Boundedness of an integral operator Let $K_n \in L^1([0,1]), n \geq 1$ and define a linear map $T$ from  $L^\infty([0,1]) $to sequences by 
$$ Tf = (x_n), \;\; x_n =\int_0^1 K_n(x)f(x)dx$$ 
Show that $T$ is a bounded linear operator from $L^\infty([0,1]) $to $\ell^\infty$ iff
$$\sup_{n\geq 1} \int_0^1|K_n(x)| dx \lt \infty$$ 
My try: 
$(\Leftarrow)$ 
$$\sup_n |x_n| = \sup_n  |\int_0^1 K_n(x)f(x) dx| \leq \sup_n\int_0^1 |K_n(x)f(x)| dx \leq \|f\|_\infty \sup_n\int_0^1 |K_n(x)|dx $$
 $(\Rightarrow)$
 I can't get the absolute value right. I was thinking uniformed boundedness and that every coordinate can be written with help of a linear functional. But then I end up with $\sup_{\|f\| = 1} |\int_0^1 K_n(x) f(x) dx | \leq \infty$. Can I choose my $f$ so that I get what I want?

Yes, you can choose $f$ as you want. If $T$ is bounded then
$$
\exists C>0\qquad\left\vert \int_0^1K_n(x)f(x)\,dx\right\vert\leq C \Vert f\Vert_{L^\infty}\qquad \forall f\in L^\infty\quad \forall n\in\mathbb{N}.
$$
Fix $m\in\mathbb{N}$, if we take $f=\text{sign}(K_m)\in L^\infty$ then
$$
\int_0^1 \vert K_m(x)\vert\,dx\leq C.
$$
Repeat this construction for every $m\in\mathbb{N}$ and you obtain
$$
\sup_{n\in\mathbb{N}}\int_0^1 \vert K_n(x)\vert\,dx\leq C<+\infty.
$$
