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Let be $f:I\rightarrow{\mathbb{R}}$ and $g:J\rightarrow{\mathbb{R}}$ two locally Lipschitz functions, we define $G:I\times{J}\rightarrow{\mathbb{R}}$ such that $G(x,y)=f(x)\cdot{g(y)}$ , is this a locally lipschitz function?

I think that this is false, but I haven't found any counterexample.

Thanks.

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1 Answer 1

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Indeed it is true. Let $[a,b]\subset I$, $[c,d]\subset J$, let $L_f, L_g$ be Lischitz constants for $f$ and $g$ respectively in $[a,b]$ and $[c,d]$. Since $f$ and $g$ are bounded respectively on the compact sets $[a,b]$ and $[c,d]$ (say by a constant $C$), you have that $$ \begin{split} &|f(x_1) g(y_1) - f(x_2) g(y_2)| \leq |f(x_1) g(y_1) - f(x_2) g(y_1)| + |f(x_2) g(y_1) - f(x_2) g(y_2)| \\ & \leq C |f(x_1) - f(x_2)| + C |g(y_1) - g(y_2)| \leq C [ L_f |x_1 - x_2| + L_g |y_1 - y_2|] \\ & \leq K \| (x_1, y_1) - (x_2, y_2)\|, \quad \forall (x_1, y_1), (x_2, y_2) \in [a,b]\times[c,d], \end{split} $$ with $K = 2 C \max\{L_f, L_g\}$.

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  • $\begingroup$ Why are f and g bounded? $\endgroup$
    – mathlife
    Dec 3, 2017 at 11:14
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    $\begingroup$ A continuous function is always bounded on a compact set (Weierstrass' theorem). $\endgroup$
    – Rigel
    Dec 3, 2017 at 11:15
  • $\begingroup$ But f and g can be no-continous $\endgroup$
    – mathlife
    Dec 3, 2017 at 11:16
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    $\begingroup$ A locally Lipschitz function is always continuous. $\endgroup$
    – Rigel
    Dec 3, 2017 at 11:16
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    $\begingroup$ This function is not Lipschitz continuous, as you can see considering the difference $|f(1/2 + 1/n) - f(1/2 - 1/n)|$. $\endgroup$
    – Rigel
    Dec 3, 2017 at 11:27

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