@MikeMiller: Yeah, but I don't understand why it's so important in $p$-harmonic analysis. If we have that two numbers $p$ and $q$ are conjugate Hölder exponents, then $u$ is $p$-harmonic, and $v$ is $q$-harmonic. It gives us that $\langle\nabla u,\nabla v\rangle=0$, but I don't see any intuitive reason why the exponents have to be Hölder conjugate. I can make the computation, sure, but other than that I don't know.
I have tried sleeping while the party of italians were playing a lourd mafia card game thing in the next room then I pretended to watch and be interested only to steal their deck of cards and hide it
@OskarTegby I hadn't paid much attention to stuff in here, and somehow I thought you were taking linear algebra or something... so I was surprised to see p-Laplacians.
With the assumption that we're done with $y-x>1$, we can say that there is a natural $n$ such that $y-x>\frac{1}{n}$. @TedShifrin That intuitively means that we're also done for $y-x>0$. Because $\frac{1}{n}$ can be sufficiently small.
@OskarTegby I think that it probably all comes back to Holder's inequality, and the point being that if you want to bash two things against each other and integrate, the best you can do is when the exponents are comnjugate.
@OskarTegby If $g \in L^q(\Bbb R)$, then the map $g^\vee: L^p(\Bbb R) \to \Bbb R$ is given by $g^\vee(f) = \int gf$. That this is continuous is Holder.
It's hard to focus, my mom and sister are making beats and saying "Go EE." while my little brother stomps in the middle of the floor. They won't stop singing ;-;
@Zee: Do you mean that I'm crazy by letting myself being distracted by them, or by wanting to not be distracted by them? Sorry for being off-topic. I'll stop this discussion after I've received the answer.
