$C^1$ process and infinite variation process Let $f\in C^1(\mathcal{R}_+,\mathcal{R})$. 
Could $f$ be an infinite variation process ?
 A: $C^1$ functions functions always have bounded variation.
Here is the complete answer :
continuous derivative ($C^1$ functions) 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.
Proof :
The continuous function $|f′|$ has its greatest value $M$ on the closed interval  $[a,\ b]$ :
$$|f'(x)|\leq M,\ \ \forall x\in [a,\ b] $$
Let $D$ be an arbitrary partition of $[a,\ b]$, with the points :
$$x=a < x_1 < ... < x_n = b$$
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 :
$$S_D=\sum_{i=1}^n\mid f(x_i)-f(x_{i-1}) \mid = \sum_{i=1}^n \mid f'(\xi_i) \mid (x_i-x_{i-1}) \leq M\sum (x_i-x_{i-1})=M (b-a)$$
Thus the total variation satifies :
$$\sup_D\{all\ S_D’s\}\leq M(b−a)<∞,$$
whence $f$ is of bounded variation on the interval $[a,\ b]$.
Reference : http://planetmath.org/continuousderivativeimpliesboundedvariation
