Any vector space can be made into a unital associative algebra, called functional-theoretic algebra, by defining products in terms of two linear functionals. In general, it is a non-commutative algebra. It becomes commutative when the two functionals are the same.
Let AF be a vector space over a field F, and let L1 and L2 be two linear functionals on AF with the property L1(e) = L2(e) = 1F for some e in AF. We define multiplication of two elements x, y in AF by
x ⋅ y=L1(x)y+L2(y)x-L1(x)L2(y)e.
So, AF forms an associative algebra with unit e and is called a functional theoretic algebra(FTA).
Suppose the two linear functionals L1 and L2 are the same, say L. Then AF becomes a commutative algebra with multiplication defined by
x ⋅ y=L(x)y+L(y)x-L(x)L(y)e.
X is a nonempty set and F a field. FX is the set of functions from X to F.
If f, g are in FX, x in X and α in F, then define
(f+g)(x)=f(x)+g(x)
and
(\alphaf)(x)=\alphaf(x).
With addition and scalar multiplication defined as this, FX is a vector space over F.
Now, fix two elements a, b in X and define a function e from X to F by e(x) = 1F for all x in X.
Define L1 and L2 from FX to F by L1(f) = f(a) and L2(f) = f(b).
Then L1 and L2 are two linear functionals on FX such that L1(e)= L2(e)= 1FFor f, g in FX define
f ⋅ g=L1(f)g+L2(g)f-L1(f)L2(g)e=f(a)g+g(b)f-f(a)g(b)e.
Then FX becomes a non-commutative function algebra with the function e as the identity of multiplication.
Note that
(f ⋅ g)(a)=f(a)g(a)and(f ⋅ g)(b)=f(b)g(b).
Let C denote the field ofComplex numbers. A continuous function γ from the closedinterval [0, 1] of real numbers to the field C is called acurve. The complex numbers γ(0) and γ(1) are, respectively,the initial and terminal points of the curve. If they coincide, thecurve is called a loop. The set V[0, 1] of all the curves is avector space over C.
We can make this vector space of curves into analgebra by defining multiplication as above.Choosing
e(t)=1,\forall\in[0,1]
{\alpha} ⋅ {\beta}={\alpha}(0){\beta}+{\beta}(1){\alpha}-{\alpha}(0){\beta}(1)e
We illustratethis with an example.
Let us take (1) the line segment joining the points (1, 0) and (0, 1) and (2) the unit circle with center at theorigin.As curves in V[0, 1], their equations can be obtained as
f(t)=1-t+itandg(t)=\cos(2\pit)+i\sin(2\pit)
Since
g(0)=g(1)=1
f(0)=1
f(1)=i
Now, we get two f-products
f ⋅ gandg ⋅ f
(f ⋅ g)(t)=[-t+\cos(2\pit)]+i[t+\sin(2\pit)]
(g ⋅ f)(t)=[1-t-\sin(2\pit)]+i[t-1+\cos(2\pit)]
Observe that
f ⋅ g ≠ g ⋅ f
f(0)g(0)=1andendsat f(1)g(1)=i.