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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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What norm do you use for the intersection? The usual way to norm $X\cap Y$ is by $\|\cdot \|_X+\|\cdot \|_Y$. Then of course, the identity map $X\cap Y\to Y$ is a continuous embedding. – user53153 Jan 14 '13 at 22:02
I want the norm of $W^{1,N}$. – Tomás Jan 14 '13 at 22:06
Then you don't win anything by asking $f$ to be bounded. Recall that $C^\infty_c$ is dense in $W^{1,N}$. If a dense subset embeds continuously, so does the entire space. – user53153 Jan 14 '13 at 22:10
Im confused now. If I take a function $u\in W^{1,N}\cap L^\infty$, is there a constant $C$ not depending on $u$ such that $$\|u\|_{\infty}\leq C\|u\|_{1,N}$$ – Tomás Jan 14 '13 at 22:22
up vote 3 down vote accepted

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 $u\in X$.

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