论文标题
补偿紧凑性：最佳弱拓扑的连续性
Compensated compactness: continuity in optimal weak topologies
论文作者
论文摘要
对于$ l $ - 均匀的线性差分运算符$ \ MATHCAL {a} $的恒定等级，我们研究含义$ v_j \ rightharpoonup v $ in $ x $中的$ x $和$ \ nathcal {a} a} v_j \ rightArrow \ rightArrow \ rightArrow \ rightArrow \ rightArrow \ rightarc {a} a} v $ f（v）$ in $ z $，其中$ f $是$ \ mathcal {a} $ - quasiaffine函数和$ \ rightsquigarrow $表示适当类型的弱收敛类型。这里$ z $是本地$ l^1 $ -type空间，要么是$ \ mathscr {m} $的测量值，要么$ l^1 $，或者hardy space $ \ mathscr {h}^1 $; $ x，\，y $是$ l^p $ -type空间，我们的意思是Lebesgue或Zygmund空间。我们选择$ x，\，y，\，z $的每种选择的条件很清晰。在$ f（v）$不是本地集成函数的情况下，也给出了类似的语句，而是将其定义为分布。在这种情况下，我们还证明了$ \ mathscr {h}^p $ - 序列$（f（v_j））_ j $，适用于适当的$ p <1 $，而新的融合会导致$（v_j）$ is $（v_j）$ is $ \ nathcal {a} $ - a} $ - a} $ - free and Order和fiep and Order of prop of houlder { $ w^{ - β，s} $。这些hölder空间的选择很清晰，正如明确的反例的构建所示。这些结果中的一些是新的，即使是分销雅各布人。
For $l$-homogeneous linear differential operators $\mathcal{A}$ of constant rank, we study the implication $v_j\rightharpoonup v$ in $X$ and $\mathcal{A} v_j\rightarrow \mathcal{A} v$ in $W^{-l}Y$ implies $F(v_j)\rightsquigarrow F(v)$ in $Z$, where $F$ is an $\mathcal{A}$-quasiaffine function and $\rightsquigarrow$ denotes an appropriate type of weak convergence. Here $Z$ is a local $L^1$-type space, either the space $\mathscr{M}$ of measures, or $L^1$, or the Hardy space $\mathscr{H}^1$; $X,\, Y$ are $L^p$-type spaces, by which we mean Lebesgue or Zygmund spaces. Our conditions for each choice of $X,\,Y,\,Z$ are sharp. Analogous statements are also given in the case when $F(v)$ is not a locally integrable function and it is instead defined as a distribution. In this case, we also prove $\mathscr{H}^p$-bounds for the sequence $(F(v_j))_j$, for appropriate $p<1$, and new convergence results in the dual of Hölder spaces when $(v_j)$ is $\mathcal{A}$-free and lies in a suitable negative order Sobolev space $W^{-β,s}$. The choice of these Hölder spaces is sharp, as is shown by the construction of explicit counterexamples. Some of these results are new even for distributional Jacobians.
