Equality condition in Minkowski's inequality for $L^{\infty}$

Question

I am trying to find out when equality holds in Minkowski's inequality for $L^{\infty}$ (i.e. a necessary and sufficient condition for equality). I did a search and there was a discussion for the case where $1<p<\infty$ but not when $p=\infty$ so I am hoping to get some ideas or for someone to point me to a source where this is discussed.

I will list a couple of observations I made while working this out (though I'm not sure whether I'm right with these):

1. If $\mu(\{x:|f(x)|\geq\|f+g\|_{\infty}-\|g\|_{\infty}\})=0$ (or with $f$ and $g$ interchanged), then I have the reverse inequality.

2. If I pick $a,b$ such that $\|f\|_{\infty}\leq a<\|f\|_{\infty}+\varepsilon$, $\|g\|_{\infty}\leq b<\|g\|_{\infty}+\varepsilon$, $\mu(\{x:|f(x)|>a\})=0$, $\mu(\{x:|g(x)|>b\})=0$, and for all $c<a+b$ I have $\mu(\{x:|f(x)+g(x)|>c\})>0$, then I also have the reverse inequality.

Answer 1

There is no simple condition for equality here. For example if $f=\chi_{(0,3)},g=\chi_{(1,2)}$ then $||f+g||_\infty =||f||_\infty +||g||_\infty$.But $ f $ and $ g $ are not multiples of each other. I hope this example convinces you that one cannot write down simple N & S conditions for equality in Minkowski's inequality.

Answer 2

If $f$, $g\in L^\infty$, then$$||f+g||_\infty=||f||_\infty+||g||_\infty\\ \Longleftrightarrow \forall \epsilon>0,~\mu(\{x:|f(x)|>||f||_\infty-\epsilon\}\cap\{x:|g(x)|>||g||_\infty-\epsilon\}\cap\{x:f(x)g(x)\geq0\})>0.$$ That is, the essential supremum of $f$ and $g$ is reached at the same place and the sign is the same.

Proof$\quad$ "$\Longrightarrow$"$\quad$Suppose $\exists~\epsilon_0>0$, such that $$\mu(\{x:|f(x)|>||f||_\infty-\epsilon_0\}\cap\{x:|g(x)|>||g||_\infty-\epsilon_0\}\cap\{x:f(x)g(x)\geq0\})=0.$$ Then if $|g|>||g||_\infty-\epsilon_0$ and $fg\geq0$, we have $|f|\leq||f||_\infty-\epsilon_0$ a.e.. Thus $$|f+g|\leq|f|+|g|\leq||f||_\infty-\epsilon_0+||g||_\infty=||f+g||_\infty-\epsilon_0\quad{\rm a.e.}.$$ This is obviously false since $\epsilon_0>0$.

"$\Longleftarrow$"$\quad$$\forall \epsilon>0$, \begin{equation*} \begin{aligned} &\quad~~\mu(\{x:|f(x)+g(x)|>||f||_\infty+||g||_\infty-2\epsilon\}\\ &\geq\mu(\{x:|f(x)|>||f||_\infty-\epsilon\}\cap\{x:|g(x)|>||g||_\infty-\epsilon\}\cap\{x:f(x)g(x)\geq0\})\\ &>0. \end{aligned} \end{equation*} Hence $||f+g||_\infty\geq||f||_\infty+||g||_\infty$. And according to Minkowski's inequality, $||f+g||_\infty\leq||f||_\infty+||g||_\infty$, therefore $$||f+g||_\infty=||f||_\infty+||g||_\infty.$$