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Let us be given a vector field $v: C \subset \mathbb R^n \to \mathbb R^n$ that has the special structure given by $$ v(x) = \alpha(x) \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} $$ with a scalar field $\alpha: C \to \mathbb R$. We assume also that $C$ is a compact subset of $\mathbb R^n$.

I want to determine a good Lipschitz constant for this vector field, i.e. find $L$ such that for all $x,y \in C$ we have $$ \| v(x)-v(y) \| \le L \| x- y\|. $$ Now due to the very specific structure of the vector field $v$ I immediately had the idea that $L$ could be given $$ L = \max_{x \in C} \| \nabla \alpha (x) \|. $$ Could someone confirm this please? If this turns out to be correct, how would one prove this rigorously? Thanks.

samsa44
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The way to relate derivative to Lipschitz constant is to consider restrictions to line segments, and apply the mean value theorem to them. However, this raises the issue of convexity of $C$, since in general the line segment will not lie in $C$.

Assuming $C$ is convex

Then $L = \max_{x \in C} \| Dv (x) \|$, where $\|\cdot\|$ means the operator norm of the matrix $D v$. To find $D v$, use the product rule: $$ D (\alpha(x) x) = (\nabla \alpha(x) )\otimes x +\alpha(x) I \tag1 $$ where $\otimes $ is the outer product. Then use the triangle inequality: $$ \|D (\alpha(x) x)\|\le \| (\nabla \alpha(x) )\| \|x\| + |\alpha(x)| $$ and take the supremum over $x\in C$.

Assuming $C$ is quasiconvex

Same as above, but with $L \le M\max_{x \in C} \| Dv (x) \|$ where $M$ is the constant of quasiconvexity.

General $C$

You can pass to the convex hull of $C$, and apply the first part. The supremum of derivative will be taken over the convex hull.

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