For $p \ge 1$, $\nabla u \in L^p$ implies $u \in L^p_{\mathrm{loc}}$. At a glance I didn't see how the assumption $p \ge 1$ enters into the proof of that Theorem on p2 of Maz'ya's Sobolev Spaces; maybe it does not.

@IlyaBogdanov Oh, indeed. The combination of negation and changes of direction was just a bit too much for me, it seems. What we want is that with decreasing distance, $\alpha$ also decreases. Which means that $\alpha$ is increasing; in other words (since we don't need it to be strictly increasing), non-decreasing. And we cannot generally make it so if it converges to 1 when distances go to zero.

@IlyaBogdanov I think the argument does work the way it is. Can you be more precise about where you see a problem?

@IlyaBogdanov you're right, that's what I meant. We can make it non-increasing by assumption.

@BillJohnson Thank you, I understand now (the typo had confused me greatly). Indeed it's clear that this gives a norm $\phi$ that is equivalent to the canonical $\ell^2$ norm yet $\phi(e_n - e_m) = 4$ whenever $n \ne m$ and $\phi(e_n) = 2$ for every $n$. That's a very pleasantly simple example.

Interesting. So if I understand correctly, not even the stronger version, where the supremum in the definition of $\beta(B_X) = 2$ is attained (which is what one actually has in $\ell^1$, $\ell^\infty$, etc.) suffices to guarantee non-reflexivity.

As much as that's an interesting paper and I'm thankful for the reference: I don't think this answers any of my questions.

One has $\beta(B_{\ell^p}) = 2^{1/p}$ for $1 \le p < \infty$. Thus, $\lim_{p \to \infty} \beta(B_{\ell^p}) = 1$, but $\beta(B_{\ell^\infty}) = 2$.