In this answer, I am confused by the following step. $$\int \hat{f} \hat{g}=\frac{1}{p} \int \hat{f}^p + \frac{1}{q} \int \hat{g}^p \iff \hat{f} \hat{g} = \frac{1}{p} \hat{f}^p + \frac{1}{q} \hat{g}^p \quad \text{a.e.}$$
The $\impliedby$ implication follows by integrating, but why does the $\implies$ implication hold?
Young's inequality states that
$$ab \leq \frac{a^p}{p} + \frac{b^q}{q}$$
whenever $a,b \geq 0$ and $\frac{1}{p} +\frac{1}{q}=1$. This means that
$$\frac{1}{p} \hat{f}^p + \frac{1}{q} \hat{g}^q - \hat{f} \hat{g} \geq 0.$$
Therefore
$$\int \left( \frac{1}{p} \hat{f}^p + \frac{1}{q} \hat{g}^q - \hat{f} \hat{g} \right) =0$$
implies
$$\frac{1}{p} \hat{f}^p + \frac{1}{q} \hat{g}^q - \hat{f} \hat{g} = 0 \qquad \text{a.e.}$$
(Recall: If $\int f(x) \, dx=0$ for some $f \geq 0$, then $f=0$ almost everywhere.)