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?