I'm learning about functions of bounded variation and need some help with this problem :
Show that the function $ f $ is of bounded variation on $ [0,\pi] $ and find it's total variation. $$ f(x) = \cos^2(x) - 1, \;x \in [0,\pi] $$
Here's my attempt :
$ f'(x) = 2\cos(x) \frac{d}{dx} \cos(x) = 2\cos(x)(-\sin(x)) = -2\sin(x)\cos(x) = -\sin(2x) $
The function $ f $ is differentiable on $ [0,\pi] $ and $ \forall x \in [0, \pi] $ we have :
$$ \lvert f'(x) \rvert = \lvert -\sin(2x) \rvert \le 1 $$
Since the derivative of $ f $ is bounded on $ [0,\pi] $ this implies that $ f \in BV[0, \pi] $.
To calculate the total variation of $ f $ we need to determine the critical points of $ f $ :
$$ f'(x) = 0 \iff -\sin(2x) = 0 \Rightarrow x = 0 \; \text{or} \; x = \frac{\pi}{2} $$
but I don't know how to continue from here to find $ V_0^\pi f$ .
First, consider the following partition of $[0,\pi]$: $\{0,\pi/2,\pi\}$. By definition, \begin{equation} V_0^{\pi}f\geq\vert f(\pi)-f(\pi/2)\vert + \vert f(\pi/2)-f(0)\vert=1+1=2 \end{equation} Next, notice $f'(x)=-\sin(2x)$ is negative in $[0,\pi/2]$ and positive in $[\pi/2,\pi]$. So $f$ is decreasing in $[0,\pi/2]$ and increasing in $[\pi/2,\pi]$. Then, for any partition $\{a_0=0,a_1,...,a_n=\pi\}$ of $[0,\pi]$, if $a_{j+1}\leq \pi/2$, then \begin{equation} \vert f(a_{j+1})-f(a_j)\vert = -(f(a_{j+1})-f(a_j)) = f(a_j)-f(a_{j+1}) \end{equation} This implies the sum of the terms of the partition below $\pi/2$ (that is, where $f$ is decreasing) is given by \begin{equation} \sum_{j=1}^{k_1}\vert f(a_{j+1})-f(a_j)\vert = \sum_{j=1}^{k_1}(f(a_j)-f(a_{j+1})) = f(0)-f(a_{k_1+1}) \end{equation} for $k_1$ such that $a_{k_1+1}\leq \pi/2\leq a_{k_1+2}$. Likewise, the sum of the terms of the partition above $\pi/2$ (where $f$ is increasing) is given by \begin{equation} \sum_{j=k_1+2}^{n}\vert f(a_{j+1})-f(a_j)\vert = \sum_{j=k_1+2}^{n}f(a_{j+1})-f(a_j) = f(\pi)-f(a_{k_1+2}) \end{equation} Finally, \begin{eqnarray} \vert f(a_{k_1+2})-f(a_{k_1+1})\vert & \leq & \vert f(a_{k_1+2})-f(\pi/2)\vert + \vert f(\pi/2)-f(a_{k_1+1})\vert\\ & = & f(a_{k_1+2}) - f(\pi/2) + f(a_{k_1+1}) - f(\pi/2) \end{eqnarray} Then \begin{eqnarray} \sum_{j=1}^{n}\vert f(a_{j+1})-f(a_j)\vert & = &\sum_{j=1}^{k_1}f(a_{j+1})-f(a_j)+\vert f(a_{k_1+2})-f(a_{k_1+1})\vert+\sum_{j=k_1+2}^{n}f(a_{j+1})-f(a_j)\\ & = & f(0)-f(a_{k_1+1})+\vert f(a_{k_1+2})-f(a_{k_1+1})\vert+f(\pi)-f(a_{k_1+2}) \\ & \leq & f(0)-f(\pi/2)+f(\pi)-f(\pi/2)\\ & = & 2 \end{eqnarray} This shows $V_0^{\pi}f\leq 2$. We conclude $V_0^{\pi}f=2$.