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.

There is an exercise in Stephen Abbott's Understanding Analysis that states:

Exercise 5.3.7. (b) Show that the function $$g(x)=\begin{cases} x/2+x^2\sin(1/x)&\text{ if }x\neq0\\\ 0&\text{ if }x=0 \end{cases}$$ is differentiable on $\mathbb{R}$ and satisfies $g'(0)\geq0$. Now, prove that $g$ is not increasing over any open interval containing $0$.

First of all, I know that for $x=0$, $$g'(0)=\lim_{x\to0}\frac{x/2+x^2\sin(1/x)}{x}=\lim_{x\to0}\left[\frac{1}{2}+x\sin\left(\frac{1}{x}\right)\right]=\frac{1}{2}\geq0,$$ and for $x\neq0$, $$g'(x)=\frac{1}{2}+2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right).$$ Hence, $$g'(x)=\begin{cases} 1/2+2x\sin(1/x)-\cos(1/x)&\text{ if }x\neq0\\\ 1/2&\text{ if }x=0, \end{cases}$$ and $g$ is differentiable on $\mathbb{R}$.

However, I do not know how to formally show that if $(a,b)$ is an open interval containing $0$, then $g$ is not increasing on it; the idea I have is to keep in mind that as $x$ approaches $0$, it oscillates 'faster,' and you can thus always find two different points $x,y\in(a,b)$ such that $g'(x)>0$ and $g'(y)<0$. Is this a valid assertion? If so, how can I go about showing it? Thanks in advance.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Yes that would work. Since $g$ is differentiable on $(a,b)$ it would be increasing on $(a,b)$ if and only if $g'\ge0$ over $(a,b)$. So, you only need to find a point $c$ in $(a,b)$ where $g'(c)<0.$

Try a point $c$ where $\cos(1/c)=1$. These points would be of the form ${1\over 2n\pi}$. We have $$\textstyle g'({\textstyle{1\over 2n\pi}})= {1\over2}+ {1\over n\pi}\cdot\sin(2n\pi) -\cos(2n\pi)={1\over2}+0-1={-1\over2}. $$

As you can select $n$ so large that ${1\over 2n\pi}$ is in $(a,b)$, you are done.

share|improve this answer

Yes, this is the right way. In your expression for $g'(x)$, you see that the term $\cos(1/x)$ oscillates infinitely many times between $-1$ and $+1$ on any open interval containing zero, and because of this fact, such an interval contains infinitely many little subintervals where $g'>0$ and infinitely many where $g'<0$.

share|improve this answer

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.