Calculate total variation of g on a given interval.

Question

I am dealing with the following function: $$g(x) = \left\{ \begin{array}{lr} 1+\sin(x) & -\frac{\pi}{4} < x < \frac{\pi}{4} \\ -\frac{1}{2} & otherwise \end{array} \right. $$ I wish to calculate the Var of g $[-\frac{\pi}{4},\frac{\pi}{4}]$. The points of discontinuity are obviously the endpoints of my interval. I'm not sure how to start. Do I start with taking the derivative? I'm looking at examples, but they're all polynomials. This seems more complicated.

Answer 1

Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.

In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have $$ g'(x)= \begin{cases} \cos(x), &-\frac{\pi}{4}<x<\frac{\pi}{4}\\ 0, &\text{ otherwise } \end{cases} $$ and we have $\|g'(x)\|_{L^1(R)}=\sqrt{2}$.

The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-\pi/4$ is $3/2-\sqrt{2}/2$ and at $x=\pi/4$ is $3/2+\sqrt{2}/2$ and hence the total jump is $3$.

Thus, the total variation is $\sqrt{2}+3$.