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Here $\lvert d(x,a)-d(x,b)\rvert_{\infty}=\sup_{x\in X}\lvert d(x,a)-d(x,b)\rvert$ is the supremum norm, $d$ is a metric and $a, b\in X$. I'm stuck on showing the equality in the title. I can get the direction $$ \lvert d(x,a)-d(x,b)\rvert_{\infty}\leq d(a,b), $$ but not the other one. Or is there a way to directly show the equality?

  • What is $d$? A metric? What are $a,b$ ? Elements of $X$? – xyz Jan 23 '24 at 10:55
  • Yes. $d$ is a metric and $a, b \in X$. –  Jan 23 '24 at 10:56
  • Not sure what equality you want to prove... Is it that $d(a,b)=\sup_{x \in X} |d(x,a)-d(x,b)|$? is it something else? – PierreCarre Jan 23 '24 at 10:57
  • The equality in the title is the one I'm trying to prove - edited the post accordingly. –  Jan 23 '24 at 11:01
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    $\lvert d(x,a)-d(x,b)\rvert_{\infty}=\sup_{x\in X}\lvert d(x,a)-d(x,b)\rvert$ doesn't make sense. – stange Jan 23 '24 at 11:14
  • I left out some context that I didn't think was necessary given the question, but $d(x,a)-d(x,b)$ is a function into a normed space with the supremum norm. More specifically $f_a\colon X \to \mathbb{R}, \ f_a = d(x,a)-d(x,x_0)$ for a fixed $x_0\in X$ and $x\in X$. Then we look at $\lvert f_a(x)-f_b(x)\rvert_{\infty}=\lvert d(x,a)-d(x,b)\rvert_{\infty} $. –  Jan 23 '24 at 11:19
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    It should be $\vert f_a - f_b \vert_\infty = \ldots$, as $f_a(x)$ is a value, and not a function anymore. – stange Jan 23 '24 at 12:41

1 Answers1

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Assuming that $d$ is a metric and that the elemets $a,b \in X$, we have that

$$ d(b,a) - d(b,b) = d(b,a) = d(a,b). $$

Then, by definition of supremum, it follows that

$$ d(a,b) = d(b,a) - d(b,b) \leqslant \sup_{x \in X} |d(x,a) - d(x,b)| = |d(x,a) - d(x,b)|_\infty. $$

xyz
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  • Thanks, this is exactly how I was visioning it in my head but couldn't quite write it down formally. –  Jan 23 '24 at 11:10