Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks

share|improve this question
    
I'm guessing your $x = 0$ and $x \neq 0$ conditions are mixed up. –  Javier Badia 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
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.