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I was hoping for help with the following problem.

Let X be the set of all bounded functions existing in B(X). For $f_1,f_2∈B(X)$ I am to show that $d({f_1}, {f_2}) = sup{[f_1(x) − f_2(x)| : x∈X, n ∈ N}$ is a metric space.

I know that all bounded functions of real numbers is called bounded if there is a number $M∈R$ such that $|f_1|\le M$ for all $n∈N$. I also know to prove that $X$ is a metric space, you need to show that $d$ is in fact a metric, meaning that it satisfies the three axioms:

  1. For all $f,g \in X$, $d(f,g) \geq 0$ and $d(f,g)=0$ if and only if $f=g$.
  2. For all $f,g \in X$, $d(f,g)=d(g,f)$.
  3. For all $f,g,h \in X$, $d(f,h) \leq d(f,g)+d(g,h)$.

I don't know how to show these properties valid for $d({f_1}, {f_2}) = sup{[f_1(x) − f_2(x)| : x∈X, n ∈ N}$ and could really use some help. Thanks.

finkleberrys
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  • 1 is easy as supremum of absolute values of $f_1-f_2$ has to be $\geq 0$, and obviously supremum is 0 iff functions are identical. Similarly 2 follows from commutative property of addition. 3 requires some justification, i.e. $|f_1-f_3| \leq |f_1-f_2|+|f_2-f_3|$ for all evaluation points $x \in X$, so $|f_1-f_3| \leq sup...+sup...$ for all $x \in X$ and then apply sup again to yield the result. Also, where is $n \in N$ used here? – nabu1227 Mar 24 '21 at 14:23
  • I think I can nail the first two, but the third is hard to do. how would I apply sup again? – finkleberrys Mar 25 '21 at 10:36
  • Well $|f_1(x)-f_3(x)| \leq |f_1(x)-f_2(x)|+|f_2(x)-f_3(x)|$ for all evaluation points $x \in X$, so by definition $|f_1(x)-f_3(x)| \leq \sup|f_1(x)-f_2(x)|+\sup|f_2(x)-f_3(x)|$ for all $x \in X$ by upper bound definition of sup. Now since sup is the least upper bound of a set, $\sup|f_1(x)-f_3(x)| \leq \sup|f_1(x)-f_2(x)|+\sup|f_2(x)-f_3(x)|$ again by definition – nabu1227 Mar 25 '21 at 11:18

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