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In page 16 of this book, the author state:

For $1\leq p <\infty$, consider the normed space of all smooth functions $\phi \in \mathbb{R}^n$ such that $$ \|\phi\|_{1,p} = \|\phi\|_p + \|\nabla \phi\|_p < \infty, $$ and denote its completion in this norm by $H^{1,p}$. Thus a function $u$ is a member of $H^{1,p}$ if

(i) There are smooth functions $\phi_i$ that converge to $u$ in $L^p$;

(ii) The gradients $\nabla \phi_i$ converge in $L^p$ to some vector-valued function $\mathbb{v}$.

The completion of a space should consist of some Cauchy sequence; see this for instance. I understand (i) hold might because that $C^{\infty}$ should be dense in $H^{1,p}$ by the definition of complation.

But how can I understand (ii) holds. I think the above question should not be understood with the help of the Sobolev space which fristly defined via distribution.

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  • $\begingroup$ Can you check that if $\{\phi_n\}$ is a Cauchy sequence for $\|\cdot\|_{1,p}$, then both $\{\phi_n\}$ and $\{\nabla \phi_n\}$ are Cauchy sequences for $\|\cdot\|_p$ in their respective natural ambient space? Then you should be able to conclude by Fischer's Theorem $\endgroup$
    – Didier
    Commented Jul 10 at 9:31
  • $\begingroup$ @Didier You can see my answer. $\endgroup$
    – tianJ
    Commented Jul 10 at 11:18

1 Answer 1

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This can be done as follows:

Denote $\hat{H}^{1,p}$ to be the standard complete space of $\hat{C}^{\infty}$, where $$ \hat{C}^{\infty} = \{\phi \in C^{\infty}: \|\phi\|_p + \|\nabla \phi\|_p <\infty\}. $$ The space $\hat{H}^{1,p}$ consists of equivalence Class of Cauchy sequence $[\underline{\phi}]$. Then we denote $H^{1,p}$ by $$ H^{1,p} = \{\text{$u \in L^p$ : $u$ is the limit of a sequence of smooth functions $\{\phi_i\}$, besides $\nabla \phi_i$ converges to a vector function $\mathbb{v}$}\} $$ Then there is a isomorphism between $\hat{H}^{1,p}$ and $H^{1,p}$ as follows $$ \mathcal{j}([\underline{\phi}]) = u_0, $$ where $u_0$ is the limit point (this is well-defined) of any Cauchy sequence in $[\underline{\phi}]$.

Now, everything is clear.

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  • $\begingroup$ Actually, the terminology saying that $H^{1,p}$ consists of functions is not completely valid. Its elements are not actual functions (like $L^p$ spaces do not really consist of functions), but really are equivalent classes. We just consider them as functions because that is how we want to see them. Really, they are just limits of actual functions in the lense we chose to consider (the Sobolev norm). $\endgroup$
    – Didier
    Commented Jul 10 at 12:19
  • $\begingroup$ @Didier It is welcome to add a precise explanation to my answer or write another answer. $\endgroup$
    – tianJ
    Commented Jul 11 at 7:36

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