Showing an operator is continuous When $C[a,b]$ is the space of all real-valued continuous function defined on $[a,b]$ endowed with the uniform norm, for $F: C[a,b] \to C[a,b]$ given by $F(x)(t) = \int_a^t x(s)ds$ to show continuity can I argue that for all $t$ it holds that $$\lim_{n \to \infty} F(x_n)(t) = \lim_{n \to \infty} \int_a^t x_n(s)ds =  \int_a^t x(s)ds = F(x)(t)$$ when $x_n$ converges to $x$ uniformly?
 A: This argument is not sufficient to show that $F$ is continuous since you've only shown that $Fx_n \to Fx$ pointwise but you need uniform convergence of $Fx_n$ to $Fx$. You could try to add some details to the above argument to conclude that the convergence is in fact uniform in $t$.
However, I feel I should point out that for linear operators on normed spaces, it is often much easier to check that the operator is bounded to see that it is continuous. This means you would check that
$$\|Fx\|_\infty \leq C \|x\|_\infty$$ for some $C$. This is easier since
$$|Fx(t)| = \big| \int_a^t x(s) ds \big | \leq \int_a^t \|x\|_\infty ds \leq (b-a) \|x\|_\infty.$$
A: You can prove it using the inequalities:
$\vert F(x_n)(t)-F(x)(t)\vert  =\vert \int_a^t (x_n(s)-x(s))\vert \leq \int_a^t \vert x_n(s)-x(s) \vert ds \leq (b-a)\cdot \Vert x_n-x\Vert_\infty$
From which you'll obtain not only  point-wise convergence (which is what you asked about), but also uniform convergence of $F(x_n)$ to $F(x)$, which is the metric on $C[a,b]$.
