If I know that $f \in L^p(\mathbb{R})$, then what can I say about $f^{\frac{r-p}{r}}$? That is, what space does it it live in and why?
The reason I ask is I don't understand how the Generalized Holder Inequality was used in the proof here. The relevant section is: $$ \int \left({\left\vert{ f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right)^{1/r} \cdot \left\vert{ f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} \cdot \left\vert{ g \left({y}\right)}\right\vert^{ (r-q)/r} ~\mathrm d y \le \underset{I} {\underbrace{ \left\Vert{ \left({ \left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q }\right)^{1/r} }\right\Vert_r} } \cdot \underset{II}{\underbrace{\left\Vert{\left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} }\right\Vert_{\textstyle \frac{pr}{r-p} } } } \cdot \underset{III}{\underbrace{\left\Vert{\left\vert{ g \left({y}\right)}\right\vert^{(r-q)/r} }\right\Vert_{\textstyle \frac{qr}{r-q} } } } $$
I can't see how the particular norms for the terms $I$, $II$ and $III$ were arrived at given only that we know $f \in L^p$ and $g \in L^q$.
Edit: I have also just noticed that in another proof for the same inequality that the numerator and denominator are switched in the norms of $I, II,$ and $III$. The other proof is here and is the same up until it states:
$$ \int_{\mathbb{R}^n}(|f(y)|^{p}|g(x-y)|^{q})^{\frac{1}{r}}|f(y)|^{1 -\frac{p}{r}}|g(x-y)|^{1 -\frac{q}{r}}dy \le ||(|f(y)|^{p}|g(x-y)|^{q})^{\frac{1}{r}}||_{L^r}|||f(y)|^{1 -\frac{p}{r}}||_{L^{\frac{r-p}{rp}}}||g(x-y)|^{1 -\frac{q}{r}}||_{L^{\frac{r-q}{rq}}} $$ So is one of these proofs incorrect or is $L^\frac{pr}{r-p}$ equivalent to $L^\frac{r-p}{rp}$?