Showing $A(x) = \int_{0}^{x}f(t)dt$ is convex if $f(x)$ is increasing (Alternative method)

Question

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!

Answer 1

The estimation $f(\alpha t)\le f(t)$ on $[0,b]$ is too inaccurate, you lose a lot of information about the function being increasing in the interval $[\alpha b\ b]$.

One way to prove the original inequality of convexity (if you would like to use the definition and do not want to relay on other convexity criteria like the derivative is increasing etc) is the following. I'll drop the integrand as it is the same for all integrals below. So let $0\le a<c=\lambda b+(1-\lambda)a<b\le 1$. The following inequalities are equivalent. $$ \int_0^c\le\lambda\int_0^b+(1-\lambda)\int_0^a\iff \int_a^c\le\lambda\int_a^b=\lambda\int_a^c+\lambda\int_c^b\iff (1-\lambda)\int_a^c\le\lambda\int_c^b $$ Now define $\Delta=b-a$ and divide both sides by $\lambda(1-\lambda)\Delta$ to get yet another equivalent inequality $$ \frac{1}{\lambda\Delta}\int_a^cf(t)\,dt\le\frac{1}{(1-\lambda)\Delta}\int_c^bf(t)\,dt. $$ Note that $\lambda\Delta=c-a$ and $(1-\lambda)\Delta=b-c$, i.e. the last inequality is simply $$ \text{the mean value of $f$ on $[a,c]$ }\le\text{ the mean value of $f$ on $[c,b]$}. $$ It is true for increasing $f$.

Answer 2

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.

Answer 3

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$The last inequality is not necessarily true.

Suppose the last inequality holds for any increasing $f$ and $0\leqslant a \leqslant b \leqslant x_0$, $0 \leqslant α \leqslant 1$. For any increasing and bounded $f$ given, suppose $|f(x)| \leqslant M$ and define$$ F(x) = f(x) - 4M x_0, \quad \forall 0 \leqslant x \leqslant x_0 $$ then $F$ is increasing on $[0, x_0]$. But for $\dfrac{1}{2} \leqslant α < 1$ and $\dfrac{x_0}{2} \leqslant b \leqslant x_0$,\begin{align*} &\peq \int_{{\textstyle\frac{α}{1 - α}} b}^a F((1 - α)t) \,\d t - \int_0^a F(t) \,\d t\\ &= \int_{{\textstyle\frac{α}{1 - α}} b}^a f((1 - α)t) \,\d t - \int_0^a f(t) \,\d t - 4\left( a - \frac{α}{1 - α}b \right) M + 4aM\\ &= \int_{{\textstyle\frac{α}{1 - α}} b}^a f((1 - α)t) \,\d t - \int_0^a f(t) \,\d t + \frac{4αb}{1 - α} M\\ &\geqslant -2M x_0 + \frac{4αb}{1 - α} M \geqslant 0. \end{align*}