Cosine as a difference of Increasing Functions

First, we are going to state and prove the following simple Lemma.

Lemma: If a function $g(x)$ is strictly increasing on the interval $[a,b]$ and $g(x)=g_1(x)-g_2(x)$ for all $x$ in the interval $[a,b],$ where $g_1$ and $g_2$ are both strictly increasing on the given interval. Then $g_1(b)>g_1(a)+g(b)-g(a).$

Proof of the Lemma: Of course, $g_2(x)=g_1(x)-g(x)$ for all $x$ in $[a,b],$ and $g_2$ is strictly increasing on that interval. Then $g_2(b)>g_2(a).$ Using that $g_2=g-g_1,$ we get that $g_1(b)-g(b)> g_1(a)-g(a).$ Therefore, $g_1(b)>g_1(a)+g(b)-g(a).$ So, the lemma is proved.

Now, let solve the problem. The existence of the functions $f_1$ and $f_2$ follows from the fact that the function $f$ is a bounded variation function. We are going to start proving that $f_1(4\pi) >5.$

1) $f_1(\pi)>f_1(0)=1,$ due to the fact that $f_1$ is increasing on the interval $[0. \pi].$

2) $f_1 (2\pi)> f_1(\pi)+2$ , using the fact that $f$ is increasing on $[\pi, 2\pi],$ and the Lemma

3) $f_1(3\pi)> f_1(2\pi),$ due to the fact that $f_1$ is increasing on the interval $[2\pi, 3\pi].$

4) $f_1(4\pi)> f_1(3\pi) +2,$ using the fact that $f$ is increasing on $[3\pi, 4\pi],$ and the Lemma

Combining these four inequalities, we obtain that $f_1(4\pi) >5.$

Now, we will prove that $c=5.$ Since we already have that $f_1(4\pi) >5$ is always true, this proves that $c\geq 5.$ Then to prove that $c$ cannot be greater than 5, it would be enough to construct two families of strictly increasing functions $f_{1,t}$ and $f_{2,t}$ (parametrized by a positive number $t$ ) defined on the interval $[0, 4\pi]$ such that $f(x) = f_{1,t}(x)-f_{2,t}(x)$ and such that $\lim_{t->0^+}{f_{1,t}(4\pi)}=5.$

The rest of the proof explains the construction of these two families of functions. We start our construction, defining the function $\bar {f}_1$ in the following way:

$\bar {f}_1(x) = \left\{ \begin{array}{ll} 1, & \quad 0\leq x \leq \pi \\ \cos x +2, & \quad \pi\leq x \leq 2\pi \\ 3, & \quad 2\pi\leq x \leq 3\pi \\ \cos x+4, & \quad 3\pi\leq x \leq 4\pi \end{array} \right.$ and we define $\bar {f}_2(x)=f(x)- \bar {f}_1(x).$ These two functions that we have defined are non-decreasing on the interval $[0, 4\pi]$ and $f=\bar{f}_1-\bar{f}_2,$ $\bar{f}_1(0)=1, \bar{f}_2(0)=0.$ Now for any positive real number $t,$ we define $f_{1,t}(x)= \bar{f}_1(x) +tx$ and $f_{2,t}(x)= \bar{f}_2(x) +tx.$ It is obvious that these two functions are strictly increasing and $f(x) = f_{1,t}(x)-f_{2,t}(x),$ $f_{1,t}(0)=1, f_{2,t}(0)=0.$ Besides that, $\lim_{t->0^+}{f_{1,t}(4\pi)}=\bar{f}_1(4\pi)= 5.$ Therefore, the answer to this question is $\boxed{5}.$

$f_1(2\pi)=f_2(2\pi)+cos(2\pi)=f_2(2\pi)+1>f_2(\pi)+1=f_1(\pi)-cos(\pi)+1=f_1(\pi)+2>f_1(0)+2$

$f_1(4\pi)=f_2(4\pi)+cos(4\pi)=f_2(4\pi)+1>f_2(3\pi)+1=f_1(3\pi)-cos(3\pi)+1=f_1(3\pi)+2>f_1(2\pi)+2$

$f_1(4\pi)>f_1(2\pi)+2>f_1(0)+2+2=5$

A non-rigorous approach

$f_1$ strictly increases $\implies f_1^\prime>0\implies -\sin(x)+f_2^\prime>0\implies f_2^\prime(x)>\sin(x)$

$f_2$ strictly increases $\implies f_2^\prime>0$

$\implies f_2^\prime>\max(0,\sin(x))$

Visualizing, $f_2$ increases by at least the area under one sine 'hump', 2, every $2\pi$ increase in $x$

$f_1=f+f_2$ , intuitively, shows that minimizing $f_2$ minimizes $f_1$

Wherefore, as there are two sine 'humps' between $x=0$ and $x=4\pi$ and $f_1(0)=1$ , $f_1(4\pi)>1+2\times2$