Bounded Variation Riddle

We are going to prove that the minimum value of $g_{1}(\pi)$ exists and it is equal to 1. We are going to divide our proof into three parts.

Part 1.

Since $g(x)$ is a function of bounded variation on $[0,\pi]$ , it can be expressed as $g_{1} - g_{2}$ where $g_{1}$ and $g_{2}$ are non-decreasing. This justifies that this representation is possible.

Part 2.

Let us defined the following functions. $\overline g_1(x)=\left\{ \begin{array}{ll} \sin(x), & 0\leq x \leq \frac{\pi}{2} \\ 1, & \frac{\pi}{2}\leq x \leq{\pi} \\ \end{array} \right.$

and

$\overline g_2(x)=\left\{ \begin{array}{ll} 0, & 0\leq x \leq \frac{\pi}{2} \\ 1-\sin(x), & \frac{\pi}{2}\leq x \leq{\pi}. \\ \end{array} \right.$

Because of the definitions, both $\bar{g}_{1}$ and $\bar{g}_{2}$ are non-decreasing, $\bar{g}_{1}(x)-\bar{g}_{2}(x)=g(x),$ $\bar{g}_{1}{(0)}= \bar{g}_{2}{(0)} = 0$ , and of course $\bar{g}_{1}{(1)} = 1.$

Lets assume $g_1$ and $g_2$ are whichever non-decreasing functions such that $g = g_{1} - g_{2}$ on $[0,\pi]$ and $g_{1}(0) = g_{2}(0) = 0$ . Since, these functions are non-decreasing and their values at $0$ is $0$ , then their values at any $x$ in the interval $[0,\pi]$ is non-negative.

We will prove that $g_{1}(x) \geq \bar{g}_{1}{(x)}$ for all $x$ in $[0, \pi]$ . We can divide the proof into two parts.

Case when $0 \leq x \leq \frac{\pi}{2}.$

Assume that $0\leq x\leq \frac{\pi}{2},$ then $g_{1}(x) - g_{2}(x) = \bar{g}_{1}{(x)} - \bar{g}_{2}{(x)},$ and hence $\bar{g_{1}}{(x)} = g_{1}(x) - g_{2}(x).$ Since $g_{2}(x) \geq 0$ for all $x$ in $[0,\frac{\pi}{2}]$ then $\bar{g_{1}}{(x)} \leq g_{1}{(x)}$ .

Case when $\frac{\pi}{2} \leq x \leq \pi$

Since $g_{1}(\frac{\pi}{2})- g_{2}(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1$ , then $g_{1}(\frac{\pi}{2}) - 1 = g_{2}(\frac{\pi}{2})$ , but $g_{2}(\frac{\pi}{2}) \geq 0$ Therefore, $g_{1}(\frac{\pi}{2}) \geq 1.$ Since $g_{1}(\frac{\pi}{2}) \geq 1$ and $g_{1}$ is non-decreasing then $g_{1}(x) \geq 1$ for all $x$ in $[\frac{\pi}{2}, \pi].$ So, $g_{1}(x) \geq \bar{g_{1}}{(x)}$ when $[\frac{\pi}{2}, \pi].$

Using the results obtained in both case discussed above, we get that $g_{1}(x) \geq \bar{g_{1}}{(x)}$ for all $x$ in $[0, \pi].$

Part 3.

In particular, $g_{1}(\pi) \geq \bar{g_{1}}{(\pi)} = 1$ and then 1 is the minimum of $\{g_{1}(\pi)|g_{1}(x) - g_{2}(x)=g(x)$ , $g_{1}$ and $g_{2}$ are non-decreasing on the interval $[0, \pi]$ and $g_{1}(0) = g_{2}(0) = 0\}.$ So, the answer to the problem is $\boxed{1}.$