0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.