$C_{[a,b]} \rightarrow \mathbb{R}$, $x(t) \mapsto f(x) = \int_a^b x(t)dt$ continuous?

Question

Problem: Prove that $f:C_{[a,b]} \rightarrow \mathbb{R}$, $x(t) \mapsto f(x) = \displaystyle\int_a^b x(t)dt$ is continuous.

My opinion:

Let $x, x' \in C_{[a,b]}$, so they are bounded and we can choose a $\delta >0$ such that $d(x,x') < \delta$. To prove that $f$ is continuous we have to prove $\forall \epsilon > 0, \exists \delta > 0 : d(x,x') < \delta \Rightarrow d(f(x),f(x') < \epsilon$

My queston: with $f(x), f(x')$, how can we find a $\epsilon > 0$ respectively? Thank all!

Answer 1

Hint: $$ \left|f(x)-f(x')\right|=\left|\int_a^b x(t)-x'(t) \, dt \right| \leq \int_a^b \left| x(t)-x'(t) \right| \,dt \leq (b-a)\sup_{t \in [a,b]} \left| x(t)-x'(t) \right|. $$

Answer 2

You have to define $d(x,x')$ in order to be able to find your $\epsilon$ accordingly.

For example if you define $$d(x,x')= \max _{t\in [a,b]} \{ |x(t)-x'(t)|\}$$ then $$|f(x)-f(x')|\le (b-a)d(x,x')$$

so you can find your $\delta$ if an $\epsilon$ is given.