use for convexity is: $f$ is convex if $$ f(\theta x_a + (1-\theta) x_b) \leq \theta f(x_a) + (1-\theta) f(x_b). $$

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We know that if $f$ is differentiable in some interval, say $I$, then $f$ is concave upward in $I$ iff $$f(x)<\frac{x_2-x}{x_2-x_1}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)$$ for all $x_1,x,x_2\in I$ such that $x_1<x<x_2$. Use this fact.

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Hey Babak, I see that if I let x be the weighted average of x2 and x1 I have my result. However, can you tell me where you got that iff for f concave upward? – Alex Feb 11 '13 at 9:08
@Alex: You know this fact is a stared problem I think in R.A.Silver man's Calculus book. – S. Snape Feb 11 '13 at 9:31
Okay thank you, but I have another question.. what exactly is the difference between concave upward and convex? Because on wikipedia it states the two interchangeably. If that is the case, then then are we not using concave up to prove convexity (which is circular logic)? Or is concave up defined as an increasing first derivative? – Alex Feb 11 '13 at 9:47
@Alex: The main idea, I think, is the same as i.a.m noted below. In fact the similar deduction is true for concavity downward whit < replaced by >. – S. Snape Feb 11 '13 at 10:11
Great help you are here at M.S.E.+1 ! – amWhy Feb 12 '13 at 2:59

If $f''>0$ for all $x$, then $f$ is concave up, and if you look at the line passing throw any two points on the curve, the line will be above the curve, which implies $f$ is convex.

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Doesn't this implicitly use Taylor's theorem (and an additional assumption that $f$ is analytic) to construct the local approximation of $f$? – user7530 Feb 11 '13 at 6:59
@user7530, No I'm just using the fact that $f$ is twice differentiable and the second derivative is positive, which implies that $f$ is concave up, and if you write the equation of any line passing two points on the curve you will get the RHS of your inequality, and the LHS is f(....) and the inequality holds since the line is always above the curve by the concavity of f. – i.a.m Feb 11 '13 at 7:04
I'm sorry I don't quite undertsand the last line. I understand your concave up argument, but what do you mean by "by the concavity of f".. f is not concave. – Alex Feb 11 '13 at 7:08
@Alex, f is concave up since f''>0. – i.a.m Feb 11 '13 at 7:18
@Alex imagine the curve of x^2 then any line joining two point on this curve will be above the curve since x^2 is concave up, this is the same argument is used for the general curve f, since it is concave up since the second derivative is positive. – i.a.m Feb 11 '13 at 7:21

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