Let $s: \mathbb{R} \rightarrow [a,b]$ be the saturation function, i.e., $s(x) = a$ if $x \leq a$, $s(x)=x$ if $a < x < b$, $s(x) = b$ if $x \geq b$.

Consider the vector saturation function $S: \mathbb{R}^n \rightarrow [a,b]^n$ defined as $$ S(x) = \left[ \begin{matrix} s(x_1) \\ \vdots \\ s(x_n) \end{matrix} \right].$$

$S$ is Lipschitz continuous with Lipschitz constant $1$.

Prove or disprove that $S$ is Lipschitz continuous with Lipschitz constant $1$ in any weighted Euclidean space, i.e., for any $A$ positive definite, $$ \left\| S(x) - S(y)\right\|_A \leq \left\| x-y\right\|_A$$ for all $x,y \in \mathbb{R}^n$, where the norm $\left\| z \right\|_A$ is defined as $\sqrt{z^\top A z}$.

  • $\begingroup$ Isn't enough to use the fact that all norms in $\mathbb R^n$ are equivalent? This gives you Lipschitz continuity. The Lipschitz constant will not be 1 in general, I suppose: that is indeed related to the constants appearing in the equivalence of norms... Have I misunderstood anything? $\endgroup$ – Romeo Feb 22 at 12:01
  • $\begingroup$ I agree that $S$ is Lipschitz continuous with norm $A$. The claim is that the Lipschitz constant (with norm $A$) is $1$, given the special structure of the function. $\endgroup$ – user693 Feb 22 at 13:58

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