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Suppose $\alpha$ and $\beta$ are real numbers and $\beta$ $\gt 0$. We define the function $f$ on $[-1,1]$ by $$f(x)=x^\alpha \sin(|x^{-\beta}|), x \neq0 $$ $$f(x)= 0, x=0.$$ Prove that
a. $f$ is continuous iff $\alpha \gt 0$.

b. $f'(0)$ exists iff $\alpha \gt 1$.

c. $f'$ is bounded iff $\alpha \ge 1+\beta$.

d. $f'$ is continuous iff $\alpha \gt 1+\beta$.

[You can use the standard properties of trig functions and their derivatives.]

I've been able to come up with something for a. and b. but I don't know how to do c. or d.

I'll post my a. and b.

a. $f$ is continuous iff $\forall ({x_n}) \rightarrow 0$ for $x_n \neq 0$. Then $x^\alpha\sin(|x^{-\beta}|) \rightarrow 0$ as $n \rightarrow \infty$.

I considered $x_n =\frac{1}{2n\pi + \frac{\pi}2} \gt 0$

$x_n \rightarrow 0$ as $n \rightarrow 0$, hence $\alpha \gt 0$. $\alpha \neq 0$ because then $x_n^\alpha =1$. $\alpha \not \lt 0 $ because then $x_n^\alpha \rightarrow \infty$ as $n \rightarrow \infty$.

It is easy to see that $f$ is continuous on $[-1,1] \setminus \{0\}$. We find that $$ -|x^\alpha| \le x^\alpha \sin (|x|^{-\beta}) \le |x^\alpha|$$ (because sin varies between $-1$ and $1$). $|x^\alpha| \rightarrow 0$ as $x \rightarrow 0$ since $\alpha \gt 0$. Therefore $f$ is continuous everywhere.

b. A function needs to be continuous everywhere on $[-1,1]$ in order to be differentiable there. In the previous part we proved that $\alpha \gt 0$, so we know that $\alpha$ has to be at least this.

$f'(0)$ exists iff $x^{\alpha-1} \sin (|x|^{-\beta}) \rightarrow 0$ as $x \rightarrow0$. We see that is does, if $\alpha \gt 1$. Therefore $f'(0)=0$ and exists.

Is what I have correct? And how would I do parts c and d?


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I'm guessing your $x = 0$ and $x \neq 0$ conditions are mixed up. – Javier Apr 10 '13 at 18:34
Oops! I'll fix that! Thanks – randi Apr 10 '13 at 19:08
@RSalimi - the existence and value of the derivative at 0 has nothing to do with the value in (0,1). That's basically the whole point. – Sharkos Apr 11 '13 at 13:49
@Sharkos,thanks,I thought That was assumed $f^{\prime}$ is continuous. – R Salimi Apr 11 '13 at 15:59

Yes, you're basically right with a., b. though perhaps you should note that b. is working by the definition of the derivative at 0 rather than some ad-hoc reason, and then that the condition you need is just $x^{\alpha-1}\to \text{const}$, not 0.

For the next two parts, you will want to actually compute the derivative away from $x=0$, and consider what happens as you approach the origin. The first part just requires the use of the product rule and so on to get the most singular behaviour; then you need to check whether your derivative at the origin fits in smoothly.

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Yeah a friend and I figured out how to do this problem last night. I'll post what I got in a little bit. Also thanks for the critique. I have a super harsh grader so anything you see is appreciated. – randi Apr 11 '13 at 13:44

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