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?
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1$\begingroup$ derivative of a product of two functions will give you a hint. $\endgroup$– Behnam EsmayliMar 4, 2017 at 17:07
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1$\begingroup$ But I don't know if I can differentiate the product? What about $f_{i}(x)=|x|$? $\endgroup$– zorro47Mar 4, 2017 at 17:15
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$\begingroup$ I said, a hint. See my answer below. $\endgroup$– Behnam EsmayliMar 4, 2017 at 17:29
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1 Answer
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\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 $f_i$.
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$\begingroup$ So convexity has nothing to do with the proof. $\endgroup$– zorro47Mar 4, 2017 at 17:45