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发表于 2025-3-21 19:49:22
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起草
发表于 2025-3-21 20:38:20
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alcohol-abuse
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老巫婆
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multiply
发表于 2025-3-23 00:30:57
Springer-Verlag Berlin Heidelberg 1990
无瑕疵
发表于 2025-3-23 03:37:43
https://doi.org/10.1007/978-3-642-47891-8If a ∈ L∞ and 1 <p< ∞ ,then the operator.is obviously bounded and .It is called the . L. generated by the function a. For f ∈ L. and .∈ L.(1/.+1/.=1), write .It is clear that is equal to the th Fourier coefficient of a. The following proposition shows that every bounded operator with such a property is a multiplication operator.
Atheroma
发表于 2025-3-23 09:22:14
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