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Mean value inequality?by quasar987
Tags: inequality 
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#1
Jun1008, 02:09 PM

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The statement of the mean value inequality (MVI) is as follows:
"Let A be an open convex subset of R^n and let f:A>R^m be continuously differentiable and such that Df(x)(y)<=My for all x in A and y in R^n (i.e. the family [itex](Df(x))_{x \in A}[/itex] is uniformly lipschitz of constant M on R^n). Then for any x_1, x_2 in A, we have f(x_2)f(x_1)<=Mx_2x_1." If m=1, then this is just the mean value theorem (MVT) plus the triangle inequality. But otherwise, the MVT applied to each component of f separately only leads f(x_2)f(x_1)<=mMx_2x_1. So the proof suggested by the book I'm reading is that we write f(x_2)f(x_1) using the fondamental theorem of calculus (FTC) as [tex]f(x_2)f(x_1)=\int_0^1\frac{d}{dt}f(x_1+t(x_2x_1))dt=\int_0^1Df(x_1+t(x_2x_1))(x_2x_1)dt[/tex] and then use the triangle inequality for integrals to get the result. But notice that the integrand is an element of R^m. So by the above, they certainly mean [tex]f(x_2)f(x_1)=\sum_{j=1}^me_j\int_0^1Df_j(x_1+t(x_2x_1))(x_2x_1)dt[/tex] which does not, to my knowledge, allows for a better conclusion than f(x_2)f(x_1)<=mMx_2x_1. Am I mistaken? Thanks! 


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