Suppose that $f\geq0$ is an element of $L^2 \cap L^3 \cap L^4$, and moreso that $||f||_2^2 = ||f||_3^3 = ||f||_4^4$. If the measure of the whole space is finite, I want to show that $f = \chi_A$ a.e. for some measurable $A$.

My attempt:

Write $S = \{x : f(x) \neq 1\}$ and let $A = S^c$. I want to show that $S$ has measure zero. As $S\cup A$ is the whole space, we can partition the integrals as so, to see that

$$\int_S |f|^2 = \int_S |f|^3 = \int_S |f|^4$$.

I'm not sure how to proceed from here.

  • 1
    $\begingroup$ You don't want to show $S$ has measure zero, you want to show that $\{ x : f(x) \not \in \{ 0,1 \} \}$ has measure zero. To do that you might try starting by assuming that $f$ is simple. $\endgroup$ – Ian Oct 17 '17 at 21:08

Hint: $$\int(f^2-f)^2=\int f^4+f^2-2f^3=\int f^4+\int f^2-2\int f^3=0$$ This tells you that $f^2-f=0$ a.e.

  • $\begingroup$ @copper.hat Thank you!! $\endgroup$ – tattwamasi amrutam Oct 17 '17 at 21:14
  • 1
    $\begingroup$ Nice. It does not seem that we need to assume the measure to be finite. $\endgroup$ – Ramiro Oct 18 '17 at 12:05
  • $\begingroup$ Yeah. I think so. $\endgroup$ – tattwamasi amrutam Oct 18 '17 at 14:09

(Turns out this doesn't answer your question, and the claim originally made in this answer is wrong.)

Nonetheless: Your claim would hold for probability measures if any two $ L^p $ norms agreed, even though this is not what you asked.

Indeed, by Jensen's inequality, the function $\|f\|_p $ is strictly convex unless $ f $ is almost constant. The proof of the required strict Jensen inequality is elementary (https://www.statlect.com/fundamentals-of-probability/Jensen-inequality )

  • $\begingroup$ That can't be right as stated, because clearly $\| \chi_A \|_p^p=m(A)$, which is not a strictly convex function of $p$, and $\chi_A$ is not almost constant if $0<m(A)<1$ (taking for granted your restriction to probability measures). $\endgroup$ – Ian Oct 17 '17 at 22:41
  • $\begingroup$ More importantly, your claim itself is false: a random variable which is equal to $1/2$ with probability $8/9$ and $2$ with probability $1/9$ is a counterexample, in that it has the same first and second moments. $\endgroup$ – Ian Oct 17 '17 at 22:50
  • $\begingroup$ In fact I think your claim is quite dramatically false, in that I think there is always a nonnegative vector $\begin{bmatrix} p \\ q \end{bmatrix}$ and a $c>0$ such that $\begin{bmatrix} x_1^{p_1} & x_2^{p_1} \\ x_1^{p_2} & x_2^{p_2} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} c \\ c \\ 1 \end{bmatrix}$, whenever $0<x_1<x_2,1 \leq p_1<p_2$. $\endgroup$ – Ian Oct 17 '17 at 22:58
  • $\begingroup$ @Ian you are right. It is the $L^p$ norm itself that is strictly convex, not the $L^p$ norm to the power $p$. This means that my first claim is actually correct for probability measures (equality of $L^p$ norms suffices), but that is not what the OP asked for $\endgroup$ – Bananach Oct 18 '17 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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