Theorem. If the real function $f$ has continuous derivative on the interval $[a,\,b]$ , then on this interval,

• $f$ is of bounded variation,
• $f$ can be expressed as difference of two continuously differentiable monotonic functions.

Proof. $1^{\underline{o}}$ . The continuous function $|f'|$ has its greatest value $M$ on the closed interval $[a,\,b]$ , i.e. $$|f'(x)| \leqq M \quad \forall x \in [a,\,b].$$ Let $D$ be an arbitrary partition of $[a,\,b]$ , with the points $$x_0 = a < x_1 < x_2 < \ldots < x_{n-1} < b = x_n.$$ Consider $f$ on a subinterval $[x_{i-1},\,x_i]$ . By the mean-value theorem, there exists on this subinterval a point $\xi_i$ such that $f(x_i)-f(x_{i-1}) = f'(\xi_i)(x_i-x_{i-1})$ . Then we get $$S_D := \sum_{i=1}^n|f(x_i)-f(x_{i-1})| = \sum_{i=1}^n|f'(\xi_i)|(x_i-x_{i-1}) \leqq M\sum_{i=1}^n(x_i-x_{i-1}) = M(b-a).$$ Thus the total variation satisfies $$\sup_{D}\{\mbox{all }S_D\mbox{'s}\} \leqq M(b-a) < \infty,$$ whence $f$ is of bounded variation on the interval $[a,\,b]$ .

$2^{\underline{o}}$ . Define the functions $G$ and $H$ by setting $$G := \frac{|f'|+f'}{2}, \quad H := \frac{|f'|-f'}{2}.$$ We see that these are non-negative and that $f' = G-H$ . Define then the functions $g$ and $h$ on $[a,\,b]$ by $$g(x) := f(a)+\int_a^xG(t)\,dt, \quad h(x) := \int_a^xH(t)\,dt.$$ Because $G$ and $H$ are non-negative, the functions $g$ and $h$ are monotonically nondecreasing. We have also $$(g-h)(x) = f(a)\!+\!\int_a^x(G(t)\!-\!H(t))\,dt = f(a)\!+\!\int_a^xf'(t)\,dt = f(x),$$ whence $f = g-h$ . Since $G$ and $H$ are by their definitions continuous, the monotonic functions $g$ and $h$ have continuous derivatives $g' = G$ , $h' = H$ . So $g$ and $h$ fulfil the requirements of the theorem.

Remark. It may be proved that each function of bounded variation is difference of two bounded monotonically increasing functions.