Weak Convergence in Sobolev Spaces
Suppose $\Omega\subset\R^n$ and donote $W^{1,p}:=W^{1,p}(\Omega)$ be the sobolev space for some $1< p< +\infty$. Recall that $f_i\in W^{1,p}$ convergent weakly to $f\in W^{1,p}$, if for any $\phi$ in the dual space of $W^{1,p}$, we have $\inner{f_i,\phi}\to\inner{f,\phi}$, denote as $f_i\weakto f$. This is distinguished by strongly convergence, as we use the dual normal instead of $W^{1,p}$ normal.
命题 1. If $f_i\weakto f$ in $W^{1,p}$, then $f_i\to f$ in $L^p$.
证明 . Since $W^{1,p}$ is a Banach space, and $f_i\weakto f$, then $f_i$ is uniformly bounded in $W^{1,p}$ by Banach-Steinhaus Theorem. Since $W^{1,p}$ is compactly embedding into $L^p$, we have, by passing to subsequence, $f_i\to g$ in $L^p$ for some $g\in L^p$.
For any $\phi\in C_0^\infty$, define a $T_\phi$ as
$$
T_\phi(f):=\int_\Omega f\phi,\quad\forall f\in W^{1,p},
$$
then $T_\phi\in (W^{1,p})’$, i.e., in the dual space of $W^{1,p}$. Thus, by weakly convergence, we have
$$
T_\phi(f_i)\to T_\phi(f)\Longleftrightarrow \int_\Omega f_i\phi\to\int_\Omega f\phi.
$$
On the other hand, $f_i\to g$ strongly in $L^p$, thus by Holder inequality, we have
$$
\int_\Omega f_i\phi\to\int_\Omega g\phi.
$$
In conclusion, we have
$$
\int_\Omega f\phi=\int_\Omega g\phi,
$$
and we finish the argument by the density of $C_0^\infty$ in $L^p$.
For any $\phi\in C_0^\infty$, define a $T_\phi$ as
$$
T_\phi(f):=\int_\Omega f\phi,\quad\forall f\in W^{1,p},
$$
then $T_\phi\in (W^{1,p})’$, i.e., in the dual space of $W^{1,p}$. Thus, by weakly convergence, we have
$$
T_\phi(f_i)\to T_\phi(f)\Longleftrightarrow \int_\Omega f_i\phi\to\int_\Omega f\phi.
$$
On the other hand, $f_i\to g$ strongly in $L^p$, thus by Holder inequality, we have
$$
\int_\Omega f_i\phi\to\int_\Omega g\phi.
$$
In conclusion, we have
$$
\int_\Omega f\phi=\int_\Omega g\phi,
$$
and we finish the argument by the density of $C_0^\infty$ in $L^p$.
发表回复