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Let (X,d) be a metric space. f: X $\rightarrow \mathbb{R}$ ist called "Katetov map" iff $\forall x, y, \in X : |f(x)-f(y)| \leq d(x,y) \leq f(x) + f(y)$. The set of all Katetov-maps on X is denoted by E(X).

If f $\in$ E(X) and S $\subset$ X are such that $f(x) = \inf \{f(s) + d(x,s) : s \in S\}$, we say that f is "supported by S".

Show that if f, g $\in$ E(X) are both supported by S $\subset$ X it holds that $d(f,g)= \sup_{s \in S} |f(s)-g(s)| $ where d denotes the sup-metric.

Apparently this is an easy fact, but I am failing to prove it. Can someone help me please?

Antigone
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1 Answers1

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Since I only need the statement for finite sets S, I am going to show how I proved that:

By applying the inequality $|\min\limits_{1\leq s \le1 n} a_s - \min\limits_{1\leq s \le1 n} b_s| \leq \max\limits_{1\leq s \le1 n}|a_s - b_s|$ we see that $\forall x$ it holds: $|f(x)-g(x)| = |\min\limits_{s \in S} \{f(s) + d(x,s)\} - \min\limits_{s \in S} \{g(s) + d(x,s)\}| \leq \max\limits_{s \in S} |\{f(s)-g(s)\}|$ and therefore $d(f,g) = \sup\limits_{x \in X} |f(x)-g(x)| \leq \max\limits_{s \in S} |\{f(s)-g(s)\}|$. The other inequality is trivially true.

Antigone
  • 115