Let f=(f1,ldots,fn):[a,b]rightarrowmathbbRnf=(f1,ldots,fn):[a,b]rightarrowmathbbRn be a continuously differentiable function. (See the comments above for an explanation as to why the hypotheses have been strengthened.)
For 1leqileqn1leqileqn, let
Li=maxxin[a,b]|f′i(x)|Li=maxxin[a,b]|f′i(x)|,
so that, by the Mean Value Theorem, for x,yin[a,b]x,yin[a,b],
|fi(x)−fi(y)|=|f′i(c)||x−y|leqLi|x−y||fi(x)−fi(y)|=|f′i(c)||x−y|leqLi|x−y|.
Then, taking the standard Euclidean norm on mathbbRnmathbbRn,
|f(x)−f(y)|2=sumni=1|fi(x)−fi(y)|2leq(sumni=1L2i)|x−y|2|f(x)−f(y)|2=sumni=1|fi(x)−fi(y)|2leq(sumni=1L2i)|x−y|2,
so
|f(x)−f(y)|leqsqrt(sumni=1L2i)|x−y||f(x)−f(y)|leqsqrt(sumni=1L2i)|x−y|.
Thus we can take
L=sqrtsumni=1L2iL=sqrtsumni=1L2i.
Since all norms on mathbbRnmathbbRn are equivalent -- i.e., differ at most by a multiplicative constant -- the choice of norm on mathbbRnmathbbRn will change the expression of the Lipschitz constant LL in terms of the Lipschitz constants LiLi of the components, but not whether f is Lipschitz.
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