differential equation/Lipschitz.

Question

Show that $\sqrt{1 + f^2} \in Lip([a, b])$, $\forall f \in Lip([a, b])$.

I have an exam tommorrow and i can't wrap my head around this problem. So I was thinking to start with the definiton: $|f(x) - f(y)| \le L|x-y|$ , but doesn't get me anywhere.

Answer 1

By continuity $\exists M > 0$ such that $|f(x)| \leq M$ for every $x \in [a,b]$ $$|1 + f(x)^2 - (1 + f(y)^2 )| = |f(x)^2 - f(y)^2| = |(f(x) - f(y))(f(x) + f(y))| \leq 2ML|x - y|$$ so $1 + f^2 \in Lip([a,b])$. The square root is Lipschitz on any compact subinterval of $(0,+\infty)$, since the image of $1 + f^2$ is a compact subinterval of $(0,+\infty)$ we have: $$\Big|\sqrt{1 + f(x)^2} - \sqrt{1 + f(y)^2}\Big| \leq K |1 + f(x)^2 - (1 + f(y)^2 )| \leq 2KML|x - y|$$ Here $K$ is the Lipschitz constant of the square root on the image of $1 + f^2$