# Uniform distance from a discontinuous function is continuous

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

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 $$F$$.

Answered by Nathanael Skrepek on November 22, 2021

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