If $||f||_2^2 = ||f||_3^3 = ||f||_4^4$, then $f$ is a characteristic function almost everywhere.

Question

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.

Answer 1

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.

Answer 2

(Turns out this doesn't answer your question, and the claim originally made in this answer is wrong.)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