Problem:
Let $g(x)=x^{-2}sin(x^{-2})$ for $x\in [-1,1]-{0}$ and $g(0)=0$.
Show that $g$ is not of bounded variation.
Questions regarding to these kind of problems
There are some theorems which allow me to check if something is of bounded variation, but Im not sure how to use them to prove that a function is not of bounded variation, for example:
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If $g$ is differentiable at $[a,b]$ and there exists a constant $M>0$ such that $|g'(x)|\leq M$, then $g$ is of bounded variation. Now, if I can prove that g(x) is differentiable but not bounded for some $M>0$ that means $g$ is not of bounded variation? I dont think so cause I can think of some counterexample.
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If $g$ is differentiable over $[a,b]$ and $g\in L^{1}([a,b])$, then g is bounded variation and the total variation is given by $\int_{a}^{b} |g'(x)|dx$. The problem here is that we can come up with some functions of bounded variation which are not integrable in its domain. Right, I havent been able yet but I think its possible haha.
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The other theorem is Jordan's theorem but Im not even gonna try to use it in here.
Now, how can I prove that something is not of bounded variation then? D:
Thanks so much for your help, have a great week 😀
Best Answer
Any function of bounded varaition is bounded. Consider the points $x_n=\frac 1 {\sqrt {(4n+1)/2}}$. Note that $|g(x_n)|=\frac {(4n+1)\pi} 2$. It follows that $g$ is not bounded, hence not of bounded variation.