# Thread: derivative and bouned variation

1. ## derivative and bouned variation

Let $F(x)=x^2\sin(\frac{1}{x})$ and $G(x)=x^2\sin(\frac{1}{x^2})$ with domain $[-1,1]$ (when $x=0$ both functions are said to be equal to 0).
I can show that both $F$ and $G$ are differentiable at every point. However, I don't know how to show that $F$ is of bounded variation while $G$ is not. I am only allowed to use the definition in my proof.

Any hints as where to start would be greatly appreciated.

2. Originally Posted by putnam120
Let $F(x)=x^2\sin(\frac{1}{x})$ and $G(x)=x^2\sin(\frac{1}{x^2})$ with domain $[-1,1]$ (when $x=0$ both functions are said to be equal to 0).
I can show that both $F$ and $G$ are differentiable at every point. However, I don't know how to show that $F$ is of bounded variation while $G$ is not. I am only allowed to use the definition in my proof.

Any hints as where to start would be greatly appreciated.

What hints do you want? You can use only the definition so begin writing the sums for F,G over the given interval and try to discover something interesting (I really don't know)...perhaps the fact that sin x is bounded will help?

Tonio

3. Originally Posted by putnam120
Let $F(x)=x^2\sin(\frac{1}{x})$ and $G(x)=x^2\sin(\frac{1}{x^2})$ with domain $[-1,1]$ (when $x=0$ both functions are said to be equal to 0).
I can show that both $F$ and $G$ are differentiable at every point. However, I don't know how to show that $F$ is of bounded variation while $G$ is not. I am only allowed to use the definition in my proof.
The function F is monotonic in each of the intervals $\left[\frac1{\bigl(n+\tfrac12\bigr)\pi}, \frac1{\bigl(n-\tfrac12\bigr)\pi}\right]$, in which the variation is less than $\frac{2}{\bigl(n-\tfrac12\bigr)^2\pi^2}$.

The function G is monotonic in each of the intervals $\left[\frac1{\sqrt{\bigl(n+\tfrac12\bigr)\pi}}, \frac1{\sqrt{\bigl(n-\tfrac12\bigr)\pi}}\right]$, in which the variation is greater than $\frac{2}{\bigl(n+\tfrac12\bigr)\pi}$.

4. Originally Posted by Opalg
The function F is monotonic in each of the intervals $\left[\frac1{\bigl(n+\tfrac12\bigr)\pi}, \frac1{\bigl(n-\tfrac12\bigr)\pi}\right]$, in which the variation is less than $\frac{2}{\bigl(n-\tfrac12\bigr)^2\pi^2}$.

The function G is monotonic in each of the intervals $\left[\frac1{\sqrt{\bigl(n+\tfrac12\bigr)\pi}}, \frac1{\sqrt{\bigl(n-\tfrac12\bigr)\pi}}\right]$, in which the variation is greater than $\frac{2}{\bigl(n+\tfrac12\bigr)\pi}$.
Thanks, that makes things really clear. However, how did you know to find those intervals?

5. Originally Posted by putnam120
Thanks, that makes things really clear. However, how did you know to find those intervals?
The function sin x oscillates between –1 and +1, and it takes those values at odd multiples of π/2. So the function G(x) oscillates between $-x^2$ and $+x^2$, attaining those values at the points where $1/x^2$ is an odd multiple of π/2. However, I was wrong to say that G(x) is monotonic in the intervals between those points. Fortunately, that does not affect the argument that I gave to show that G(x) is not of bounded variation. It is still true that the variation of G(x) in each of those intervals is at least $2/((n+\tfrac12)\pi)$, and the sum of those numbers diverges.

For F(x), the situation is more serious, and the argument that I gave before does not work without some modification. What you can say is that the turning points of F(x) occur at the points where $\tan1/x = 1/(2x)$ (easy calculus calculation). There is exactly one such point in each of the intervals $\left[\frac1{\bigl(n+\tfrac12\bigr)\pi}, \frac1{\bigl(n-\tfrac12\bigr)\pi}\right]$. So F has only one turning point in each such interval. The variation of F in that interval is therefore at most twice the maximum difference between $-x^2$ and $+x^2$ in the interval, namely $\frac4{\bigl(n-\tfrac12\bigr)^2\pi^2}$. That's sufficient to show that F is of bounded variation.