Lipschitz Continuity-Function

Question

I cannot prove that $A\cos(ax+b)$, where $A$, $a$, $b$ are real numbers is Lipschitz continuous. Am I wrong for trying to do so using the definition?

Answer 1

If the function is Lipschitz continuous, there must be some constant $K$ such that the function satisfies $$ |A \cos(ax_1+b)-A\cos(ax_2+b)|\leq K|x_1-x_2| $$ for all $x_1$ and $x_2$. The RHS is bounded by the maximum possible difference, which is $2A$ (if $ax_1+b$ and $ax_2+b$ are separated by a multiple of $\pi$). Therefore, the RHS will be $2A$ if $$ (ax_1+b)-(ax_2+b)=n\pi $$ $$ x_1-x_2=\frac{n\pi}{a}, \ n\geq1. $$ Therefore, our inequality is $$ K \geq \frac{2Aa}{n\pi} $$ If you chose $K = 2Aa/\pi$ you can ensure the inequality. Therefore, since we found $K$, the function is Lipschitz continuous.