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 $u\in L^\infty(U)$. 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".

I am wondering about an intuitive / deeper reason as to what goes wrong. 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.

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.

I am assuming here that $U\subseteq\mathbb{R}^n$ is open and bounded with $C^1$-boundary. Then for $u\in W^{k,p}(U)$ 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 $k=n/p$ (related to the Sobolev conjugate $p^*\to \infty$ as $p\to n$).

I have seen a counterexample, but am wondering about a more intuitive / deeper reason as to what goes wrong. 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 dimension seems to arise in geometry/topology often.

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 $u\in L^\infty(U)$. 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".

I am wondering about an intuitive / deeper reason as to what goes wrong. 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.