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I would like to prove a real $C^{\infty}$(polynomial) multivariable function $F : (a_1,a_2,...a_n) \rightarrow (b_1,b_2,...b_n) $ is lipchitz of parameter $l$

is it sufficient to prove the norm of dominant eigenvalue of $Jacobian(F)$ is less than $l$.

In other words, is the "spectral radius of Jaobian matrix is less than 1" a sufficient condition for $F$ to be lipschitz of paramter $l$?

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  • $\begingroup$ Consider $F(x,y) = (500y, 0)$. $\endgroup$ Apr 16, 2018 at 13:30
  • $\begingroup$ @NateEldredge yes, it is a good counter example, what if i change the the condition to all coefficients in Jacobian matrix inferior than $l$? $\endgroup$
    – SC_thesard
    Apr 16, 2018 at 13:55
  • $\begingroup$ Then it is true, and the proof is very easy in that case. $\endgroup$ Apr 16, 2018 at 14:02
  • $\begingroup$ English note: most people would write "less than", not "inferior than". $\endgroup$ Apr 16, 2018 at 14:03
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    $\begingroup$ (Oh, with a factor of $\sqrt{n}$ that is.) What you really want is a bound on the operator norm of the Jacobian. $\endgroup$ Apr 16, 2018 at 14:24

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As Nate's example shows, the eigenvalues aren't enough, but the Lipschitz constant is bounded by the square root of the largest eigenvalue of $J^*J$ or $J J^*$. This is the largest singular value of the Jacobian. The singular value decomposition decomposes $J$ as $UDV$, where $U$ and $V$ are orthogonal matrices and $D$ is a diagonal matrix whose entries are the singular values of $J$, so the largest singular value is the length of the major axis of the image of the unit ball under $J$.

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  • $\begingroup$ Thank you can you recommand some papers with proof of this result? $\endgroup$
    – SC_thesard
    Apr 16, 2018 at 14:31
  • $\begingroup$ what happens if $Jacobian(F)$ is diagonalisable? i think in this case my intuition should be correct as the spectral radius coincides with the subdivison norm of the matrix (sup(||Mx||/||x||))? $\endgroup$
    – SC_thesard
    Apr 16, 2018 at 14:33
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    $\begingroup$ Even if it's diagonalizable, you still need the singular value decomposition. If the matrix is symmetric, then the eigenvectors are orthogonal and the operator norm is equal to the largest eigenvalue. Proving that the SVD gives the Lipschitz constant should be an easy exercise -- just consider $\|UDVa\|$ when $a$ is a unit vector. $\endgroup$ Apr 16, 2018 at 21:55
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It looks to me that $F$ is Lipschitz iff each of the components $b_1,\dotsc, b_n$ is Lipschitz. If one of the components, say $b_1$, has degree $\geq 2$ it cannot be Lipschitz. Indeed, in this case $\newcommand{\bR}{\mathbb{R}}$

$$\sup_{a\in\bR^n}|\nabla b_1(a)|=\infty. $$

Given a constant $L>1$ choose $a\in \bR^n$ such that $|\nabla b_1(a)|>L$. Set $h:=\nabla b_1(a)$. Then

$$\lim_{t\to 0} \frac{1}{t}(b_1(a+th)-b_1(a)= |h|^2>L|h| $$

so that for $t>0$ sufficiently small

$$\frac{1}{t|h|} |b_1(a+th)-b_1(a)|> L.$$

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