【摘 要】
：
Given a function(φ)on R,the Hausdorff operator H(φ)f is defined by(H(φ))(x)=∫1/|t|f(x/t)(φ)(t)dtR.(1)E.Liflyand asked the following problem([1] problem 3):
【机 构】
：
ZhejiangUniversity
【出 处】
：
Harmonic Analysis and Applications(2014调和分析及其应用学术会议)
论文部分内容阅读
Given a function(φ)on R,the Hausdorff operator H(φ)f is defined by(H(φ))(x)=∫1/|t|f(x/t)(φ)(t)dtR.(1)E.Liflyand asked the following problem([1] problem 3):prove or disprove that the real Hardy space H1(R)can be characterized via Hausdorff operators in the sense that(1)for an integrable f is in H1(R),then f is necessarily in H1(R).If not(which is very probable)a counterexample should be given.The first aim of this paper is to show that the answer of this problem is negative.We also give a negative answer to the similar problem in which the Hausdorff operator is in higher dimensional case:HΦ,Af(x)=∫Φ(u)f(xA(u))duRn,where A = A(u)=(aij(u))n i,j=1 is the n×n matrix with the entries aij(u)being measurable functions of u.The second aim of this paper is to study the another problem raised by E.Liflyand([1] problem 4).
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