Determining Bounded Variation I want to check if $f(x) = \cos(\frac{x}{2})$, $x \in [0,2\pi]$, is of bounded variation.
I am following the definition here: http://www.math.ubc.ca/~feldman/m321/variation.pdf
However, i'm not sure how to apply it for an exercise. How do I speak about the partitions of my functions generally?
 A: You can prove that the variation is finite for any partition $P=\{0=x_0<x_1<\cdots <x_n=2\pi\}$ of $[0,2\pi]$ as follows: the variation on the partition $P$ is
$$
V_P=\sum_{k=0}^n|f(x_{k+1})-f(x_k)| = \sum_{k=0}^n| \cos (x_{k+1}/2)-\cos (x_k/2)|.
$$
Since $\cos x$ is differentiable, we can apply the mean value theorem on all the terms of the sum to get
$$
\sum_{k=0}^n| \cos (x_{k+1}/2)-\cos (x_k/2)|\leq \frac{1}{2}\sum_{k=0}^n|\sin \xi_n||x_{k+1}-x_k|\leq \frac{1}{2}\sum_{k=0}^n|x_{k+1}-x_k| =\pi,
$$
where $\xi_k\in [x_k/2,x_{k+1}/2]$ for all $k$. Since the estimate above has been obtained for an arbitrary partition $P$, we conclude that for any other partition $P'$ the same estimate holds, so the total variation of $f$ in $[0,2\pi]$ is
$$
V=\sup_{P} V_P\leq \pi.
$$
Another way of finding the total variation (actually we didn't find it in the previous argument, but only estimated by a finite value) is the following: for $g$ differentiable in $[a,b]$, the total variation of $g$ is
$$
V=\int_a^b|g'(x)|\, dx.
$$
Then, in this case we would get $V=\int_0^{2\pi}\big|\frac{1}{2}\sin \frac{x}{2}\big|\, dx=2\leq \pi$.
For the sake of completeness, I would like to mention that whenever you are asked to prove that a function is of bounded variation, you must be able to prove the finiteness of the variation $V_P$ for an arbitrary partition $P$. The only case when you will care about the form of the partitions is when you have to prove that a function is not of bounded variation. In this case you will have to find a sequence of partitions $P_n$ that make the variation tend to infinity as $n\to \infty$. For example, consider
$$
h(x)=\begin{cases} 0 &\mbox{if } x=0\\
x^2\cos(\pi/x^2) & \mbox{if } x>0, \end{cases} x\in [0,1].
$$
The function $h$ is not of bounded variation, and to prove it you can use the sequence of partitions given by $x_0=0$, $x_k=1/\sqrt{n+1-k}$, $k=1,2,\ldots n$.
