Showing that $x\sin(1/x)$ has local maximum and local minimum in $(-1/m,1/m)$ for any positive integer $m\geq 1$.

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be given by $$f(x)=\begin{cases}x\sin(\frac{1}{x}), \quad x\neq 0\\ 0, \quad x=0 \end{cases}$$ Show that $f$ has a local maximum and a local minimum in $(-\frac{1}{m},\frac{1}{m})$ for every positive integer $m\geq 1$.

My idea was to consider the set $A=[-\frac{1}{m},\frac{1}{m}]$. Since $A$ is compact, and $f$ is continuous on $A$, by the extreme value theorem, $f$ attains a maximum and a minimum in $A$. Then I wanted to show by Rolle's theorem that the extremum values cannot be attained at the end-points. This is evident if $f$ is differentiable on $(-\frac{1}{m},\frac{1}{m})$ but I can show that for $x\neq 0$, $f$ is differentiable since $f(x)=a(x)\cdot b(c(x))$ where $a(x)=x,b(x)=\sin x, c(x) = 1/x$ which are all "well-behaved" away from $0$ and hence $f$ is differentiable by application of the product rule and chain rule. The problem arises at $x=0$ where $f$ is not differentiable. So this approach fails.

Any hints on an alternative method to prove this?

Answer 1

The derivative of $f$ (away from $x=0$, where $f$ is not differentiable) is

$$f'(x) = \sin \left( \frac 1x\right) - \frac 1x \cos \left( \frac 1x\right).$$

Thus $f'(x) = 0$ if

$$ \tan \left( \frac 1x \right) = \frac 1x.$$

Then it is easy to see graphically that there are infinitely many critical point (say, let $y = 1/x$, then you are looking for the intersection of the two graphs $z = \tan y$ and $z=y$).

Rigorously that can be proved by intermediate value theorem: in the interval

$$ 2n\pi < y < 2n \pi + \frac \pi 2,\ \ n\neq 0,$$

the function $g = \tan y - y$ has

$$ g(2n\pi ) = -2n\pi <0, \ \ \lim_{y \to 2n\pi + \frac \pi 2} g(y) = +\infty,$$

thus there is $y^+_n$ in the interval $(2n\pi, 2n\pi + \pi/2)$ so that $g(y_n^+) = 0$. So there is a sequence $\{x_n^+ = 1/y_n^+\}$ so that $x_n^+ \to 0$ and $f'(x_n^+) = 0$.

The second derivative of $f$ is

$$f''(x) = -\frac{1}{x^3} \sin \left( \frac 1x\right).$$

Since $y_n^+ = 1/x_n^+ \in (2n\pi, 2n\pi + \pi/2)$, $\cos (y_n^+) >0$ and thus

$$ f''(x_n^+) <0, \ \ \ \forall n.$$

Hence $x_n^+ $ is a local maximum.

As a result, for all $m>0$, one can find $n$ large enough so that $|x_n^+|< 1/m$, hence there is a local maximum in $(-1/m, 1/m)$.

One can similarly find a sequence of local minimum tending to zero, by examining $g$ in the intervals $((2n + 1)\pi, (2n+1) \pi + \pi/2)$. I will leave the details to you.

Answer 2

Let's consider an arbitrary half neighborhood of $0$, say $(0,\epsilon)$

In the open interval, at points of extrema we must have: $f'(x)=0\implies \sin (\frac 1x)-\frac 1x\cos \frac 1x=0$

$f'(x)=(\tan (\frac 1x)-\frac 1x)\cos \frac 1x$

By Archimedean property, there exists $k\in \mathbb N\cup\{0\}$ such that $I_k=\Big(\frac 1{(2k+1)\pi+\frac\pi 2},\frac1{(2k+1)\pi-\frac\pi 2} \Big)\subset (0,\epsilon)$

By IVT, you may verify that $f'(x)=0$ has a solution $s_1$ on $I_{k1}=\Big(\frac 1{(2k+1)\pi},\frac1{(2k+1)\pi-\frac\pi 2} \Big)$ and $s_2$ on $I_{k2}=\Big(\frac 1{(2k+1)\pi+\frac\pi 2},\frac 1{(2k+1)\pi}\Big)$

$f''(x)=-\frac{\sin \left(\frac{1}{x}\right)}{x^3}$ is negative for $s_1$ and positive for $s_2$ whence it follows that $s_1$ is a point of local maxima and $s_2$;a point of local minima.

The same analysis as above is valid for the other half of the interval $(-\epsilon, \epsilon)$ because $f$ is even. And since $\epsilon\gt 0$ is arbitrary, the result follows from Archimedean property of real numbers.