Given the following equation for $\epsilon \in (0, 1]$: $$ x = y - \epsilon \sin y $$ Prove that it defines a unique continuous function: $$ y = f(x) $$

Here is a sketch I've worked out so far. The idea is based on the fact that if a function is continuous and monotone then it must have a monotone continuous inverse function.

To follow that idea I'm going to remap the values of $x$ and $y$. Say we have a function: $$ f(x) = x - \epsilon \sin x $$

We know that the identity function $g(x) = x$ is continuous. But $\sin x$ is continuous as well, hence by the theorem for the sum of continuous functions we may state that $f(x)$ is continuous as well.

To proceed we need to somehow show that $f(x)$ is monotonically increasing for $x\in[0, +\infty)$. We might consider only this interval because $f(x)$ is odd hence symmetric with respect to the origin: $$ f(-x) = -x - \epsilon \sin(-x) = -x + \epsilon \sin x = - (x - \epsilon\sin x) = -f(x) $$

The only thing left is to show $f(x)$ is monotonically increasing for $x\in[0, +\infty)$, which I couldn't accomplish. Once this is done we may state that $f(x)$ is monotonically increasing, is continuous hence it has a monotonically increasing continuous inverse function, which finishes the proof if we swap $x$ and $y$ again.

There is a limitation though, I'm not allowed to use derivatives. This problem is from the section about the continuity of a function, before the definition of derivatives. Also this function has already been under the microscope here, but from a different perspective.

How does one rigorously show $f(x)$ is increasing? It feels like I could utilize $x \ge \sin x$ for $x\in \Bbb R^+$ somehow, but not sure. Also, it's not clear to me how $\epsilon$ is involved. Also is the overall idea valid? Here is the graph of the function. Thank you!


Hint: To show that $f$ is monotonic, it suffices to use the fact that $|\sin(a) - \sin(b)| \leq |a - b|$.

One way to show that this inequality holds is to use the sum-to-product identity. In particular: $$ |\sin(a) - \sin(b)| = 2\cdot \left|\sin\left(\frac{a-b}{2}\right)\right| \cdot \left|\cos\left(\frac{a+b}{2}\right)\right|\\ \leq 2\cdot \frac{|a-b|}{2} \cdot 1 = |a-b| $$

  • $\begingroup$ I've been able to proceed using your hint, thank very much! $\endgroup$ – roman Jun 28 at 17:44

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