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 A_F be a vector space over a field F, and let L₁ and L₂ be two linear functionals on A_F with the property L₁(e) = L₂(e) = 1_F for some e in A_F. We define multiplication of two elements x, y in A_F by

${\displaystyle x\cdot y=L_{1}(x)y+L_{2}(y)x-L_{1}(x)L_{2}(y)e.}$

It can be verified that the above multiplication is associative and that e is the identity of this multiplication.

So, A_F 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 A_F becomes a commutative algebra with multiplication defined by

${\displaystyle x\cdot y=L(x)y+L(y)x-L(x)L(y)e.}$

X is a nonempty set and F a field. F^X is the set of functions from X to F.

If f, g are in F^X, x in X and α in F, then define

${\displaystyle (f+g)(x)=f(x)+g(x)\,}$

and

${\displaystyle (\alpha f)(x)=\alpha f(x).\,}$

With addition and scalar multiplication defined as this, F^X 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) = 1_F for all x in X.

Define L₁ and L₂ from F^X to F by L₁(f) = f(a) and L₂(f) = f(b).

Then L₁ and L₂ are two linear functionals on F^X such that L₁(e)= L₂(e)= 1_F For f, g in F^X define

${\displaystyle f\cdot g=L_{1}(f)g+L_{2}(g)f-L_{1}(f)L_{2}(g)e=f(a)g+g(b)f-f(a)g(b)e.}$

Then F^X becomes a non-commutative function algebra with the function e as the identity of multiplication.

Note that

${\displaystyle (f\cdot g)(a)=f(a)g(a){\mbox{ and }}(f\cdot g)(b)=f(b)g(b).}$

Let C denote the field of Complex numbers. A continuous function γ from the closed interval [0, 1] of real numbers to the field C is called a curve. The complex numbers γ(0) and γ(1) are, respectively, the initial and terminal points of the curve. If they coincide, the curve is called a loop. The set V[0, 1] of all the curves is a vector space over C.

We can make this vector space of curves into an algebra by defining multiplication as above. Choosing ${\displaystyle e(t)=1,\forall \in [0,1]}$ we have for α,β in C[0, 1],

${\displaystyle {\alpha }\cdot {\beta }={\alpha }(0){\beta }+{\beta }(1){\alpha }-{\alpha }(0){\beta }(1)e}$

Then, V[0, 1] is a non-commutative algebra with e as the unity.

We illustrate this 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 the origin. As curves in V[0, 1], their equations can be obtained as

${\displaystyle f(t)=1-t+it{\mbox{ and }}g(t)=\cos(2\pi t)+i\sin(2\pi t)}$

Since ${\displaystyle g(0)=g(1)=1}$ the circle g is a loop. The line segment f starts from :${\displaystyle f(0)=1}$ and ends at ${\displaystyle f(1)=i}$

Now, we get two f-products ${\displaystyle f\cdot g{\mbox{ and }}g\cdot f}$ given by

${\displaystyle (f\cdot g)(t)=[-t+\cos(2\pi t)]+i[t+\sin(2\pi t)]}$

and

${\displaystyle (g\cdot f)(t)=[1-t-\sin(2\pi t)]+i[t-1+\cos(2\pi t)]}$

See the Figure.

Observe that ${\displaystyle f\cdot g\neq g\cdot f}$ showing that multiplication is non-commutative. Also both the products starts from ${\displaystyle f(0)g(0)=1{\mbox{ and ends at }}f(1)g(1)=i.}$

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