# What goes wrong for the Sobolev embeddings at $k=n/p$?

For $u\in W^{k,p}(U)$, where $U\subseteq\mathbb{R}^n$ is open and bounded with $C^1$-boundary, we have the celebrated Sobolev inequalities:
If $k < n/p$ then $u\in L^q(U)$ for $q$ satisfying $\frac{1}{q}=\frac{1}{p}-\frac{k}{n}$,
If $k > n/p$ then $u$ lies in a particular Hölder space.

It is also known that this doesn't work for the borderline case $k=n/p$ (which is related to the Sobolev conjugate $p^*\to \infty$ as $p\to n$), although one would expect/hope for $u\in L^\infty(U)$.
**An exception as Denis mentions: it works for $(k,p)=(n,1)$ via the fundamental theorem of calculus.
**As a counterexample, the function $u(x)=\log\log(1+\frac{1}{|x|})$ with domain $U=int(\mathbb{D}^n)$ lies in $W^{1,n}$ but not in $L^\infty$ (for $n>1$). Instead, apparently the result we end up with is that the functions lie in "the space of functions with bounded mean oscillation".

1) Is there an intuitive / deeper reason as to what goes wrong?
2) Is there some sort of geometric reason why the Sobolev embedding theorem has to fail for $k=n/p$? I've been told that this critical case seems to arise often in geometry/topology.

• actually, the embedding holds true in one critical case, namely $p=1$, $k=n$. Mar 9, 2013 at 7:45
• I'm not sure I quite appreciate exactly what is being asked. You lay out two cases and then say that what is essentially just a third case is an example of the other two not working. You say that the Sobolev embedding theorem "fails" or "goes wrong" when $k=n/p$, but one might say that it is simply neither of the two cases you lay out at the start. Nothing "fails", it just happens to be its own special case. Mar 11, 2013 at 23:12

I'll take a stab. In the following we consider the case $W^{1,n}$ in $\mathbb{R}^n$. My short answer is that under rescaling by factor $\lambda$, derivatives scale by $\lambda$ and volumes by $\lambda^{-n}$, so integrating derivatives to the $n$ won't change under rescaling. The following examples illustrate how this affects embeddings.
As for no Holder continuity, look at a smooth bump function $\varphi$ supported on $B_1$ with $|\nabla \phi| < 2$. The rescalings $\varphi(x/\epsilon)$ have arbitrarily bad modulus of continuity, but bounded $W^{1,n}$ norm, since (key point) the derivative to the $n$ (~$\epsilon^{-n}$) grows exactly like the volume of support (~$\epsilon^{n}$) decays. This says that we cannot control the modulus of continuity by the $W^{1,n}$ norm. (As expected, these functions have unbounded $W^{1,p}$ norm for $p > n$.)
As for not embedding into $L^{\infty}$, the way I would try to see how things could go wrong is take a function $\psi$ positive, supported on $B_2$, with $\psi \equiv 1$ on $B_1$ and $|\nabla \psi| < 2,$ and add dyadic rescalings together. Consider $$u(x) = \sum_{i} h_i\psi(2^{i}x)$$ for some $h_i$ we will choose to give bounded $W^{1,n}$ norm but unbounded height of $u$. Note that $|\nabla (h_{i}\psi(2^{i}x))|$ grows like $h_i2^{i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{1,n}$ norm we want $$\sum_{i} h_i^{n} < C.$$ Again, the key point is that volume decays with the same power that the derivatives of rescalings to the $n$ grows. To give unboundedness we just want $$\sum_{i} h_i = \infty.$$ The canonical example of such a sequence is $h_i = 1/i$. Ultimately this is just the same example as you gave since $\sum_{i=1}^k 1/i$ ~ $\log(k)$ ~ $\log\log(2^k)$ is the size of $u$ at $r = 2^{-k}$, but it shows how this example naturally arises.
• Thank you for this very insightful answer. Can I just ask how one pinpoints $\log \log(1+1/|x|)$ exactly from what you have said. I understand that $\sum_{i=1}^k 1/i \sim \log \log(2^k)$, but why do we specifically choose $1+1/|x|$? Aug 15, 2023 at 1:01