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Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that: $$2\cdot f(x)-\sin(f(x))=x, \forall x\in \mathbb{R}$$ Prove that $\lim_\limits{x\to 0}{f(x)}=0$.

I think I need to use the sandwich theorem, so I have found its first part, but not the second as follows:

$\forall x\in \mathbb{R}$ and $\forall f(x)\in \mathbb{R}$, it is: $$\sin(f(x)) \leq f(x) \Rightarrow 2\cdot f(x)-x \leq f(x) \Rightarrow f(x) \leq x$$

So, now I have to figure out how to have a perfect sandwich theorem application. Any hint? (I haven't yet been taught derivatives and differentiability)

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  • $\begingroup$ Hint: $$\left|2f(x)-\sin f(x)\right|\geqslant\left|f(x)\right|$$ $\endgroup$ – Did Jul 28 '15 at 10:14
  • $\begingroup$ $\sin(-\pi)=0>-\pi$, so your statement is not entirely correct. However it is correct for all positive $f(x)$. $\endgroup$ – wythagoras Jul 28 '15 at 10:15
  • $\begingroup$ @Did I don't understand how to use this... $\endgroup$ – Jason Jul 28 '15 at 10:21
  • $\begingroup$ Well, don't you know the limit of the LHS when $x\to0$? Hence the limit of the RHS is... $\endgroup$ – Did Jul 28 '15 at 10:23
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    $\begingroup$ As I wrote, this was a hint (which you were supposed to develop, not ask that every substep be proven to you). To prove the inequality, one can either follow @Scounged's suggestion, or note that $u:y\mapsto2y-\sin y$ is such that $u(0)=0$ and $u'(y)\geqslant1$ for every $y$ hence $u(y)\geqslant y$ if $y\geqslant0$ and $u(y)\leqslant y$ if $y\leqslant0$. $\endgroup$ – Did Jul 28 '15 at 11:45
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Hint:

Write the identity as: $$f(x)=\frac{x+\sin f(x)}2$$ Now, apply the triangle inequality and that $|\sin t|\le |t|$ for any $t\in\Bbb R$: $$0\le|f(x)|\le\frac{|x|+|f(x)|}2$$

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  • $\begingroup$ So, after that $$\left| f(x) \right| \leq \left| x \right| \Leftrightarrow -\left| x \right| \leq f(x) \leq \left| x \right|$$ And then using the sandwich theorem it is easily proven that $\lim_\limits{x\to 0}{f(x)}=0$. $\endgroup$ – Jason Jul 29 '15 at 8:07
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Suppose the contrary. Then there exists $\varepsilon>0$ such that, for all positive integers $n$, there is $x_n$ with $0<|x_n|<\frac{1}{n}$ and $|f(x_n)|>\varepsilon$.

The given identity yields $$ 2f(x_n)-\sin f(x_n)=x_n $$ Now it's not difficult to show that $|2t-\sin t|\ge|t|$ (from the standard inequality $\sin t\le t$ when $t\ge0$, equality holding only for $t=0$). So we have $$ |x_n|=|2f(x_n)-\sin f(x_n)|\ge |f(x_n)|>\varepsilon $$ which is a contradiction as soon as $1/n<\varepsilon$.

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Hint: not quite what you had in mind, but it certainly suffices to show that the function $g(x)=2x-\sin(x)$ has a continuous inverse.

In order to show this, it suffices to note that $g$ is differentiable with $g'(x)>0$

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  • $\begingroup$ I haven't yet "formally" been taught about derivatives and differentiability, although I know them from mathematic contests... $\endgroup$ – Jason Jul 28 '15 at 10:29
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If you assume the contrary, i.e $f(x)\rightarrow a\neq 0$, then for small enough $|x|<\delta$ you have that $|f(x)-a|<\epsilon$, so $f(x)\neq 0$ for such $x$ (choose $\epsilon$ small enough, so that $0\notin (a-\epsilon,a+\epsilon)$) and you can divide by $f(x)$ both sides of the equation. Letting $x\rightarrow 0$ you get $2-\frac{sin(a)}{a}=0\Rightarrow 2a=sin\,a$ which has only the zero solution.

Note: $f(x)$ can not tend to $\infty$.

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    $\begingroup$ But why should $f$ have a limit at $0$ to begin with? $\endgroup$ – egreg Jul 28 '15 at 10:37
  • $\begingroup$ I think you caught me here $\endgroup$ – Svetoslav Jul 28 '15 at 10:38
  • $\begingroup$ maybe it can be justified. Will think, but if you see it, please post it. $\endgroup$ – Svetoslav Jul 28 '15 at 10:40
  • $\begingroup$ @egreg Aha,firstly it is not needed to divide by $f(x)$. Then I can say like this: $f(x)$ either tends to $\infty$ which is impossible, or has some compression points (not sure for the notion). Then, from the equation, it follows that each compression point should be equal to $0$. $\endgroup$ – Svetoslav Jul 28 '15 at 10:50
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    $\begingroup$ @ Jason In my previous comment I justified it, but I didn't edit my answer. My idea was that you do not need to divide by $f(x)$. Just let $x\rightarrow 0$. Then $f$ either tends to $\infty$ or it has some compression points. If it is only one compression point, then it is the limit. But this doesn't matter here. From the equation it follows that each compression point should be equal to $0$ (by choosing the corresponding subsequence of $x\rightarrow 0$). Therefore the uniqueness of all compression points $\Rightarrow$ $lim f(x)=0$ $\endgroup$ – Svetoslav Jul 29 '15 at 10:20

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