Continuity of a map in a metric space

Question

Let $C^0([a,b])$ denote the space of continuous function $f:[a,b]→\Bbb R$. Define $ d(f,g)= \sup_{[a,b]}|f-g| $. We define $F:C^0([a,b])→\Bbb R$ to be $F(f)=\int_a^b f$. I want to show that $F$ is a continuous map.

As always, take sequence ${f_n}$ s.t $d(f_n,f)→ 0$. Then I want to show that $F(f_n)→F(f)$. And this is pretty simple as well. Then I am done with it? I thought I need to use Lipschitz to prove its continuity？

Answer 1

So fix $\epsilon > 0$. You wanna find $\delta > 0$ such that if $d(f, g) \leq \delta$, then $|F(f) - F(g) | < \epsilon$. We may note that $F$ is linear, so $|F(f) - F(g)| = |F(f - g)|$. Now find an upper bound on $|F(f - g)|$ in terms of $d(f, g)$.

Answer 2

Well, $F$ is indeed Lipschitz, for if $f$ and $g$ are real continuous functions on $[a,b]$ one has $$|F(f) - F(g)| \leqq \int_a^b |f(t)-g(t)|dt \leqq (b-a)\sup_{t \in [a,b]} |f(t)-g(t)| = (b-a)d(f,g).$$ But you don't need this to prove continuity, although it allows you to conclude that $F$ is uniformly continuous, which is something stronger. Your argument is sufficient to prove that $F$ is continuous at each $f \in C^0[a,b]$.