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Let $\Omega\subset\mathbb{R}$ be a bounded domain and $p\in (1,\infty)$. Take $s\in (0,1)$ and define $$W^{s,p}(\Omega)=\left\{u\in L^p(\Omega):\ \int_\Omega\int_\Omega\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}<\infty\right\}$$

Its is know that $W^{s,p}(\Omega)$ is a Banach space with the norm $$\tag{1}\|u\|_{s,p}=\|u\|_p+\left(\int_\Omega\int_\Omega\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}\right)^{1/p}$$

Moreover, there exist a unique linear bounded operator $T: W^{1,P}(\Omega)\to W^{1-1/p,p}(\partial\Omega)$ (called Trace operator) with the property that $Tu=u_{|\partial\Omega}$ if $u\in C^\infty(\overline{\Omega})$. Also $T$ is surjective, hence, we can define on $W^{s,p}(\partial\Omega)$ the norm $$\|u\|=\inf\{\|v\|_{1,p}:\ v\in W^{1,p}(\Omega),\ Tv=u\}$$

$W^{s,p}(\partial\Omega)$ is a Banach space with the norm $\|\cdot\|$. This can be seen for example by noting that $\|\cdot\|$ is just the quotient norm associated with the space $\frac{W^{1,p}(\Omega)}{W_0^{1,p}(\Omega)}=W^{1-1/p,p}(\partial\Omega)$ (isomorphism). My questions is: is $\|\cdot\|$ and $\|\cdot\|_{s,p}$ equivalents?

Remark: This question is somehow related to this one

Remark 2: One can see that the infimum in $\|u\|$ is achieved by some $u_0$. This is true because we are minimizing a weakly sequentially lower semi continuous, coercive and convex functional on a closed convex subset of $W^{1,p}(\Omega)$. Therefore, by the continuity of $T$, we have that $$\|Tu_0\|_{s,p}=\|u\|_{s,p}\leq c \|u\|,\ \forall\ u\in W^{1-1/p,p}(\partial\Omega)$$

It remains only to prove the inverse inequality...

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I think so. There might be a direct argument, but an indirect could be the following:

Taking a look on the assertion of When are two norms equivalent on a Banach space?. First, $W^{s,p}(\Gamma)$ is a Banach space w.r.t. both norms. If we have a sequence $\{x_k\}$ converging in one of the norms towards $x$, then it also converges in $L^p(\Gamma)$ towards $x$. Hence, the limit is independent of the norm. Now, you can use the above assertion to check the equivalence.

Edit: Your edit also shows that the limits are equivalent.

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  • $\begingroup$ Yes, your are right. With my edit, the equivalence follows by an application of the Open Mapping Theorem. Thank you. $\endgroup$ – Tomás Sep 7 '13 at 20:39

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