# Calculate total variation on $[0,2\pi]$ of $f(x)$

Calculate total variation on $$[0,2\pi]$$ of $$f(x)$$, where

\begin{align*} f(x) = \begin{cases} \sin(x), & \text{if x \in [0,\pi)} \\ 2, & \text{if x= \pi}\\ \cos(x), & \text{if x \in (\pi,2\pi]} \end{cases} \end{align*}

My idea Let $$P$$ a partition of $$[0,2\pi]$$, where exists $$r$$ such that $$x_r=\pi$$. Then,

$$\sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|=\sum_{i=1}^{r-1} |f(x_{i})-f(x_{i-1})|+|f(x_r)-f(x_{r-1})|\\ +|f(x_{r+1})-f(x_r)|+ \sum_{i=r+2}^{n} |f(x_{i})-f(x_{i-1})| \\= \sum_{i=1}^{r-1} |\sin(x_{i})-\sin(x_{i-1})|+|2-\sin(x_{r-1})|+|\cos(x_{r+1})-2|\\+ \sum_{i=r+2}^{n} |\cos(x_{i})-\cos(x_{i-1})|$$ but how can I bound that sums?, these sums converges and I didn´t know it?, and what happen if the point $$\pi$$ is not in the partition? But the real problem is calculate the $$\mathrm{sup}\sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|$$ and I suppose that is $$2$$, could you guide me?. Thanks, for your help.

• The mean value theorem may be useful! Apr 21, 2020 at 18:31

Total variation is additive over subintervals as $$V_a^b(f) = V_a^c(f)+ V_c^b(f)$$ for $$a < c < b$$.

Thus, $$V_0^{2\pi}(f) = V_{0}^{\pi/2}(f)+V_{\pi/2}^{2\pi}(f) = 1 + V_{\pi/2}^{2\pi}(f)$$ since the sine function is monotone increasing on $$[0,\pi/2]$$ and $$V_{0}^{\pi/2}(f) = |\sin(\pi/2) - \sin(0)|=1$$.

Taking a partition $$\frac{\pi}{2} = x_0 < x_1< \ldots < x_{r-1} < x_r = \pi < x_{r+1} < \ldots we have, using the monotonicity of $$f$$ on $$[0,\pi/2)$$ and $$(\pi,2\pi]$$,

$$\sum_{j=1}^n |f(x_j) - f(x_{j-1})|\\ = \sum_{j=1}^{r-1}|\sin(x_j)- \sin(x_{j-1})| + |f(x_r) - \sin(x_{r-1})|+ |\cos(x_{r+1}) - f(x_r)| + \sum_{j=r+1}^{n}|\cos(x_j)- \cos(x_{j-1})|\\ = 1 - \sin(x_{r-1}) + |2 - \sin(x_{r-1})| + |2 - \cos (x_{r+1})| +1 - \cos(x_{r+1})$$

The supremum over partitions is approached as $$x_{r-1} \to \pi$$ and $$\sin(x_{r-1}) \to 0$$ from the left and $$x_{r+1} \to \pi$$ and $$\cos(x_{r+1}) \to -1$$ from the right. Thus,

$$V_{0}^{2\pi}(f) = 1 + V_{\pi/2}^{2\pi}(f) = 1+8 = 9$$

Note that the sum over a partition that excludes the single point $$\pi$$ is always dominated by a sum over a partition with $$\pi$$ included, since

$$|f(x_{r+1}) - f(x_{r-1})| \leqslant |f(x_{r+1}) - f(x_{r})|+ |f(x_{r}) - f(x_{r-1})|$$