MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
actually, the embedding holds true in one critical case, namely $p=1$, $k=n$. – Denis Serre Mar 9 '13 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. – Spencer Mar 11 '13 at 23:12
up vote 11 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.