Continuity of linear operator but with a composition I'm trying to show that the following functionals are bounded, but i'm a bit lost on how much the composition affects the norm: 


*

*$f(x) = \int_{0}^{1} x(\sqrt{t}) dt$

*$f(x) = \lim_{n\rightarrow \infty}\int_{0}^{1} x({t}^n) dt$
where $x\in C([0,1],\mathbb{R})$ and $\|x\| = \max_{t \in [0,1]} |x(t)|$ (sup norm).
So when i compute the norm for instance in the first case: 
$\|f\| = \sup_{\|x\|=1} |f(x)| \leq \sup_{\|x\|=1} \int_{0}^{1} | x(\sqrt{t})| dt \leq \sup_{\|x\|=1} \int_{0}^{1} \| x\| dt = 1$ as  $ | x(\sqrt{t})| \leq  \max_{{t} \in [0,1]} |x(\sqrt{t})| = \max_{{y} \in [0,1]} |x(y)| = \|x\| $.
Is something wrong with this approach?  
Also as norm is continuous in the second case the limit wouldn't affect the result in this approach.
 A: 1.You have correctly proved that $||f|| \le 1$ (hence $f$ is bounded). I will only add that the inverse inequality $||f|| \ge 1$ also holds. To prove it take $x_0 \equiv 1$. Then $||x_0|| = 1$ and $f(x_0) = 1$. Therefore $||f|| = \sup\limits_{||x|| = 1} |f(x)| \ge |f(x_0)| = 1$. So, we proved that $||f|| = 1$.
2.Here first of all you have to prove that the sequence $f_n(x) = \int\limits_0^1 x(t^n)dt$ converges somewhere for all $x \in C[0,1]$ before considering norm of its limit $f$. If you know that the limit exists you can conclude that $||f|| \le 1$ since $||f_n|| \le 1$ (exactly as you said it follows from continuity of norm). But it is easier to just compute the limit. 
It is known that for $t \in [0,1)$ sequence $t^n$ converges to $0$ and therefore for all continuous functions $x(t)$ sequence $x(t^n)$ converges to $x(0)$ for all $t \in [0,1)$. Therefore by Lebesgue dominated convergence theorem (since $x$ is continuous it is bounded) $f_n(x)$ converges to $x(0)$. Therefore $f(x) = x(0)$. Now it is easy to see that $f$ is bounded and $||f|| = 1$.
