Finding continuous funcions satisfying the condition $|f(x)-\sin(x)|=|x|$

Question

The problem asks to determine all functions defined and continuous over the set of reals that satisfy : $$|f(x)-\sin(x)|=|x|$$ for all reals $x$.

It's immediate that for any real $x$, we either have $f(x) = x + \sin(x)$ or $f(x)=-x+\sin(x)$. I wonder how continuity can help us finish in this case as for instance the IVT helps us to prove that the only continuous functions satisfying $(f(x))^2) = 1 $ for all reals are the two constant functions $ x \longmapsto 1$ and $x \longmapsto -1$.

Thanks for any ideas.

Answer 1

The only place you can switch between $\sin(x) - x$ and $\sin(x) + x$ while maintaining continuity is $x=0$. Thus there will be four solutions.

More formally, let $A = \{x: f(x) = \sin(x)-x\}$ and $B = \{x: f(x) = \sin(x) + x\}$. $A$ and $B$ are closed sets whose union is $\mathbb R$ and whose intersection is $\{0\}$. Use the fact that $(0,\infty)$ and $(-\infty, 0)$ are each connected.

Or if you want to use IVT: suppose $x, y > 0$ and $f(x) - \sin(x) = x > 0$ while $f(y) - \sin(y) = -y < 0$. IVT would say $f(z) - \sin(z) = 0$ for some $z$ between $x$ and $y$, which is impossible. Similarly if $x, y < 0$.

Answer 2

Hint There are four possibilities for such a function: \begin{align*} f(x) & = x + \sin x \\ g(x) & = -x+\sin x\\ h(x) & = \begin{cases}-x+\sin x , & \text{ if } x \leq 0\\ x+\sin x , &\text{ if } x>0\end{cases}\\ k(x) & = \begin{cases}x+\sin x , & \text{ if } x \leq 0\\ -x+\sin x , &\text{ if } x>0\end{cases} \end{align*} At $x=0$ both the possible functions $x+\sin x$ and $-x+\sin x$ coincide but elsewhere IVT can help.