# Is this a smooth approximation to the $\ell$-infinity distance actually a quasi-metric?

The $$\|\cdot\|_{\infty}$$-norm on $$\mathbb{R}^n$$ for $$n\in \mathbb{Z}^+$$ is not a smooth function. However, I came across this post which essentially says that a pointwise approximation to the maximum function, and therefore $$\|\cdot\|_{\infty}$$ is $$m_{\lambda}(x) = \sum_{i=1}^n\,\frac{e^{\lambda\, x_i}x_i}{\sum_{j=1}^n\,e^{\lambda\, x_j}}$$ where $$x=(x_i)_{i=1}^n\in \mathbb{R}^n$$ is arbitrary.

Fix $$\lambda>0$$ and consider the map $$d_{\lambda}:\mathbb{R}^n\times \mathbb{R}^n\rightarrow [0,\infty)$$ defined by $$d_{\lambda}(x,y) = m_{\lambda}(|x-y|)$$ where $$|z|:=(|z_i|)_{i=1}^n$$ for any $$z\in \mathbb{R}^n$$.

Does $$d_{\lambda}$$ define a quasi-metric on $$\mathbb{R}^n$$?

It certainly satisfies positivity, symmetry and $$d_{\lambda}(x,y)=0$$ if and only if $$x=y$$ by its not obvious that it should satisfy a relaxed triangle inequality: ie $$d_{\lambda}(x,y) \le C\big(d_{\lambda}(x,z)+d_{\lambda}(z,y)\big)$$ for every $$x,y,z\in \mathbb{R}^n$$ and some $$C\ge 1$$ independent of $$x,y,z$$.

• Isn't $C$ just the lipschitz constant of $m_\lambda$? Jun 22, 2023 at 1:30
• @Mark I can't see why, I initially thought so but... Jun 22, 2023 at 1:34
• Oh, the lipschitz constant instead bounds the quality of the approximation $d_\lambda(x,y) \leq C \lVert x-y\rVert_\infty$. Jun 22, 2023 at 2:47
• hmm but then would we not need a lower-bound also, of the form $\|u-z\|_{\infty}\lesssim d_{\lambda}(z,u)$ (for $u,z\in\{x,y,z\}$) to get the conclusion (here I hide absolute constants by $\lesssim$)? Jun 22, 2023 at 3:10

$$\newcommand\R{\mathbb R}$$By rescaling, without loss of generality $$\lambda=1$$.
So, the question becomes the following: is there a real $$C$$ not depending on $$u=(u_1,\dots,u_n)\in\R_+^n$$, $$v=(v_1,\dots,v_n)\in\R_+^n$$, and $$w=(w_1,\dots,w_n)\in\R_+^n$$ such that for all such $$u,v,w$$ $$w\le u+v\implies m(w)\le C(m(u)+m(v)) \tag{1}\label{1},$$ where $$w\le u+v$$ means $$w_i\le u_i+v_i$$ for all $$i\in[n]:=\{1,\dots,n\}$$ and $$m(u):=\dfrac{\sum_{i\in[n]}e^{u_i}u_i}{\sum_{i\in[n]}e^{u_i}}$$?
Clearly, $$m(u)\le\max u:=\max_{i\in[n]}u_i$$ for $$u\in\R_+^n$$. Also, by (say) Chebyshev's sum inequality, $$m(u)\ge\bar u:=\frac1n\sum_{i\in[n]}u_i$$ for $$u\in\R_+^n$$. So, for all $$u,v,w$$ in $$\R_+^n$$ such that $$w\le u+v$$, $$m(w)\le\max w\le\max u+\max v\le n\bar u+n\bar v\le n m(u)+n m(v),$$ so that \eqref{1} holds with $$C=n$$.