The space of Fourier-Haar multipliers.

The Haar system constitutes an unconditional basis for a separable rearrangement invariant (symmetric) spaceEif and only if the multiplier determined by the sequence<InlineEquation ID=”IE1”><EquationSource Format=”MATHTYPE”><![CDATA[ % MathType!MTEF!2!1!+- % feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqefm0B1jxALjhiov2D % aebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8qqaq % Fr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qq % Q8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacq % aH7oaBdaWgaaqcbaAaamXvP5wqonvsaeHbfv3ySLgzaGqbciab-5ga % Ujab-TgaRbWcbeaakiabg2da9maabmaabaGaeyOeI0IaaGymaaGaay % jkaiaawMcaamaaCaaaleqabaGae8NBa4gaaOGaaiilaiab-TgaRjab % g2da9iaaicdacaGGSaGaaGymaiaacYcaaaa!4E50! ]]></EquationSource><EquationSource Format=”TEX”><![CDATA[$$ \lambda _{nk} = \left( { - 1} \right)^n ,k = 0,1, $$]]></EquationSource></InlineEquation>forn= 0 andk= 0, 1, . . . , 2n forn>1, is bounded inE. If the Lorentz space ?(?) differs fromL1 andL8 then there is a multiplier with respect to the Haar system which is bounded in ? (?) and unbounded inL8 andL1. [ABSTRACT FROM AUTHOR]