# Second derivative positive $\implies$ convex

In proof of the following theorem;

If $f$ has a second derivative that is non-negative (positive) over an interval then $f$ is convex (strictly convex). $f$ is in real number space.,

the book I refer, uses Taylor series expansion but disregards terms of order 3 and above. So I'm not convinced of the correctness of the proof, which I paste below. Is there a way to bound the terms of order 3 and above in the follow proof? I think bounding the error of higher order terms is important in many cases. So would really appreciate a clear answer. Thanks a lot.

• The proof uses Taylor's theorem. – Jonathan Y. Oct 3 '13 at 21:15
• Note that we need $f''$ to be continuous in order to use the Lagrange form of the remainder. Hence this proof only works for the special case when $f$ is $C^2$. For the more general case where only assume that $f''$ exists (and is positive), we need something along the lines of Daron's answer down below. – Oskar Henriksson Jan 3 '18 at 17:40

To expand on Jonathan Y's comment, note that the argument of $f''$ in $(2.73)$ is $x^*$ and not $x_0$. The proof simply states that $x^*$ lies between $x$ and $x_0$. It turns, that you can pick such an $x^*$ such that $(2.73)$ is exact (that is, there a no higher order terms to begin with). Check out the Jonathan Y's link, look for the "Lagrange form" of the remainder, $R_k$, and plug in $k=1$.
• please comment on how to chose $x^*$ so that all the higher order terms vanish except the first three terms in the Taylor expansion. Thank you – Frank Moses Jun 23 '16 at 4:10
If $f$ is not convex it means that there is some interval $[a,b]$ where the line segment joining $f(a)$ and $f(b)$ is not always above the graph. Lets translate and rescale the argument of the function to make $a=0$, $b=1$. We are allowed to do this and it makes the algebra cleaner. Now there exists $c \in [0,1]$ such that $f(c)$ is above the line. By the mean value theorem there exists a point in $[0,c]$ where the derivative of $f$ is equal to the average rate of change between $f(0)$ and $f(c)$. But the average rate of change between $f(0)$ and $f(c)$ is greater than the slope of the line from $f(0)$ to f(1). Since $f'$ is increasing we have that after $c$, the function will be increasing too steeply to intersect the line from above at $f(1)$. This contradiction implies the function is convex.
• Note that this does not require reasoning by contradiction: in a nutshell, $cf(1) + (1-c)f(0) - f(c) ={}$ $c[f(1)-f(c)] - (1-c)[f(c)-f(0)] ={}$ $c(1-c)[f'(\beta)-f'(\alpha)]\geq 0$ with $\alpha\in [0,c]$, $\beta\in [c,1]$. – Alex Shpilkin Jan 31 '19 at 2:22