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continuous derivative implies bounded variation
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 $f^{{\prime}}$ has its greatest value $M$ on the closed interval $[a,\,b]$, i.e.
$f^{{\prime}}(x)\leqq M\quad\forall x\in[a,\,b].$ 
Let $D$ be an arbitrary partition of $[a,\,b]$, with the points
$x_{0}=a<x_{1}<x_{2}<\ldots<x_{{n1}}<b=x_{n}.$ 
Consider $f$ on a subinterval $[x_{{i1}},\,x_{i}]$. By the meanvalue theorem, there exists on this subinterval a point $\xi_{i}$ such that $f(x_{i})f(x_{{i1}})=f^{{\prime}}(\xi_{i})(x_{i}x_{{i1}})$. Then we get
$S_{D}:=\sum_{{i=1}}^{n}f(x_{i})f(x_{{i1}})=\sum_{{i=1}}^{n}f^{{\prime}}(% \xi_{i})(x_{i}x_{{i1}})\leqq M\sum_{{i=1}}^{n}(x_{i}x_{{i1}})=M(ba).$ 
Thus the total variation satisfies
$\sup_{{D}}\{\mbox{all }S_{D}\mbox{'s}\}\leqq M(ba)<\infty,$ 
whence $f$ is of bounded variation on the interval $[a,\,b]$.
$2^{{\underline{o}}}$. Define the functions $G$ and $H$ by setting
$G:=\frac{f^{{\prime}}+f^{{\prime}}}{2},\quad H:=\frac{f^{{\prime}}f^{{% \prime}}}{2}.$ 
We see that these are nonnegative and that $f^{{\prime}}=GH$. Define then the functions $g$ and $h$ on $[a,\,b]$ by
$g(x):=f(a)+\int_{a}^{x}G(t)\,dt,\quad h(x):=\int_{a}^{x}H(t)\,dt.$ 
Because $G$ and $H$ are nonnegative, the functions $g$ and $h$ are monotonically nondecreasing. We have also
$(gh)(x)=f(a)\!+\!\int_{a}^{x}(G(t)\!\!H(t))\,dt=f(a)\!+\!\int_{a}^{x}f^{{% \prime}}(t)\,dt=f(x),$ 
whence $f=gh$. Since $G$ and $H$ are by their definitions continuous, the monotonic functions $g$ and $h$ have continuous derivatives $g^{{\prime}}=G$, $h^{{\prime}}=H$. 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.
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