Showing a function $\mathbb R \to \mathbb R^2$ is bilipschitz

Question

Problem:

Is the function $f\colon \mathbb R \to \mathbb R^2$, defined by $f(x)=(x, \cos x)$, $M$-bilipschitz for some $M\geq 1$? (Using the standard metric)

So far:

By the MVT we have $\lvert \cos x -\cos y \rvert \leq \lvert x-y\rvert$ for all $x,y\in \mathbb R$, so we easily get that $$d(f(x),f(y))=\sqrt{(x-y)^2+(\cos x-\cos y)^2} \leq \sqrt{2(x-y)^2}= \sqrt{2}d(x,y).$$ My problem is in showing the lower bound $1/\sqrt{2}$, as intuitively this should indeed be a sensible lower bound, and thus $f$ should be $\sqrt{2}$-bilipschitz. I'm completely blanking on bounding the distance from below. Or am I in fact wrong in assuming this function really is bilipchitz?

Answer 1

We always have $|f(x)-f(y)|= \sqrt{(x-y)^2 + (\cos x-\cos y)^2}\ge |x-y|$, so the function is indeed bilipschitz onto its image.