Outgrowths of Hardy's Inequality
Outgrowths of Hardy's Inequality
复制标题
哈代不等式的产物
DOI：
10.1090/conm/412/07766
复制
发表时间：
2006
影响因子：
4.8
通讯作者：
J. Goldstein
中科院分区：
文献类型：
--
作者：
W. Arendt;G. Goldstein;J. Goldstein
(U(A)/)(X)=A*/(Ax)，f e L 2(R)，Xer.则U(A)在L(R)上是酉的，且U(A)“=UQ)。“拉普拉斯标度如A”意味着U(A)-1 AU(A)=AA对所有A>；0成立，这很容易验证。现在，我们有些正式地继续，并不对域做出准确的声明。用ĵ表示乘法的算子类似于A，在这个意义下，对于(Mp F)(X)=xf{x)，f e L 2(R)，X EURO R，我们对所有A&>0有U(A)“MPU(A)=XMP”。因此
(U (A) / ) (x) = A* / (Ax), f e L 2 (R) , xeR. Then U (A) is unitary on L (R) and U(A)" = U Q ) . "The Laplacian scales like A" means U (A) -1 AU (A) = AA holds for all A > 0, as is easy to verify. Now we proceed somewhat formally and do not make precise statements about domains. The operator representing multiplication by ĵ pscales like A, in the sense that for ( M p f ) (x) = \x\~ f{x), f e L 2 (R) ,X€R, we have U (A)" MPU (A) = XMp for all A > 0. Consequently