I’d like to record the following calculation, so that I don’t forget it (again).

(I worked out the details over spring break, promptly forgot everything when school started back up, and had to go to the trouble of recreating my argument from scratch last night; obviously I would not like to repeat the cycle unnecessarily.)

The following supercritical case of Sobolev embedding holds for \(s \in \mathbb{R}\), \(d \geq 1\), \(1 < p < \infty\): suppose $$s \ – \, \frac{d}{p} = k + \theta, \qquad k \in \mathbb{N}_0, \ \theta \in (0, 1).$$ Then \(H^{s, p}(\mathbb{R}^d)\) embeds into \(C^{k, \theta}(\mathbb{R}^d)\). (When \(k > 0\), the strategy, of course, is to trade off the integer part for actual derivatives, and use the fractional part to determine the modulus of continuity. Hence for the proof, we may assume \(k = 0\).)

The general criterion is: \(s p > d\). When \(s = 1\), this is precisely the classical statement of Morrey’s inequality.

When \(s > 1\), we can trade off regularity to gain integrability, so that every decrease of \(s\) by \(h\) units produces a decrease of \(\frac{1}{p}\) by \(\frac{h}{d}\). In particular, the quantity \(s \ – \, \frac{d}{p}\) remains invariant throughout the embedding process. With this, we can reduce to the \(s = 1\) case.

So the only remaining task is to verify this in the \(0 < s < 1\) regime, where no embeddings can reduce matters to the classical \(W^{1, p}\) case (since that is no longer going down in regularity). There are probably other ways to do this (I just realized, in the process of writing this, that the Littlewood-Paley characterization of the Hölder spaces gives a very quick way to proceed; just use Bernstein), but I wanted to record this specific calculation before I lost it for a second time.

We first observe that whenever \(s p > d\), we have the embedding of \(H^{s, p}(\mathbb{R}^d)\) into \(C_0(\mathbb{R}^d)\). Indeed, we compute $$K_N(x) = \int_{\mathbb{R}^d} e^{2 \pi i \xi \cdot x} \langle \xi \rangle^{- s} \psi(\xi / N) \, d\xi,$$ which is $$|K_N(x)| \lesssim \min(N^{d \ – \, s}, N^{d \ – \, s} (N |x|)^{- m})$$ for every \(m \geq 1\), by the Leibniz rule. It follows that $$K = K_{\leq 1} + \sum_{N > 1} K_N$$ is absolutely-convergent, and gives a function obeying $$|K(x)| \lesssim \begin{cases} |x|^{- (d \ – \, s)} & |x| \lesssim 1 \\ |x|^{- m} & |x| \gg 1 \end{cases}.$$

This is the Bessel potential to order \(s > 0\), and we see now that to guarantee \(\langle \nabla \rangle^{- s} f \in L^{\infty}\) for all \(f \in L^p\), we will want \(| \cdot |^{- (d \ – \, s)} \in L^{p’}_{\text{loc}}\). That is, we want \(p’ (d \ – \, s) < d\), or \(1 \ – \, \frac{s}{d} < 1 \ – \, \frac{1}{p}\), and from this we see the source of the restriction on \(s\) and \(p\).

Then we get $$\|f\|_{L^{\infty}(\mathbb{R}^d)} \lesssim \|f\|_{H^{s, p}(\mathbb{R}^d)},$$ and this a priori inequality and density gives the initial claim.

We now wish to sharpen this inequality: take \(\lambda > 0\), and insert \(f_{\lambda}\) given by \(f_{\lambda}(x) = f(\lambda x)\) into the bound $$\|f\|_{L^{\infty}(\mathbb{R}^d)} \lesssim \|f\|_{L^p(\mathbb{R}^d)} + \||\nabla|^s f\|_{L^p(\mathbb{R}^d)};$$ the left-hand side is unaffected, while the right becomes $$\lambda^{- \frac{d}{p}} \|f\|_{L^p(\mathbb{R}^d)} + \lambda^s \lambda^{- \frac{d}{p}} \||\nabla|^s f\|_{L^p(\mathbb{R}^d)}.$$ Optimizing over \(\lambda\) gives $$\|f\|_{L^{\infty}(\mathbb{R}^d)} \lesssim \|f\|_{L^p(\mathbb{R}^d)}^{1 \ – \, \frac{1}{s} \frac{d}{p}} \||\nabla|^s f\|_{L^p(\mathbb{R}^d)}^{\frac{1}{s} \frac{d}{p}}.$$

Now observe that $$\|\tau_h f \ – \, f\|_{L^p(\mathbb{R}^d)} \leq 2 \|f\|_{L^p(\mathbb{R}^d)},$$ while $$\|\tau_h f \ – \, f\|_{L^p(\mathbb{R}^d)} \leq |h| \|\nabla f\|_{L^p(\mathbb{R}^d)};$$ the first is obvious, and the second, of course, is a familiar calculus inequality. In particular, complex interpolation in the \(H^{s, p}\) spaces (see my previous post on the subject), or even just (a valid) Stein interpolation with \(T_z = \langle \nabla \rangle^{- z}\), gives $$\|\tau_h f \ – \, f\|_{L^p(\mathbb{R}^d)} \lesssim |h|^s \|f\|_{H^{s, p}(\mathbb{R}^d)},$$ for all \(0 \leq s \leq 1\). To get the desired dependence, we redefine \(f\) to be \(f_{\lambda}\) and \(h\) to be \(h / \lambda\), to get $$\lambda^{- \frac{d}{p}} \|\tau_h f \ – \, f\|_{L^p(\mathbb{R}^d)} \lesssim \lambda^{- s} |h|^s (\lambda^{- \frac{d}{p}} \|f\|_{L^p(\mathbb{R}^d)} + \lambda^s \lambda^{- \frac{d}{p}} \||\nabla|^s f\|_{L^p(\mathbb{R}^d)}),$$ and for \(s > 0\), sending \(\lambda \to \infty\) gives the expected relationship.

Now we insert $$\|\tau_h f \, – \, f\|_{L^{\infty}(\mathbb{R}^d)} \lesssim \|\tau_h f \ – \, f\|_{L^p(\mathbb{R}^d)}^{1 \ – \, \frac{1}{s} \frac{d}{p}} \||\nabla|^s (\tau_h f \ – \, f)\|_{L^p(\mathbb{R}^d)}^{\frac{1}{s} \frac{d}{p}}.$$ Now using our inequality for translations on \(L^p\), we get $$\|\tau_h f \ – \, f\|_{L^{\infty}(\mathbb{R}^d)} \lesssim (|h|^s)^{1 \ – \, \frac{1}{s} \frac{d}{p}} \||\nabla|^s f\|_{L^p(\mathbb{R}^d)},$$ which we recognize to be of the desired form. And so we obtain the embedding.