# Prove that equation $x = y - \epsilon \sin y$ defines a unique continuous function $y = f(x)$ for $\epsilon\in (0, 1]$.

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: $$|\sin(a) - \sin(b)| = 2\cdot \left|\sin\left(\frac{a-b}{2}\right)\right| \cdot \left|\cos\left(\frac{a+b}{2}\right)\right|\\ \leq 2\cdot \frac{|a-b|}{2} \cdot 1 = |a-b|$$