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I am reading Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis, and also Evans' PDE. I am confused about what $||\nabla u||_{W^{1,p}(\Omega)}$ precisely means. In these two books it is often written, like for example in the Sobolev, Gagliardo and Nirenberg inequality, which states that:

$$ || u||_{p^*} \leq C ||\nabla u||_p $$

The book defines $\nabla u=(u_{x_1},..., u_{x_n})$, so it is a vector. But the book also defines the L^p (And Sobolev $W^{1,p}$) spaces as those formed by functions from $\mathbb{R}^n$ to $\mathbb{R}$, and the norm by integrating that function.

Is there a standard meaning for $||\nabla u||_{W^{1,p}(\Omega)}$?

Because, for example, I have checked in the proof of $ || u||_{p^*} \leq C ||\nabla u||_p $ that it means that $||u||_p$ is bounded by the product of the $L^p$ norms of the derivatives of $u$. But if we check the proof of Proposition 9.3, we have, at a certain step, for $h \in \mathbb{R}^n$:

$$\int_0^1 |\nabla u(x+t)|^p dt$$

How does one interpret that integral? I understand taking the absolute value of a vector could mean take the absolute value of every component in that vector, but integrating a vector? Am I missing something?

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Means that $\partial_{x_j} u \in L^p(\Omega)$ $\forall j=1,...,n$, i.e. $\nabla u \in L^p(\Omega) \times \cdot \cdot \cdot \times L^p(\Omega)$.

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  • $\begingroup$ And $\int_0^1 |\nabla u(x+t)|^p dt$? Does it mean $\int_0^1 |\partial_{x_i}u(x+t)|^p dt$? $\endgroup$ – D1X May 14 '16 at 22:15
  • $\begingroup$ $\forall i=1...n$, yes, It is just a notation. $\endgroup$ – user288972 May 14 '16 at 22:17
  • $\begingroup$ Understood. Thanks. $\endgroup$ – D1X May 14 '16 at 22:19
  • $\begingroup$ I don't get the inequality, since it doesn't take in account the (possibly unbounded) integration constant. what does mean $\|u\|_{p^*}$ ? or we consider that $\Omega = \mathbb{R}^n$ in that case the integration constant is constrained to be $0$ for $\|u\|_p$ to be finite ? @D1X $\endgroup$ – reuns May 14 '16 at 23:44
  • $\begingroup$ It has not been said, but I think it is assumed that $\Omega$ is limited $\endgroup$ – user288972 May 15 '16 at 0:07

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