Is there a relation between order relation and norm in L1 space on Banach Algebra?
First of all, I am going to assume that your reference to Banach Algebra means that you are thinking of an L1 on a locally compact group with convolution. *, as multiplication. The usual order is that f>=g if f(t)>=g(t) almost everywhere. The firrst, simple, connection between the order and algebra structures is that if f>=0 and g>=0 then f*g>=0. A more powerful result involves the fact that L1 is a lattice and that the space of bounded linear operators on L1 is also a lattice if we define S>=T to mean that S(f)>=T(f) for all f>=0. Define T_f on L1 by T_f(g)=f*g. Then the map taking f to T_f is an isometric lattice homomorphism. This (and more) is in Satz 2.3 of Wolfgang Arendt's 1979 Tubingen dissertation "Uber das Spektrum Regularer Operatoren". I don't know of any more accessible source for this result.
@Eugene, the order is simply the (partial!) order f < g iff f(x) < g(x) except on a set of measure zero (i.e. almost everywhere).
@Dnyashwar.
Of course 0 0 and |f| < g then ||f||_1 < ||g||_1.
So far so good. The interesting fact is: if for an increasing sequence f_1< f_2 < f_3
First of all, I am going to assume that your reference to Banach Algebra means that you are thinking of an L1 on a locally compact group with convolution. *, as multiplication. The usual order is that f>=g if f(t)>=g(t) almost everywhere. The firrst, simple, connection between the order and algebra structures is that if f>=0 and g>=0 then f*g>=0. A more powerful result involves the fact that L1 is a lattice and that the space of bounded linear operators on L1 is also a lattice if we define S>=T to mean that S(f)>=T(f) for all f>=0. Define T_f on L1 by T_f(g)=f*g. Then the map taking f to T_f is an isometric lattice homomorphism. This (and more) is in Satz 2.3 of Wolfgang Arendt's 1979 Tubingen dissertation "Uber das Spektrum Regularer Operatoren". I don't know of any more accessible source for this result.
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