Is that a metric?

Question

Let $X$ be a finite set. Let $X'=\{M : X \to \mathbb{R}\}$ endowed with the norm $\|M\| = \sup_{x\in X} |M(x)|$.

Define $d:X'\times X' \to \mathbb{R}$ by $$ d(M_1,M_2) = \frac{1}{1+\alpha}\,,$$ where $$\alpha = \sup\{r>0:\|M_1-M_2\|<1/r\}\,.$$ Is $d$ a metric ? I'm asking about the triangle inequality.

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The motivation behind my question is this paper I got curious about

https://arxiv.org/abs/math/0603062

The authors define metric on page 9 as follows.

Define a marked rooted graph $(G,o,M)$ as a graph $G$ with root $o$ and mark $M:V\to \mathbb{R}$ (the authors take something more general than $\mathbb{R}$; it doesn't matter). Two marked rooted graphs are equivalent if there is an isomorphism of graphs between them conserving the root and the mark.

Let $\mathcal{G}_{\ast}$ be the set of equivalence classes of rooted connected marked graphs.

It is claimed that if we define $$ d_{loc}((G_1,o_1,M_1),(G_2,o_2,M_2)) = \frac{1}{1+\alpha} $$ where $$ \alpha=\sup\{r>0:\exists \varphi:B_{G_1}(o_1,[r]) \to B_{G_2}(o_2,[r]) \text{ graph isomorphism with } \varphi(o_1)=o_2 \text{ and } |M_2(\varphi(v))-M_1(v)|<1/r \ \forall v\in B_{G_1}(o_1,[r])\} \, , $$ then $d_{loc}$ induces a metric on $\mathcal{G}_{\ast}$. I suppose the simpler question above should gives us some clue to verify their claim.

Thanks!

Edit: added some attemps.

Answer 1

Ok, I got the answer to the more complicated claim of the authors.

Let $\alpha_{i,j} = \sup\{r>0: \exists \varphi:B_{G_i}(o_i,[r]) \to B_{G_j}(o_j,[r]) \text{ graph isomorphism with } \varphi(o_i)=o_j \text{ and } \|M_j \circ \varphi - M_i\|_{B_{G_i}(o_i,[r])}<1/r\}$.

We need to show that $\frac{1}{1+\alpha_{1,2}} \le \frac{1}{1+\alpha_{1,3}} + \frac{1}{1+\alpha_{2,3}}$.

Assume $\alpha_{1,2}=r$ and let $s>r$. Then there is no graph iso $\varphi:B_{G_1}(o_1,[s]) \to B_{G_2}(o_2,[s])$ having $\varphi(o_1)=o_2$ and $\|M_2 \circ \varphi - M_1\|<1/s$.

We have two cases : if there is no graph iso $\psi:B_{G_1}(o_1,[s]) \to B_{G_3}(o_3,[s])$ having $\psi(o_1)=o_3$, then $\alpha_{1,3} < s$. In this case, $d(G_1,G_2) = \frac{1}{1+r} = \sup_{s>r} \frac{1}{1+s} \le \frac{1}{1+\alpha_{1,3}} = d(G_1,G_3)$. Similarly, if there is no graph iso $\psi:B_{G_2}(o_2,[s]) \to B_{G_3}(o_3,[s])$ having $\psi(o_1)=o_3$, then $\alpha_{2,3} < s$, implying $d(G_1,G_2) \le d(G_2,G_3)$. In either case we get $d(G_1,G_2) \le d(G_1,G_3) + d(G_2,G_3)$.

So assume now that there exists a graph iso $\psi_1:B_{G_1}(o_1,[s]) \to B_{G_3}(o_3,[s])$ having $\psi_1(o_1)=o_3$, and a graph iso $\psi_2:B_{G_2}(o_2,[s]) \to B_{G_3}(o_3,[s])$ having $\psi_2(o_2)=o_3$. Then $\psi_2^{-1} \circ \psi_1:B_{G_1}(o_1,[s]) \to B_{G_2}(o_2,[s])$ is a graph iso with $\psi_2^{-1}\circ \psi_1(o_1)=o_2$. So by assumption, we must have $\|M_2 \circ \psi_2^{-1} \circ \psi_1 - M_1\| \ge 1/s$.

Hence, \begin{align*} \frac{1}{s} & \le \|M_1 - M_2 \circ \psi_2^{-1} \circ \psi_1\| \le \|M_1 - M_3 \circ \psi_1\| + \|M_3 \circ \psi_1 - M_2 \circ \psi_2^{-1} \circ \psi_1\|\\ & = \|M_1-M_3 \circ \psi_1\| + \|M_3 - M_2 \circ \psi_2^{-1}\| \\ & = \|M_3 \circ \psi_1 - M_1\| + \|M_3 \circ \psi_2 - M_2\| \, . \end{align*} So $\frac{1}{\|M_3 \circ \psi_1 - M_1\| + \|M_3 \circ \psi_2 - M_2\|} \le s$ and thus \begin{align*} \frac{1}{1+s} &\le \frac{1}{1+\frac{1}{\|M_3 \circ \psi_1 - M_1\| + \|M_3 \circ \psi_2 - M_2\|}} = \frac{\|M_3 \circ \psi_1 - M_1\| + \|M_3 \circ \psi_2 - M_2\|}{1+\|M_3 \circ \psi_1 - M_1\| + \|M_3 \circ \psi_2 - M_2\|} \\ & \le \frac{\|M_3 \circ \psi_1 - M_1\|}{1+\|M_3 \circ \psi_1 - M_1\|} + \frac{\|M_3 \circ \psi_2 - M_2\|}{1+ \|M_3 \circ \psi_2 - M_2\|} \\ & = \frac{1}{1+\frac{1}{\|M_3 \circ \psi_1 - M_1\|}} + \frac{1}{1+\frac{1}{\|M_3 \circ \psi_2 - M_1\|}} \, . \end{align*} Finally, we have $\alpha_{1,3} \le \frac{1}{\|M_3 \circ \psi_1 - M_1\|}$ and $\alpha_{2,3} \le \frac{1}{\|M_3 \circ \psi_2 - M_1\|}$ by definition (because the $r$ must satisfy $r<\frac{1}{\|M_j\circ \varphi - M_i\|}$). We thus showed that $\frac{1}{1+s} \le \frac{1}{1+\alpha_{1,3}} + \frac{1}{1+\alpha_{2,3}}$, and this completes the proof as before.

Answer 2

Since $X$ is finite so $d(M_1,M_2)= \frac{1}{ 1+\frac{1}{\| M_1-M_2\|_\infty}}$ We view $M_i$ as a vector in $\mathbb{R}^{|X|}$

Then $d$ is metric from direct computation and triangle inequality of $\|\ \|_\infty$