Sobolev Embedding (Case: p=N) Let $\Omega\subset\mathbb{R}^N$ be a regular bounded domain. Suppose $p=N$, then by Sobolev theorem, we have that for fixed $q\in [1,\infty)$  $$\|u\|_q\leq C\|u\|_{1,N}\ ,\forall\  u\in W^{1,N}(\Omega)$$
for some constant $C$. 
Moreover, the function  $$f(x)=\log\log\Big(1+\frac{1}{|x|}\Big)$$
is an example of a function $f\in W^{1,N}$ ($N\geq 2$) such that $f\notin L^\infty$.
My question is: Is $W^{1,N}\cap L^\infty$ continuous embedded in $L^\infty$?
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 A: First of all, the one-dimensional example does not work. The function $\log\log (1+1/t)$ does not belong to $W^{1,1}((0,1))$. Indeed, the space $W^{1,1}((0,1))$ consists of the antiderivatives of Lebesgue integrable functions on $(0,1)$, and all such functions are bounded. One can also give an explicit estimate of Poincaré type, 
$$
\sup_{(0,1)} \left|f - \int_0^1 f\right| \le \int_0^1 |f'|
$$
But in dimensions $N\ge 2$ the function $u(x)=\log\log(1+|x|^{-1})$ is indeed the standard example used to show that $W^{1,N}$ does not embed into $L^\infty$. 
The question was whether adding the assumption $u\in L^\infty$ could save the embedding. It does not, as can be seen by considering $\min(u,M)$ for arbitrarily large $M$. 
There's a more general reason for why the assumption of boundedness could not help here. The space $C^\infty_c$ is dense in $W^{1,N}$, as in any other Sobolev space with $p< \infty$. If an estimate of the form $F(u)\le C\|u\|_{X}$ holds on a dense subset of a normed space $X$, then it holds for all $u\in X$.
