We learned in class that if $f \in L^p$ and $g \in L^q$ and $\frac{1}{p}+\frac{1}{q} =1$, then $fg \in L^1$. I've been searching around for a more general formula for awhile, but it's surprisingly hard to find. I suppose that means that the obvious generalization, namely $$\frac{1}{p}+\frac{1}{q}=\frac{1}{r} \rightarrow fg \in L^r$$ isn't true. My question is simply: is this true, and if not, what's a simple counterexample?
Also, is there anything we say about $fg$ at this level of generality?
It follows from the $r=1$ case. For simplicity, assume that $f$ and $g$ are real and positive. Note that if $\frac 1 p+\frac 1 q=\frac 1 r$, then $\frac r p + \frac r q = 1$, and $f^r\in L^{\frac{p}{r}}$ and $g^r\in L^\frac{q}{r}$, so $(f^rg^r)=(fg)^r\in L^1$, but this is equivalent to $fg\in L^r$.
As Sangchul Lee pointed out in a comment, you also have a variant of Hölder's inequality in this case: $$\lVert fg\rVert_r\leq \lVert f\rVert_p\lVert g\rVert_q$$
It follows from the basic Hölder's inequality using the same trick. You can also find it on Wikipedia.