Uniform distance from a discontinuous function is continuous

Question

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?

Nathanael Skrepek · 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 $(f_n)_{ninmathbb{N}}$ 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 $F$.

Uniform distance from a discontinuous function is continuous

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