total variation of uniformly bounded function If a function is uniformly bounded and has finite variation, is the finite variation less than the uniform bound times a constant?
Thank you.

No.
Take functions of $C([0;1])$ for example. Take $f_n(0)=0$ and $f_n'(x)=1$ if $\frac{2k}{n}\leq x < \frac{2k+1}{n}$ for some integer $k$ and $f_n'(x)=-1$ otherwise. those functions are saw shaped, you can easily show that $||f_n||_\infty = \frac{1}{n}$ and the total variation of $f_n$ is always one.
Reciprocally if $f(x)=1 \; \forall x$ the total variation is $0$ but the uniform norm is 1. As you can see the total variation isn't a norm over function of bounded variation. However, $N_x: f \mapsto |f(x)|+V(f)$ (where $V(f)$ is the total variation of $f$) is in fact a norm with the following property : $N_x(f)\geq ||f||_\infty$. 
