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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.

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    $\begingroup$ 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
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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.

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  • $\begingroup$ @copper.hat Thank you!! $\endgroup$ – tattwamasi amrutam Oct 17 '17 at 21:14
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    $\begingroup$ 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
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(Turns out this doesn't answer your question, and the claim originally made in this answer is wrong.)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 )

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  • $\begingroup$ 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

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