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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

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  • $\begingroup$ For $f_x(y):=\operatorname{sgn}K(x,y)$ we have $(Tf_x)(x)=\int |K(x,y)|dy\le \|T\|_{\infty,\infty}$; taking $\sup_x$ gives the inequality. $\endgroup$
    – Pietro Majer
    Commented Jul 4, 2013 at 19:07
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    $\begingroup$ Thank you. But I don't think the inequality is true, since if we fix $x$, then $\int K(\cdot,y)f_x(y)dy$ is really an $L^{\infty}$ function whose norm is bounded by $\|T\|$, but it's not enough to show that the inequality is valid for some specific point. $\endgroup$
    – Danqing
    Commented Jul 4, 2013 at 19:22
  • $\begingroup$ It is enough: as I said, take the supremum over all x. $\endgroup$
    – Pietro Majer
    Commented Jul 5, 2013 at 6:12
  • $\begingroup$ We cannot say $\int|K(x,y)|dy\le\|T\|$, how can we take the $\sup$ to get the inequality? You have to notice that if you want to take $\sup_x$ for $(Tf_x)(x)$, then $f_x$ should be a function independent of $x$, which is the definition of the operator of $T$. $\endgroup$
    – Danqing
    Commented Jul 5, 2013 at 13:35

2 Answers 2

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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.

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  • $\begingroup$ Thank yo for you answer, but there is still one point which I don't understand and can you help me to make it clear? I agree that $$ ess\sup \int|K(\cdot,y)|dy\le ess\sup_{x}\ \sup_{\|u\|_{L^{\infty}=1}}|(Tu)(x)|, $$ but how can we get rid of the $\sup$ in the RHS since $\sup_{\|u\|_{L^{\infty}=1}}|(Tu)(x)|$ may be $\infty$ for some point $x$. $\endgroup$
    – Danqing
    Commented Jul 4, 2013 at 20:43
  • $\begingroup$ @Danqing You are right, I have withdrawn the middle part of my answer. The commutation of the essential supremum with a supremum may pose some problems. $\endgroup$
    – Bazin
    Commented Jul 5, 2013 at 16:00
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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.

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  • $\begingroup$ Thanks for your answer. I read it carefully but I still don't know how to solve my question by your hint, since I think there is a essential difference between the discrete case and continuous case. Let me make it clear and use notations $x$ and $y$ in both cases. If we fix $t$ and define $f_t(y)=\textrm{sgn } K(t,y)$, then $\|\int K(\cdot,y)f_t(y)dy\|\le \|T\|$, in particular, in the discrete case, $\int K(t,y)f_t(y)dy\le\|T\|$, therefore we get the conclusion. But obviously the reasoning doesn't work for the continuous case since $\int K(t,y)f_t(y)dy$ may be any number. $\endgroup$
    – Danqing
    Commented Jul 5, 2013 at 19:18
  • $\begingroup$ Somebody may argue that if we can take $ess\sup$ of $\int K(t,y)f_t(y)dy$ as a function of $t$, but we have to notice that we cannot exclude the case that for each fixed t, $\int K(t,y)f_t(y)dy=\infty$. Thank you for your discussion about the reverse, but there is a more direct argument where we don't need the duality. We need only to notice that $|\int K(x,y)f(y)dy|\le \|f\|_{L^{\infty}}\int |K(x,y)|dy$ for each point $x$. $\endgroup$
    – Danqing
    Commented Jul 5, 2013 at 19:26
  • $\begingroup$ I don't think you read my hint. $\endgroup$
    – Nik Weaver
    Commented Jul 5, 2013 at 22:10
  • $\begingroup$ Actually I finished the discussion until the case $K$ is a characteristic function of arbitrary measurable set, but I don't see that rest cases follow. $\endgroup$
    – Danqing
    Commented Jul 6, 2013 at 1:55

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