# Prove that the product of two Lipschitz functions is locally Lipschitz.

Let $$X$$ be a reflexive Banach space and $$f_{i}\colon X \rightarrow \mathbb{R}$$ be Lipschitz functions. Set $$f=f_{1}f_{2}$$.

How to prove that $$f$$ is locally Lipschitz?

• derivative of a product of two functions will give you a hint. Mar 4, 2017 at 17:07
• But I don't know if I can differentiate the product? What about $f_{i}(x)=|x|$? Mar 4, 2017 at 17:15
• I said, a hint. See my answer below. Mar 4, 2017 at 17:29

\begin{align} |f(x)-f(y)| {}={}& |f_1(x){}\cdot{}f_2(x)-f_1(y){}\cdot{}f_2(y)| \\ {}={}& |(f_1(x)-f_1(y))f_2(x)+f_1(y)(f_2(x)-f_2(y))| \\ {}\leq{}& |f_1(x)-f_1(y)|{}\cdot{}|f_2(x)|+|f_1(y)|{}\cdot{}|f_2(x)-f_2(y)|.\end{align}
Now, by Lip of $$f_i$$ we have a bound of $$Ld(x,y)$$ on differences. It suffices to prove local boundedness of $$f_i$$.