Define the metric $d(f,g)triangleq sup_{x in [0,1]} |f(x)-g(x)|$ on the set $operatorname{B}$ of uniformly bounded functions from the interval $[0,1]$ to $mathbb{R}$, fix $g in operatorname{B}$, and define the map $F:operatorname{B}rightarrow [0,infty)$ by $F(f):=d(g,f)$. Is the map $F$ continuous? It certainly is on the subset $C([0,1],mathbb{R})$ but what about on the rest of the space?

MathOverflow Asked by Zorn's Llama on November 22, 2021

1 AnswersIf $d$ is a metric on $B$ then the mapping $F(f) := d(g,f)$ is certainly continuous with respect to the topology induced by the metric $d$:

Let $(f_n)_{ninmathbb{N}}$ a sequence in $B$ that converges to $f in B$ w.r.t. $d$. This is equivalent to $$ d(f,f_n) to 0. $$ Hence, by the triangular inequality $$ F(f_n) = d(g,f_n) leq d(g,f) + d(f,f_n) to d(g,f) + 0 = F(f). $$ On the other hand by the reversed triangular inequality $$ F(f_n) = d(g,f_n) geq d(g,f) - d(f,f_n) to d(g,f) + 0 = F(f). $$ This means $$ F(f) leq lim_{ninmathbb{N}} F(f_n) leq F(f) $$ which implies the continuity of $F$.

Answered by Nathanael Skrepek on November 22, 2021

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