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.

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    $\begingroup$ actually, the embedding holds true in one critical case, namely $p=1$, $k=n$. $\endgroup$ Mar 9, 2013 at 7:45
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    $\begingroup$ 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. $\endgroup$
    – Spencer
    Mar 11, 2013 at 23:12

1 Answer 1


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.

  • $\begingroup$ 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 $1+1/|x|$? $\endgroup$
    – AmorFati
    Aug 15, 2023 at 1:01

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