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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?

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angryavian
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  • The exponents (on the right-hand side of both equalities) are missing, right? (I.e. $\hat{f}^p$ and $\hat{g}^q$ instead of $\hat{f}$ and $\hat{g}$, respectively.) – saz Feb 07 '15 at 15:22
  • @saz Yes, thank you for catching that. I've edited my question. – angryavian Feb 07 '15 at 15:54

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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.)

saz
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