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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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    $\begingroup$ derivative of a product of two functions will give you a hint. $\endgroup$ Mar 4, 2017 at 17:07
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    $\begingroup$ But I don't know if I can differentiate the product? What about $f_{i}(x)=|x|$? $\endgroup$
    – zorro47
    Mar 4, 2017 at 17:15
  • $\begingroup$ I said, a hint. See my answer below. $\endgroup$ Mar 4, 2017 at 17:29

1 Answer 1

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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. $\endgroup$
    – zorro47
    Mar 4, 2017 at 17:45

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