# which condition says that $f$ is necessarily bounded variation

Which of the following condition below imply that the $f:[0,1]\to\mathbb{R}$ is necessarily Bounded Variation?

1. monotone;
2. continuous and monotone;
3. has derivative on $(0,1)$;
4. bounded derivative on $(0,1)$.

I studied one result long ago that a Monotone function is of bounded variation, and The function $\cos \frac{\pi}{x}$ is continous but not a function of founded variation I can not recall the proof. for $4$ by MVT we get $|f(x)-f(y)|<K|x-y|$ where $|f'(x)|<K$ Then $\sum_{i=1}^{n}|f(x_i)-f(x_{i-1})|$ is definitely bounded and so $4$ is also true, could any one tell me what else can be said?

• If $f$ is monotone, how can you write $|f(x_i)-f(x_{i-1})|$ differently? - For 3.: what about the function you remembered was not of bounded variation? – Thomas Apr 18 '13 at 8:44

1. implies BV because due to monotonicity the sum $\sum |f(x_i)-f(x_{i-1})|$ becomes either $\sum f(x_i)-f(x_{i-1})$ or $-\sum f(x_i)-f(x_{i-1})$ and subsequently telescopes.
3. Indeed, $\cos(\pi/x)$ (defined arbitrarily at $0$) is differentiable on $(0,1)$ and does not have bounded variation, as you can check by considering partition points $\{x_1,\dots,x_n\} = \{n^{-1},(n-1)^{-1},\dots, 1/2\}$ where $n$ can be as large as you want.
4. implies $f$ is Lipschitz, and a Lipschitz function is BV because $\sum |f(x_i)-f(x_{i-1})|\le L\sum (x_i-x_{i-1})$ where the latter sum telescopes.