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Let $a$ be a non negative (positive almost everywhere) weight in $L_{loc}^1(\Omega)$, $\Omega\subseteq\mathbb{R}^n$ is open. For $\varphi\in C_c^{\infty}(\Omega)$ define $$ \Vert\varphi\Vert_a^2=\int_{\Omega} a(x)|\nabla\varphi(x)|^2\,\mathrm{d}x. $$ Then $\Vert\cdot\Vert_a^2$ is a norm on $C_c^{\infty}(\Omega)$. Let $D_0^1(\Omega;a)$ denotes the closure of $C_c^{\infty}(\Omega)$ with respect to the $\Vert\cdot\Vert_a^2$-norm.

Let $\{\varphi_n\}\in C_c^{\infty}(\Omega)$ and $\varphi_n\to \varphi$ in $D_0^1(\Omega;a)$ and also we have $\varphi_n\to \psi$ in some space $L^p(\Omega),\,1\leqslant p\leqslant\infty$.

Can we say that $\varphi=\psi$? And why?

A similar situation happens in the proof of Sobolev embedding theorem. Where we have a smooth sequence $\{\varphi_n\}$ tends to a $u\in W^{1,p}(\mathbb{R}^n)$ and then we show $\varphi_n$ tends to some $v\in L^{p^*}(\mathbb{R}^n)$ and then we say $u=v$ a.e. In this situation, my understanding is : since $\Vert\varphi_n-u\Vert_p\to 0$ and $\Vert\varphi_n-v\Vert_{p^*}\to 0$ there is a subsequence $\{\varphi_{n_{k_i}}\}$ converges almost everywhere to both $u$ and $v$, then $u=v$ a.e.

But in the $D_0^1(\Omega;a)$ case, does there also a subsequence converges to $\varphi$ a.e.? Or some other approachs?

Any advice will be appreciated!

Edit

Here is some of my thought:

  1. For $u_k\in L^p(\Omega), 1\leqslant p\leqslant\infty$, $u_k\to u$ in $L^p(\Omega)$, define $$ \langle u,\varphi\rangle=\int_{\Omega}u\varphi $$ for all $\varphi\in C_c^{\infty}(\Omega)$. We see this integral makes sense since $\varphi$ has compact support and $u$ is locally integrable. And it is a continuous linear functional on $C_c^{\infty}(\Omega)$. Furthermore $$ |\langle u_k,\varphi\rangle-\langle u,\varphi\rangle|\leqslant \int_{\Omega}|u_k-u||\varphi|\to 0 $$ by Holder's inequality. Thus we have $u_k\to u$ in $\mathcal{D}'(\Omega)$.

In Sobolev case we have $u_k$ tends to both $u$ and $v$ in $\mathcal{D}'(\Omega)$, so $u=v$ a.e.

  1. For $u_k\in D_0^1(\Omega;a)$, $u_k\to v$ in $D_0^1(\Omega;a)$, define $$ \langle u,\varphi\rangle=\int_{\Omega}a^{1/2}(x)|\nabla u|\varphi. $$ for all $\varphi\in C_c^{\infty}(\Omega)$. Then the integral makes sense by applying Holder's inequality. And it is also a linear continuous functional on $\mathcal{D}'(\Omega)$. Furthermore $$ |\langle u_k,\varphi\rangle-\langle v,\varphi\rangle|\leqslant \int_{\Omega}a^{1/2}(x)|\nabla u_k-\nabla u||\varphi|\to 0 $$ by Holder's inequality. So we have $u_k \to v$ in $\mathcal{D}'(\Omega)$, which implies $u=v$ a.e. And this answers my question why $\psi=\varphi$ a.e.

Is there something wrong in my "proof"? Or some better approaches?

Thanks!

Edit

$$ |\langle u_k,\varphi\rangle-\langle v,\varphi\rangle|\leqslant \int_{\Omega}a^{1/2}(x)|\nabla u_k-\nabla u||\varphi|\to 0 $$ seems only mean $a^{1/2}(x)|\nabla u|\to a^{1/2}(x)|\nabla v|$ in $\mathcal{D}'(\Omega)$, and this says nothing about my question I think!

Could any one help me?

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  • $\begingroup$ The answer probably depends on $a$. Can you prove that elements of $D_0^1(\Omega;a)$ actually belong to some $L^p(\Omega)$? $\endgroup$
    – Siminore
    May 4, 2012 at 11:08
  • $\begingroup$ @DavideGiraudo: Thanks! Fixed. $\endgroup$
    – Y.Z
    May 4, 2012 at 11:19
  • $\begingroup$ @Siminore: Acturally this question appears in the proof of $D_0^1(\Omega;a)\hookrightarrow L^p(\Omega)$ for some $p$. $\endgroup$
    – Y.Z
    May 4, 2012 at 11:22

1 Answer 1

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Your answer lies in the theory of distributions. Both your space $D^1_0(\Omega,a)$ and $L^p(\Omega)$ are subspaces of the space of distributions, that is to say, bounded linear functionals acting on $C^\infty_c(\Omega)$.

Now, without an embedding theorem, it is not given that just because $\varphi_n$ converges in one space, it needs to converge to another - in fact, for properly ugly cases, you can construct sequences that converge in one but not the other. Consider, for an example, the case of $\frac{\cos(nx)}{n}$ on some bounded interval. This sequence tends to 0 in any $L^p$ space as $p \rightarrow \infty$, but does not have a strong limit in $H^1$ (taking, in this case, $a(x) \equiv 1$).

However, if the same sequence converges in both spaces, then the properties of distributions say that they must converge to the same distribution.

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