TITLE: Showing $A(x) = \int_{0}^{x}f(t)dt$ is convex if $f(x)$ is increasing (Alternative method)
QUESTION [5 upvotes]: This is same question as posted here: Math.StackExchange.com but I have used a different approach to prove the question and this question is regarding that approach.
After I realise that I made a mistake in understanding definition of convex criteria, this is my second attempt:
We need to show $\forall [a,b] \subseteq [0,x]$ and $\alpha \in [0,1]$ that
$$A(\alpha b + (1-\alpha ) a) \le \alpha A(b) + (1-\alpha) A(a)$$
Left side of inequation is 
$$\int_{0}^{\alpha b + (1-\alpha ) a} f(t) dt  = \int_{0}^{\alpha b} f(t) dt + \int_{\alpha b }^{ (1-\alpha ) a} f(t)dt\\
 = \color{blue} {\alpha \int_{0}^{b}f(\alpha t) dt} + \color{green}{(1-\alpha) \int_{\tfrac{\alpha b}{1-\alpha}}^{a} f((1-\alpha)t) dt}$$
Right side of inequation is
$$\color{blue}{\alpha\int_{0}^{b} f(t) dt}+ \color{green}{(1-\alpha)\int_{0}^{a} f(t) dt}$$
While clearly we have blue parts satisfying 
$$\alpha \int_{0}^{b}f(\alpha t) dt \lt\alpha\int_{0}^{b} f(t) dt $$
I cannot see how to show the same for green
$$(1-\alpha) \int_{\tfrac{\alpha b}{1-\alpha}}^{a} f((1-\alpha)t) dt \lt_{\color{red}{??}} (1-\alpha)\int_{0}^{a} f(t) dt$$
Any hint to redeem this method is welcome!

REPLY [2 votes]: If $g:[a,b]\to \mathbb R $ is continuous, then $g$ is convex on $[a,b]$ iff $g$ is midpoint convex on $[a,b].$ (The latter property means $g( (1/2)x+ (1/2)y)\le  (1/2)g(x)+ (1/2)g(y)$ for all $x,y\in [a,b].$) I'll use this to give a simple proof of convexity in your problem.
Suppose $f$ is increasing on $[0,b]$ and define $A(x)$ accordingly. Note that $A(x)$ is continuous on $[0,b].$ Let $0\le x <y\le b.$ Let $m$ be the midpoint of $[x,y],$ i.e., $m= (1/2)x+ (1/2)y.$ Then
$$\int_0^m f = \int_0^x f + \int_x^m f = \int_0^y f - \int_m^y f.$$
Thus
$$\tag 1 \int_0^m f = (1/2)\left (\int_0^x f + \int_0^y f + \int_x^m f  -  \int_m^y f\right ).$$
But note that because $[x,m],[m,y]$ have the same length, we have $\int_x^m f  \le \int_m^y f.$ Here we've used the fact that $f$ is increasing. It follows that the right side of $(1)$ is $\le (1/2)\int_0^x f + (1/2)\int_0^y f.$ So we have shown $A(m) \le (1/2)A(x) + (1/2)A(y).$ Thus $A$ is midpoint convex, hence convex.