A Minimal Property of Monotonic Functions

             "Monotonic" has here the meaning of "increasing" or "decreasing".

      Problem 20 475 (for grade 12, published in GMB_6--1985) solved in GMB_6--1986, page 212 :

Solution

               We will show a seemingly simpler inequality :

          If $\;g\;:\;[0,1]\;\rightarrow \;\mathbb{R}$ is increasing, then

$$\int_0^1\left | g(t)-c \right | dt\;\geqslant\;\int_0^1\left | g(t)-g\left (\frac{1}{2}\right ) \right | dt\;,\;\;(\forall )c\in \mathbb{R}. \tag{1}$$

 

          If we demonstrated (1), then taking $\;g(t)=f\left (a+(b-a) \cdot t \right )$, we have $g$ is increasing if $f$ is increasing, so

 $$\int_0^1 \left | f \left ( a+(b-a) \cdot t \right )-c \right | dt \geqslant \int_0^1 \left | f \left ( a+(b-a) \cdot t \right ) -f \left ( \frac{a+b}{2} \right ) \right | dt.$$

In the above integrals we substitute

$$\begin{cases}x=a+(b-a)\cdot t \;,\;\;t=\frac{x-a}{b-a}\\t=0\;\leftrightarrow \;x=a\;,\;\;t=1\;\leftrightarrow \;x=b \end{cases}$$

and we get

 $$\int_a^b \left |f(x)-c \right | \cdot \frac{dx}{b-a}\geqslant \int_a^b \left | f(x)-f\left ( \frac{a+b}{2} \right ) \right | \cdot \frac{dx}{b-a}$$

 which multiplied by $b-a >0$ leads us to the original statement.

 

          To prove (1) we write

$$\int_0^1\left | g(x)-c \right |dx=\int_0^{\frac{1}{2}}\left |g(x)-c \right |dx+\int_{\frac{1}{2}}^1\left | g(x)-c \right |dx.\tag{2}$$

In the last integral of the (2) we replace $\begin{cases}x=u+\frac{1}{2}\\x=\frac{1}{2}\;\to u=0\\x=1\;\to u=\frac{1}{2}\end{cases}$

$\int_{\frac{1}{2}}^1\left |g(x)-c \right | dx\;\underset{x=u+\frac{1}{2}}{==}\;\int_0^{\frac{1}{2}}\left |g(u+\frac{1}{2})-c \right |du=\int_0^{\frac{1}{2}}\left |g(x+\frac{1}{2})-c \right |dx$, so (2) is written

$$\int_0^1 \left |g(x)-c \right |dx=\int_0^{\frac{1}{2}}\left | g(x)-c \right |dx+\int_0^{\frac{1}{2}}\left |g(x+\frac{1}{2})-c \right |dx. \tag{3}$$

     We have $g(x+\frac{1}{2})-g(x)\;\underset{g\nearrow}{=}\left |g(x+\frac{1}{2})-g(x) \right |=\left |\left (g(x+\frac{1}{2})-c \right )-(g(x)-c) \right |\;\;\underset{|a-b|\leqslant |a|+|b|}{\leqslant}\;\;$

 $$\leqslant \left |g(x+\frac{1}{2})-c \right |+|g(x)-c|$$

and taking the integral we get

$$\int_0^{\frac{1}{2}}\left ( g(x+\frac{1}{2})-g(x) \right ) dx \leqslant \int_0^{\frac{1}{2}} \left | g(x+\frac{1}{2})-c \right |dx+\int_0^{\frac{1}{2}} \left | g(x)-c \right |dx \underset{(3)}{=}\int_0^1 |g(x)-c|dx.$$

         So we have inequalities :

$$\int_0^1 |g(x)-c|dx \geqslant \int_0^{\frac{1}{2}}\left (g(x+\frac{1}{2})-g(x) \right )dx\;\;\;\Leftrightarrow$$

 $$\Leftrightarrow \;\int_0^1 |g(x)-c|dx \geqslant \int_0^{\frac{1}{2}}\left \{ \left [g(x+\frac{1}{2})-g(\frac{1}{2}) \right ]+ \left [g(\frac{1}{2})-g(x) \right ] \right \}dx \;\Leftrightarrow$$

 $$\Leftrightarrow \;\int_0^1|g(x)-c|dx \geqslant \int_0^{\frac{1}{2}}[g(x+\frac{1}{2})-g(\frac{1}{2})]dx+\int_0^{\frac{1}{2}}[g(\frac{1}{2})-g(x)]dx.\tag{4}$$

 The second integral in (4) becomes, with the substitution $\begin{cases}v=x+\frac{1}{2}\\x=0\leftrightarrow v=\frac{1}{2}\\x=\frac{1}{2}\leftrightarrow v=1\end{cases}$

$\int_0^{\frac{1}{2}}[g(x+\frac{1}{2})-g(\frac{1}{2})]dx\;\overset{v=x+\frac{1}{2}}{=}\;\int_{\frac{1}{2}}^1[g(v)-g(\frac{1}{2})]dv=\int_{\frac{1}{2}}^1[g(x)-g(\frac{1}{2})]dx,$

and, because on the set $[\frac{1}{2},1]$ we have $g(x)-g(\frac{1}{2})=|g(x)-g(\frac{1}{2})|$, we can't write

$$\int_0^{\frac{1}{2}}[g(x+\frac{1}{2})-g(\frac{1}{2})]dx=\int_{\frac{1}{2}}^1\left |g(x)-g(\frac{1}{2})\right |dx.$$

With the one above, and with the fact that on the set $[0,\frac{1}{2}]$ we have $g(\frac{1}{2})-g(x)=|g(\frac{1}{2})-g(x)|$, the  relationship (4) is written 

$\int_0^1|g(x)-c|dx \geqslant \int_{\frac{1}{2}}^1\left |g(x)-g(\frac{1}{2})\right |dx+\int_0^{\frac{1}{2}}\left |g(x)-g(\frac{1}{2})\right |dx=\int_0^1\left |g(x)-g(\frac{1}{2}) \right | dx,$

 and (1) is demonstrated.

 

          Let now be $f$ decreasing function. Then $-f$ is increasing and for $-f$ we apply the demonstrated property, taking the constant equal to $-c$ :

$$\int_a^b\left | (-f(x))-(-f\left (\frac{a+b}{2} \right ))\right | dx \leqslant \int_a^b\left |(-f(x))-(-c)\right |dx\;\Leftrightarrow$$

 $$\Leftrightarrow\;\int_a^b\left |f\left (\frac{a+b}{2}\right )-f(x) \right |dx\leqslant \int_a^b|c-f(x)|dx$$

 and inequality results again.

$\blacksquare$