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I have been studying Sobolev spaces and easy PDEs on those spaces for a while now and keep wondering about the norms on these spaces. We obviously have the usual norm $\|\cdot\|_{W^{k,p}}$, but some proofs also use the semi-norm $|\cdot|_{W^{k,p}}$ and I don't really understand why we make use of this semi-norm? Why is the full norm $\|\cdot\|_{W^{k,p}}$ not enough? I started arguing that the semi-norm is easier to calculate (in case one really wishes to do that), but I don't think that really is the reason, after all, I have never seen anyone really calculate these norms for proofs or anything. So are there other reasons for the use of this semi-norm?

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  • $\begingroup$ Yes, the norm is $|u|_{W^{k,p}(\Omega)} = \left(\sum_{|\alpha|=k} ||D^\alpha u||_{L^p(\Omega)}^p\right)^\frac{1}{p}$, $\|u\|_{W^{k,p}} = \left(\sum_{|\alpha|\leq k} ||D^\alpha u||_{L^p(\Omega)}^p\right)^\frac{1}{p}$. $\endgroup$ – dinosaur Jul 4 '13 at 12:52
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    $\begingroup$ Maybe you could specify for us in what situation you saw such kind of behavior? $\endgroup$ – Tomás Jul 6 '13 at 11:32
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I give a few reasons, sticking to the case $k=1$ for simplicity.

  1. Scaling behavior. If you scale the variable as $u_r(x)=u(rx)$, the homogeneous norm of $u_r$ can be expressed in terms of the homogeneous norm of $u$: namely, $|u_r|_{W^{1,p}}=r^{p-n}|u|_{W^{1,p}}$ where $n$ is the dimension. There is no such relation for the full Sobolev norm: since the derivatives of different orders scale differently, the behavior of their sum under scaling is messy.

  2. Conformal invariance. Observe that the factor $r^{p-n}$ in item 1 disappears when $p=n$. In this case, the homogeneous Sobolev norm is invariant under scaling, and (by a simple argument) under conformal maps. There are few conformal maps when $n>2$, but the conformal invariance of the Dirichlet integral $|\cdot|_{W^{1,2}}^2$ in the case $n=p=2$ is tremendously useful. The Sobolev norm $\|\cdot\|_{W^{1,2}}$ is not conformally invariant.

  3. Poincaré inequality $\|u-u_\Omega\|_{L^p}\le C\|Du\|_{L^p}$ naturally has the homogeneous seminorm as the right-hand side, since it expresses the fact that the size of gradient controls the size of function itself. If we included the full Sobolev norm on the right, the inequality would become trivial.

  4. Sobolev inequality $\|u\|_{L^{p^*}}\le C\|Du\|_{L^p}$ (where $u$ is compactly supported) naturally has the homogeneous seminorm as the right-hand side. This allows both sides to scale in the same way, which makes it possible to have $C=C(p,n)$ independent of the size of support. In fact, the sharp constant $C(p,n)$ has been known since the 1960s (Rodemich, Aubin, Talenti): see, for example, Topics in Optimal Transportation by Villani. After stating the formula for the functions $h_p$ for which the sharp Sobolev inequality turns into equality, Villani remarks:

Note that $h_p$ does not necessarily lie in $L^p$ (which has no importance whatsoever).

In a nutshell: adding $L^p$ norm of $u$ (and lower derivatives) to the homogeneous norm is a quick but dirty way to obtain a Banach space. It is often preferable to work with $|\cdot|_{W^{k,p}}$, taking the quotient by constant functions to make it a norm.

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  • $\begingroup$ wow, those are very interesting reasons, thank you very much for your answer! $\endgroup$ – dinosaur Jul 8 '13 at 20:13
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Really nice answer! One more important reason to use the seminorm: if you are trying to solve elliptic partial differential equations in unbounded domains, say, $$\Delta u=f\quad\text{in }\mathbb{R}^N,$$ where, say, $f\in L^p(\mathbb{R}^N)$, $1<p<N/2$, then using Green functions and singular integrals you can show that the second order derivatives of $u$ will be in $L^p(\mathbb{R}^N)$ but the function $u$ or its first derivatives will not be in $L^p(\mathbb{R}^N)$ in general, so $u$ will not belong to the space $W^{2.p}(\mathbb{R}^N)$.

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