Let $f : C[a,b]\to\mathbb{R}$ with $$f(x) =\int_a^b x^2(t)\mbox{d}t $$ Show that $f$ is continous. Norm in $C[a,b]$ given as $$\|x\| = \max\limits_{t\in [a,b]}x(t) $$
My idea is to show that $f$ is Lipschitz (every L. operator is continous). We need $L\geq 0$ s.t $$\forall x,y\in C[a,b]\left (|f(x)-f(y)|\leq L\|x-y\|\right ) $$ We have, by continuity in $[a,b]$, $K>0$ $$\forall x\in C[a,b], \forall t\in [a,b], (|x(t)|\leq K) $$ Looking at $f(x)$: $$|f(x)| = \left |\int_a^b x^2(t)\mbox{d}t\right |\leq \int_a^b |x^2(t)|\mbox{d}t\leq \int_a^b K^2\mbox{d}t\leq K^2|b-a| $$ but at this point I'm lost. All I can say is $\|x-y\|\geq 0$ which isn't helpful.
Am I going the right direction and/or how to proceed?
Let:
$B: C[a,b] \times C[a,b] \to \Bbb R$ be the (check that it is) bilinear form given by:
$$B(x,y) = \int_a^b x(t)y(t) dt$$
We have that:
$$|B(x,y)| \le \int_a^b |x(t)||y(t)| dt \le (b-a)\|x\|_{\infty} \|y\|_{\infty} $$
Hence, $B$ is continuous and in particular $\|B\| \le b-a$.
Now, $f(x) = B(x,x)$ is the composition of two continuous functions, hence it is continuous.