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?
1 Answer
If $d$ 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 $(f_n)_{n\in\mathbb{N}}$ 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 $$ F(f) \leq \lim_{n\in\mathbb{N}} F(f_n) \leq F(f) $$ which implies the continuity of $F$.

$\begingroup$ In fact, the argument works for any metric space. $\endgroup$ Jul 22, 2020 at 13:35

1$\begingroup$ yes, i didn't specify the space and metric on purpose. $\endgroup$ Jul 22, 2020 at 13:36