If $u_n \rightarrow u$ in $C([a, b])$, is it true that $\int u_n(x) dx \rightarrow \int u(x) dx$?

Question

Let $u_n \rightarrow u$ in $C([a, b])$.Is it true that $$ \underset{a}{\overset{(a+b)/2}{\int}} u_n(x) dx \rightarrow \underset{a}{\overset{(a+b)/2}{\int}} u(x) dx ?$$

If $u_n \rightarrow u$ in $C([a, b])$ then $u_n(x) \rightarrow u(x)$ for all $x \in [a, b]$. So, I think that $$ \left|\underset{a}{\overset{(a+b)/2}{\int}} (u_n(x) - u(x)) dx \hspace{8pt} \right| \rightarrow 0 $$ hence

$$ \underset{a}{\overset{(a+b)/2}{\int}} u_n(x) dx \rightarrow \underset{a}{\overset{(a+b)/2}{\int}} u(x) dx$$

Is it correct?

Thank you!

Answer 1

Yes, the metric in $C([a,b])$ is the supremum metric so that $$ \left| \int_a^{(a+b)2} u_n(t) \, dt - \int_a^{(a+b)2} u(t) \, dt \right| = \left| \int_a^{(a+b)2} u_n(t) - u(t) \, dt \right| \le \int_a^{(a+b)/2} |u_n(t) - u(t)| \, dt$$ and thus $$ \left| \int_a^{(a+b)2} u_n(t) \, dt - \int_a^{(a+b)2} u(t) \, dt \right| \le \frac{b-a}{2} \max_{t \in [a,b]} |u_n(t) - u(t)| \to 0.$$