MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – Pietro Majer Jul 4 '13 at 19:07
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. – Danqing Jul 4 '13 at 19:22
It is enough: as I said, take the supremum over all x. – Pietro Majer Jul 5 '13 at 6:12
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 $T$. – Danqing Jul 5 '13 at 13:35

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.

share|cite|improve this answer
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 $x$. – Danqing Jul 4 '13 at 20:43
@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. – Bazin Jul 5 '13 at 16:00

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.

share|cite|improve this answer
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. – Danqing Jul 5 '13 at 19:18
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 $x$. – Danqing Jul 5 '13 at 19:26
I don't think you read my hint. – Nik Weaver Jul 5 '13 at 22:10
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. – Danqing Jul 6 '13 at 1:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.