# Positive Second derivative and convexity

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.
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.
 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 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 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 at 7:08 @Alex, f is concave up since f''>0. – i.a.m Feb 11 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 at 7:21