Operators from $L^{\infty}$ to $L^{\infty}$ If $T$ defined as $T(f)(x)=\int K(x,y)f(y)dy$, where $K(x,y)$ is locally integrable, is bounded from $L^{\infty}$ to $L^{\infty}$, how can we show that $\|\int|K(\cdot,y)|dy\|_{L^{\infty}}\le \|T\|_{L^{\infty}\rightarrow L^{\infty}}$? 
This problem comes from Page 13 of Meyer and Coifman's Wavelets: Calderón-Zygmund and Multilinear Operators.
I asked this question on StackExchange Mathematics before but nobody answered me. https://math.stackexchange.com/questions/428449/operators-from-l-infty-to-l-infty-below-bound-of-the-norm
 A: Interesting question: obviously a sufficient condition for your operator $T$ to be a bounded endomorphism of  $L^\infty$ is that
$$
\sup_x\int\vert K(x,y)\vert dy=C<+\infty \ (\sharp).\quad\text{This implies trivially  $\vert (Tu)(x)\vert\le C\Vert{u}\Vert_{L^\infty}$.}
$$
Note that this is precisely the case for the convolution by an $L^1$ function $f$, a case for which $K(x,y)=f(x-y)$. 
Many interesting operators fail to be bounded endomorphisms of $L^1$ and of $L^\infty$ but are bounded on $L^p$ for $p\in(1,+\infty)$. This is the case in particular of the Hilbert transform (convolution with $pv(1/x)$) which sends $L^1$ into $L^1_{weak}$ and $L^\infty$ into $BMO$ by a Marcinkiewicz argument.
A: Well, the discrete version of this question (where $Tf(m) = \sum K(m,n)f(n)$ maps $l^\infty$ to $l^\infty$) isn't so hard. That should give you an intuition for why it's true. I don't see any slick way to handle the continuous case; I guess you could do it first when $K$ is the characteristic function of a rectangle, then when $K$ is a finite linear combination of such things, then when $K$ is the characteristic function of any measurable set, then when $K$ is any simple function, and then finally for general $K$.
Incidentally, the reverse inequality is easier. For $h \in L^1$ let $H$ be the bounded linear functional on $L^\infty$ given by integrating against $h$. Then for any $\epsilon > 0$ we can find $h$ with $\|h\|_1 = 1$ such that $\|H \circ T\| \geq \|T\| - \epsilon$. But $$(H \circ T)(f) = \int h(x)\int K(x,y) f(y)dydx = \int\left(\int h(x)K(x,y)dx\right)f(y)dy$$ is integration against the function $\int h(x)K(x,y)dx$, and the $L^1$ norm of this function is at most $\int\int |h(x)||K(x,y)|dxdy = \int |h(x)|\big(\int |K(x,y)|dy\big)dx \leq \|\int|K(\cdot,y)|dy\|_\infty$. At least that works if $K$ is integrable so you can change order of integration.
