Showing $\alpha (x) = x^2$ is of bounded variation on $[-2,1]$

Question

I have just started learning about bounded variation, and I am confused about how to actually show it. Now I know the definition is that the set of all sums $\Sigma_{i=1}^n |\alpha(x_i) - \alpha(x_{i-1})|$ is bounded. But how do I actually show this? My thought so far is that the widest variation is between -2 and 0. (i.e. the difference here is 4). And this shows the upper bound is 4? Am I on the right track?

Answer 1

Since the function is differentiable, we know the total variation is just the integral of the derivative's absolute value, i.e.

$$V(x^2, [-2,1])=\int_{-2}^1|2x|\,dx < \infty$$

which of course is easy as $|2x|\le 4$ on $[-2,1]$, so the integral is bounded by $12$. (You can also work out the variation if you like, it's very easy to see it is just $5$ by computing the integral directly)

Answer 2

You can use the mean value theorem: there exists points $\xi_i \in [-2,1]$ such that

$$\sum_{i=1}^n|a(x_i)-a(x_{i-1})| = \sum_{i=1}^n |a'(\xi_i)||x_i-x_{i-1}| $$ $$= 2\sum_{i=1}^n |\xi_i| |x_i-x_{i-1}|\le 4 \sum_{i=1}^n (x_i-x_{i-1}) \le 12.$$

Answer 3

One way is to use MVT:

Choose a partition of $[-2,1], P=\left \{ -2,x_2\cdots,x_{n-2},1 \right \}.$ Then there are $x_{i-1}<c_i<x_i$ such that $|\alpha(x_i)-\alpha(x_{i-1})|\le |\alpha'(c_i)(x_i-x_{i-1})|.$ But $|\alpha'(x)|\le 4$ on $[-2,1]$ so $\sum {|f(x_i)-f(x_{i-1})}|\le 4\sum |x_{i}-x_{i-1}|=4\cdot 3=12.$

Another faster way would be to note that $V_{\alpha}([-2,1])=V_{\alpha}([-2,0])+V_{\alpha}([0,1])=4+1=5.$