Prove that the function $F(f)=\int_a^bf(x)dx$ is continuous.

Question

Define $d:X\times X\to \mathbb{R}$ by $$d(f,g)=\int_a^b |f(x)-g(x)|dx$$ Let $\rho$ be the usual metric on $\mathbb{R}$. Prove that the function $F:X\to\mathbb{R}$ defined $$F(f)=\int_a^bf(x)dx$$ is continuous.

Answer 1

$$|F(f)-F(g)|=\left| \int_a^b f(x)dx - \int_a^b g(x)dx\right|=\left| \int_a^b (f(x)-g(x))dx \right| \le \int_a^b |f(x)-g(x)|dx = d(f,g)$$

which shows that $F$ is a contraction: we can take $\delta=\epsilon$ uniformly.

So $F: C([a,b]) \to \mathbb{R}$ is (uniformly) continuous.

Answer 2

I suppose that $f$ and $g$ are continuous, we have $$ \rho(F(f),F(g))=|F(f-g)|\leqslant d(f,g) $$ so that $F$ is continuous.