# 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?

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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.
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 $$\sup_{n\in\mathbb{N}}\int_0^1 \vert K_n(x)\vert\,dx\leq C<+\infty.$$