Showing two functions are a.e. equivalent in $L^3$ norm Let $X$ be a measure space with measure $\mu$. Let $f,g \in L^3(X, \mu)$. Assume we have $$||f||_{L^3} = ||g||_{L^3} = \int_{X} ||f(x)||^2g(x) \ d\mu(x) = 1$$
How can we show $g=|f|$ a.e.?
 A: Let $p=3/2$, $q=3$, and define $h(x)\equiv|f(x)|^2$ for every $x\in X$. Note that $1/p+1/q=1$. Then, Hölder’s inequality actually holds as an equality:
\begin{align*}
1&=\int_X|f(x)|^2g(x)\,\mathrm d\mu(x)\leq\int_X|f(x)|^2|g(x)|\,\mathrm d\mu(x)=\|hg\|_1\leq\|h\|_p\|g\|_q\\&=\|f\|_3^2\|g\|_3 =1.
\end{align*} This is possible only if $|h|^p$ (that is, $|f|^3$) and $|g|^q$ (that is, $|g|^3$) are “linearly dependent” functions in the sense that there exist constants $\alpha$ and $\beta$, not both zero, such that $\alpha|f(x)|^3=\beta|g(x)|^3$ for $\mu$-almost every $x\in X$.
Without loss of generality, assume that $\alpha\neq0$ and let $\gamma\equiv\beta/\alpha$. Therefore, $|f(x)|^3=\gamma|g(x)|^3$ for $\mu$-almost every $x\in X$. But $\|f\|_3=\|g\|_3=1$ forces $\gamma$ to be $1$.
Finally, the fact (see the chain of inequalities above) that $$\int_X|f(x)|^2g(x)\,\mathrm d\mu(x)=\int_X|f(x)|^2|g(x)|\,\mathrm d\mu(x)$$ implies also that $|f(x)|^2g(x)\geq 0$ for $\mu$-almost every $x\in X$. This, together with $|f(x)|^3=|g(x)|^3$ for $\mu$-almost every $x\in X$, is sufficient to conclude that $|f(x)|=g(x)$ for $\mu$-almost every $x\in X$ (in particular, $g$ is non-negative $\mu$-almost everywhere).
