【摘要】 <正> 熟知的,在富里分析中,设f（x）是2π周期可积函数(记作f∈L_2π¹或L¹)。S_n（f,x）是f的富里级数（f）的部分和。存在绝对常数A（只与p有关的）,使对任意,则这里‖·‖_p表示L^p范数。这时我们说算子列T_n:f→S_n是一致（p,p）型的。也就是从L^p到L^p的有界算子,且算子范数列有界。但T_n不是一致更多还原

【Abstract】 Let Sn(dF, x) be the partial sums of the Fou ier-Stieltjes (dF) with F(x)∈ BV[ 0 . 2π].We prove that there exists an absolute constant A such that N1(Sn((dF, x)) A ||F||v, where N1 (g) = sup ames [x |g(x) >α].In particular N1 (Sn(f, x) )<A ||f||1 for f∈ L271 (which is called uniform weak boundedness of operators Sn).The same statement is asserted concerning Fourier Stieltjes single integralsWhere F(x) ∈ BV( -00 , oo),Those are based on weak boundedness of conjugate transforms(Hilbert-Stieltjes transform) .In conclusion, we point out that there exists no absolute constant A such that.更多还原