MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Let $(X,B,\mu)$ and $(Y,C,\nu)$ be probability spaces, and let $m$ be the product measure. Let $f:X \times Y \rightarrow [0,\infty )$ be a $B \otimes C $ measurable function, $1 < p < \infty$, and define $f_x(y):=f(x,y)$.

What is the most general condition on $f$ to make sure that: $||f||_{L_p(m)}=\int{||f_x||_{L_p(\nu)}}d\mu(x)$ ?

I know that if $\int{|f_x(y)|^p}d\nu(y)$ is constant there is equality. But is it the only case?

share|cite|improve this question
up vote 8 down vote accepted

yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see here.

share|cite|improve this answer
Thank you. I wonder if under the same conditions as in the original question, is it always true that $||f||_{L_p(m)} \gt ∫||f_x||_{L_p(ν)}dμ(x)$? I think it can be proven using Tonelli's theorem and concave Jensen's inequality, but I'm not sure they are applicable here. – Shlomi May 2 '11 at 14:56

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.