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May 13, 2022 at 23:22 vote accept Akira
May 13, 2022 at 12:14 comment added Akira @DieterKadelka in the book $f:X \to \mathbb R$ is vanishing at infinity if $\forall \varepsilon>0, \exists K \subset X$ compact such that $|f(x)| \le \varepsilon$ for all $x \in X \setminus K$.
May 13, 2022 at 12:12 comment added Dieter Kadelka The problem only is: What is "vanishing at infinity", f.i. what if $X = Y = C([0,1])?
May 13, 2022 at 11:59 comment added Akira @DieterKadelka I guess to prove $\Xi \equiv 0$, it's sufficient to assume that $X,Y$ are Polish.
May 13, 2022 at 11:57 comment added Dieter Kadelka Are you sure that $X$ and $Y$ are Polish only? Maybe they are locally compact too.
May 13, 2022 at 11:42 answer added Dirk timeline score: 3
May 13, 2022 at 11:10 history asked Akira CC BY-SA 4.0