So recently one of my friends came up with this conjecture :
For an increasing, concave-up and continuous function $f$ defined on $\mathbb{R}$, the sequence $\{ a_{n} \}$ satisfies $f(a_{n}) = (n-1) \pi$ for all $n \in \mathbb{N}$. Then, $\int_{a_{n+1}}^{a_{n+2}} | \sin f(x) | \ dx > \int_{a_{n}}^{a_{n+1}} | \sin f(x) | \ dx$.
The following is his solution, assuming that $f$ is differentiable too.
Since $f$ is differentiable, $f'$ decreases and $f'$ is positive. Then,
$$\int_{a_{n+1}}^{a_{n+2}} f'(a_{n+1}) | \sin f(x) | \ dx > \int_{a_{n+1}}^{a_{n+2}} f'(x) | \sin f(x) | \ dx$$
substitute $u = f(x)$ then
$$\int_{a_{n+1}}^{a_{n+2}} f'(x) | \sin f(x) | \ dx = \int_{n \pi}^{(n+1) \pi} | \sin u | \ du$$
$$= \int_{(n-1) \pi}^{n \pi} | \sin u | \ du = \int_{a_{n}}^{a_{n+1}} f'(x) | \sin f(x) | \ dx$$
$$> \int_{a_{n}}^{a_{n+1}} f'(a_{n+1}) | \sin f(x) | \ dx$$
Therefore,
$$\int_{a_{n+1}}^{a_{n+2}} f'(a_{n+1}) | \sin f(x) | \ dx > \int_{a_{n}}^{a_{n+1}} f'(a_{n+1}) | \sin f(x) | \ dx$$
and
$$\int_{a_{n+1}}^{a_{n+2}} | \sin f(x) | \ dx > \int_{a_{n}}^{a_{n+1}} | \sin f(x) | \ dx$$
since $f'$ is positive.
He asked me if this inequality holds even if $f$ is only continuous, maybe not differentiable.
The following is my solution, which is pretty much the same.
$\color{red}{\rm NOTE}$ : You do not have to read the whole proof. The main idea is that a convex function can be divided into countable differentiable functions, since it has at most countable non-differentiable points. Separate the integral into the sum of each integral and apply the inequality above, and add up the sum then it is easily proven.
Since $f$ is convex, it has at most countable non-differentiable points. (Proof here) Let these functions be $f_{n, i}$ for $1 \le i \le r$. Let the domain of each function be $D_{n, i}$ then $D_{n, i}$ is a separation of $(a_{n}, a_{n+1})$. That is, if $i \ne j$ then their intersection is an empty set. Let $D_{n, i} = (b_{n, i-1}, b_{n, i})$. ($b_{n, 0} = a_{n}$ and $b_{n, r} = a_{n+1}$.)
Each $f'_{n, i}$ is decreasing and they are positive. Also, since every value of $f'_{n, i}$ is larger than $f'_{n, i+1}$, $f'_{n, 0} (a_{n}) > f'_{n, i} (x)$. Split the integral into $r$ integrals of each $f_{n, i}$, then apply the inequality to get :
$$\int_{a_{n+1}}^{a_{n+2}} f'_{n+1, 0} (a_{n}) | \sin f(x) | \ dx = \sum_{i=0}^{r-1} \int_{b_{n+1, i}}^{b_{n+1, i+1}} f'_{n+1, 0} (a_{n}) | \sin f_{n+1, i+1} (x) | \ dx$$
$$> \sum_{i=0}^{r-1} \int_{b_{n+1, i}}^{b_{n+1, i+1}} f'_{n+1, i+1} (x) | \sin f_{n+1, i+1} (x) | \ dx$$
Substitute each $f_{n+1, i+1}$ to $u$ then
$$\sum_{i=0}^{r-1} \int_{b_{n+1, i}}^{b_{n+1, i+1}} f'_{n+1, i+1} (x) | \sin f_{n+1, i+1} (x) | \ dx = \sum_{i=0}^{r-1} \int_{f_{n+1, i+1} \:\: (b_{n+1, i}\:)}^{f_{n+1, i+1} \:\: (b_{n+1, i+1}\:)} | \sin u | \ du$$
Since $b_{i}$ ($0 \le i \le r$) are all in $(a_{n}, a_{n+1})$,
$$\sum_{i=0}^{r-1} \int_{f_{n+1, i+1} \:\: (b_{n+1, i}\:)}^{f_{n+1, i+1} \:\: (b_{n+1, i+1}\:)} | \sin u | \ du = \int_{n \pi}^{(n+1) \pi} | \sin u | \ du = \int_{(n-1) \pi}^{n \pi} | \sin u | \ du$$
Now break down the integral into $r$ integrals of $f_{n, i}$ and substitute $u = f_{n, i} (x)$. Use the inequality $f'_{n+1, 0} (a_{n}) < f'_{n, i} (x)$ to make $f'_{n+1, 0} (a_{n})$ and add up the sum of integrals to obtain $\int_{a_{n}}^{a_{n+1}} f'_{n+1, 0} (a_{n+1}) | \sin f(x) | \ dx$. Since $f'_{n+1, 0} (a_{n})$ is positive, we get the desired inequality.
This is the real question :
The proof above has no problem if $r$ is finite, but does it hold even when $r$ is countable? If there are infinite non-differential points then I don't see the difference between $r$ being countable and uncountable. I misunderstood the lemma that there should be finite non-differentiable points, and I have no idea whether this will work for countable-but-infinite $r$.
If it is wrong or unclear to apply this solution for $r$ being infinite, how would one prove the conjecture differently? Is the conjecture even true for infinite cases?
