View Full Version : Product of linear functional
Klaus_Hoffmann
Jul15-07, 02:20 AM
Let be a set of LInear functionals U_{n}[f] n=1,2,3,4,.........
so for every n U_{n}[\lambda f+ \mu g]=\lambda U_{n}[f]+\mu U[g] (linearity)
the question is if we can define the product of 2 linear functionals so
U_{i}U_{j}[f] makes sense.
Let be a set of LInear functionals U_{n}[f] n=1,2,3,4,.........
so for every n U_{n}[\lambda f+ \mu g]=\lambda U_{n}[f]+\mu U[g] (linearity)
the question is if we can define the product of 2 linear functionals so
U_{i}U_{j}[f] makes sense.
You can define the product of linear functionals as composition, so you have an algebra of functionals.
HallsofIvy
Jul15-07, 07:35 AM
Defining an "ordinary" product, that is as the product of the results of the functionals would destroy linearity.
Unfortunately, since a "functional" maps functions to numbers, the composition of two functionals does not exist.
Unfortunately, since a "functional" maps functions to numbers, the composition of two functionals does not exist.
Of course, I didn't think about that. :uhh:
lurflurf
Jul15-07, 08:33 PM
One could procede as in multilinear algebra/tensor calculus and define an outer product.
Thus let u,v be functionals
u,v:V->F
(uv)f=(vf)u
One might say the product between two linear functionals is a bilinear functional.
matt grime
Jul16-07, 08:17 AM
You can trivially define the product (as in multiplication) of two linear functionals. It just isn't a linear functional. A linear function is in particular a C/R/F valued function, so it lies in the algebra of functions, as radou sort of said.
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