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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.

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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.

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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 $g'<0$.

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