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I have the following function: $$f:\Bbb R\to \Bbb R\text{ defined by } f(x)=\left\{\begin{matrix}0 & \text{ if } x\leq0, \\ \cos(x)\sin\left(\frac1{x}\right) &\text{ if } x>0. \end{matrix}\right. $$ I have to prove that this function has the Darboux Property.

It is quite obvious that the function is discontinuous at $0$; in fact, the right-side limit doesn't even exist.

Intuitively I can also see that for any interval of the form $I=(0, \epsilon)$, with $\epsilon$ positive, $f(I)=[-1, 1]$ but I'm not sure if my intuition is correct, nor could I manage to find any way to prove this.

Another thing that I noticed is that $f\left(\frac1{2k\pi}\right)=0$ for any $k\in \Bbb N$(and even $\Bbb Z$ I guess). That didn't lead to anything.

Any ideas/hints/solutions would be appreciated!

Sumanta
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Wolfuryo
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2 Answers2

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Take $a,b\in\Bbb R$ with $a<b$. You want to prove that if $y$ lies between $f(a)$ and $f(b)$, then there is a $c\in[a,b]$ such that $f(c)=y$. If $a>0$, this is clear, since $f$ is continuous on $(0,\infty)$. And the statement is trivial if $b<0$. So, suppose that $a\leqslant0<b$. Then $f(a)=0$ and $f(b)\in[-1,1]$. Take $n\in\Bbb N$ so large that:

  • $\displaystyle\frac1{2n\pi+\pi/2},\frac1{2n\pi-\pi/2}<b$;
  • $y$ lies between $\displaystyle\cos\left(\frac1{2n\pi+\pi/2}\right)$ and $\displaystyle-\cos\left(\frac1{2n\pi-\pi/2}\right)$.

Then, by the intermediate values theorem, there is some$$c\in\left[\frac1{2n\pi+\pi/2},\frac1{2n\pi-\pi/2}\right]$$such that $f(c)=y$ and $c\in[a,b]$.

José Carlos Santos
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Consider $g(x)=x^2\cos(x)\cos(\frac1{x}), x \ge 0$ and $0$ elswhere and note that $g$ is differentiable at zero and $g'-f=h$ which is continuous on the line so has an antiderivative $h$; this means that $f=(g-h)'$ and then it has Darboux property by the usual theorem that derivatives have such even if they are discontinuous (this is an easy application of MVT)

Conrad
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