# Lipschitz continuity of vector saturation

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 $$\sqrt{z^\top A z}$$.

• 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? – Romeo Feb 22 at 12:01
• 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. – user693 Feb 22 at 13:58