Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I want to estimate the integral $$\int k(x,y)f(y)dy$$assuming the fact that $k(x,y), f(y)$ are in $L^p, L^q$ respectively. But I want to bound the the whole integral in $L^r$, $r\in [1,\infty]$.

I tried using Hölder but I am not given any more information. Even if I could apply Hölder then I always get confused on introducing $r$. I know its quite easy but I think a small hint could do if I have to apply Hölder.


share|cite|improve this question
Is there a link between $p,q$ and $r$ or not? – Davide Giraudo Nov 22 '12 at 19:22
@Davide Giraudo : This is a question thrown out by my classmate. I think $p^{-1}+q^{-1}=1$ should be the criteria right ?? If thats the case can you give me a small hint to apply Hölder inequality ?? – Theorem Nov 22 '12 at 19:26
@Davide Giraudo : if $r=1$ then i would apply it straight forward, but if r is not $1$ i get often confused in choosing the right exponents . – Theorem Nov 22 '12 at 19:27
Maybe the relationship is $\frac 1r=\frac 1p+\frac 1q$. – Davide Giraudo Nov 22 '12 at 20:24
@DavideGiraudo : Sir, I am pretty sure i was just told that $r\in[1,\infty]$, and had no relation with $p$ and $q$. – Theorem Nov 22 '12 at 20:28

Note that if $f \in L^q(A)$ and $g \in L^q(B)$, $f(x) g(y) \in L^q(A \times B)$. Then if $k \in L^p(A \times A)$ with $1/p + 1/q = 1$, $$\left| \int_{A \times B} k(x,y) f(x) g(y)\ dx \ dy \right| \le \|k\|_p \|f\|_q \|g\|_q$$ Since this is true for every $g \in L^q(B)$, $\int k(x,y) f(x)\ dx \in L^p(B)$ with $\|\int k(x,y) f(x)\ dx \|_p \le \|k\|_p \|f\|_q$.

However, it won't work if $r \ne p$. In order to have $\int_A k(x,y) f(x)\ dx \in L^r(B)$, what you'd want is $k(\cdot,y) \in L^p$ for almost all $y \in B$ with $y \to \|k(\cdot,y)\|_p \in L^r(B)$. If $L^p(B)$ is not a subset of $L^r(B)$, this won't be true. In fact, given $h \in L^p(B) \backslash L^r(B)$ and $f\ne 0 \in L^q(A)$, take $g \in L^p(A)$ with $\|g\|_p = 1$ and $\int_A f g \ dx = \|f\|_q$, and define $k(x,y) = g(x) h(y)$. Then $k \in L^p$ with $\|k\|_p = \|g\|_p \|h\|_p$, and $\int_A k(x,y) f(x)\ dx = \|f\|_q h(y) \notin L^r(B)$.

share|cite|improve this answer

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.