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The wikipedia article on the vector valued Mean value theorem, says

For $f:\mathbb R^n \to \mathbb R^n$, if the gradient is bounded, $$ \| \nabla f \| \le M, $$ then $$ \|f(x)-f(y) \| \le M \|x-y\|. $$

What is the norm used for the gradient $\| \nabla f \|$?

I tried to look in some other references.
There the matrix-norm is mentionned.
They gave one example, where I don't understand how to get the bound of $\frac14$ on $\|\nabla f \|$.

Define $g:\mathbb R_+^2 \to \mathbb R_+^2$ with $$ g(x,y)=\left( \frac{1}{4+x+y}, \frac{1}{4+x+y} \right), $$ then, entry-wise, $$ \nabla f \le \begin{pmatrix} \frac{1}{16} & \frac{1}{16} \\ \frac{1}{16} & \frac{1}{16} \end{pmatrix}=A, $$ whence $$ \|\nabla f \| \le \|A\|=\frac14. $$

what norm reduces $\|A\|$ to $\frac14$?

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The $\ell^1$ norm satisfies it: $\|A\|_1=\sum_{i=1}^4\frac{1}{16}=\frac{1}{4}$ –  Tomás May 17 '13 at 13:19

1 Answer 1

up vote 2 down vote accepted

For finite-dimensional spaces, matrix norms are equivalent, so the choice does not matter.

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Exactly. Unless you are interested in a sharp inequality, but this is not very frequent. –  Siminore May 17 '13 at 13:46

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