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I have been working and researching on the following problem.

Find a continuous function defined on $(0,1]$ with range $(0,1)$.

I found the function

$\frac{1+(1-x)\sin\left(\frac{1}{x}\right)}{2}$.

I can prove it is continuous at its domain but I am unable to prove it is onto $(0,1)$.

I was stuck in the step, where $y$ is a constant from $(0,1)$ and $x$ is the value I desired in domain.

$\begin{align} (1-x)\sin\left(\frac{1}{x}\right)=2y-1. \end{align}$

So my question is, how to prove the function is onto?

zony_miu
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  • Choose any value in $y \in (0,1)$ and show that there are small $x$ where $0 < y< f(x)<1$ and small $x$ where $0 < f(x)<y<1$. You already have continuity so there must be an $x$ with $f(x)=y$ – Henry Apr 20 '21 at 13:12
  • @Henry Please correct my logic if I am wrong. Since this function is proved to be continuous, if y is within any two $f(x_1), f(x_2)$, then there must exists x such that f(x)=y, otherwise, the function will not be continuous. The problem now is to find $x_1,x_2$. And I don't know how. – zony_miu Apr 20 '21 at 14:04
  • try $x=\frac{1}{\pi(2n+\frac12)}$ and $x=\frac{1}{\pi(2n+\frac32)}$ for large integer $n$ – Henry Apr 20 '21 at 14:16

1 Answers1

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Since $y\rightarrow 2y-1$ is bijective from $(0,1)$ to $(-1,1)$, so it amounts to show that $f(x)=(1-x)\sin(1/x)$ is onto $(-1,1)$.

Now given any $z\in(-1,1)$, assume first that $z>0$. Let $x_{n}=1/(2n\pi+\sin^{-1}(z/2))$, then $f(x_{n})\rightarrow z/2$.

Similarly, with $y_{n}=1/(2n\pi+\sin^{-1}((z+1)/2))$, then $f(y_{n})\rightarrow(z+1)/2$.

Since $z\in(z/2,(z+1)/2)$, we have some $n$ such that $z\in[f(x_{n}),f(y_{n})]$. Now we exploit Intermediate Value Theorem to obtain some $c$ in between $x_{n}$ and $y_{n}$ such that $f(c)=z$.

A similar trick applied for $z<0$.

user284331
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