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 $x,y,z$.

  • $\begingroup$ Isn't $C$ just the lipschitz constant of $m_\lambda$? $\endgroup$ Jun 22, 2023 at 1:30
  • $\begingroup$ @Mark I can't see why, I initially thought so but... $\endgroup$ Jun 22, 2023 at 1:34
  • $\begingroup$ Oh, the lipschitz constant instead bounds the quality of the approximation $d_\lambda(x,y) \leq C \lVert x-y\rVert_\infty$. $\endgroup$ Jun 22, 2023 at 2:47
  • $\begingroup$ 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 $\lesssim$)? $\endgroup$ Jun 22, 2023 at 3:10

1 Answer 1


$\newcommand\R{\mathbb R}$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 $C=n$.

  • $\begingroup$ Thanks again Iosif ! $\endgroup$ Jun 23, 2023 at 4:13

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