TITLE: Boundedness of an integral operator
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Suppose that $\int_0^1|K_n(x)|\,\mathrm{d}x$ is unbounded, yet there exists some $M$ so that
$$
\left|\int_0^1K_n(x)f(x)\,\mathrm{d}x\right|\le M\|f\|_\infty
$$
Find $N$ so that $\int_0^1|K_N(x)|\,\mathrm{d}x\gt M$. Define
$$
f(x)=\left\{\begin{array}{}+1&\text{if }K_N(x)\ge0\\-1&\text{if }K_N(x)\lt0\end{array}\right.
$$
Then
$$
\begin{align}
\int_0^1K_N(x)f(x)\,\mathrm{d}x
&=\int_0^1|K_N(x)|\,\mathrm{d}x\\
&\gt M
\end{align}
$$
yet $\|f\|_\infty=1$. Contradiction.