addition and multiplication of functions in function space, continuous? I have a norm that works in function space of C[0,1]. How do I show that addition and multiplication of functions (C[0,1]xC[0,1]->C[0,1]) are continuous functions?
 A: 
*

*To use C[0,1] to show continuity within Function Composition you will have to take into consideration only a few concepts for concluding that (C[0,1]xC[0,1]->C[0,1]) can create a paradigm for continuous functions [within structures].

*We will assume that C[0,1] can become a Function of Several Real Variables by using a seminorm. In a Euclidean norm Vector spaces must be allowed to use (0,0) for magnitude within any cartesian. So, we have precedent for understanding how norms can function as both zero vector spaces, by using (0,1) you actually solve your own problem by assigning an Integer [at least 1].

*

*Proving Continuity and Limit:

*

*Consider the distance function of ℝᵤ, where

C[0,1] ≐ d ≑ d(x,y)=d(x₁,…,xᵤ,y₁,...yᵤ)= √((x₁-y₁)²+…+(xᵤ-yᵤ)²)

Proving Continuity and Limit:

consider the function ƒ in C[0,1] such that ƒ(0,0)=0 is otherwise defined by
      ƒ(x,y)=x²y/x²*²+y²

The functions
x→C(x,0) 
         and
y→C(0,y) 
        are both constant and equal to zero, and are therefore continuous
        
        
*Providing C[0,1] as a seminorm and concluding its use for Functional Continuity :
        
*
        
*For every real number

ε > 0 and δ > 0 
                   such that
                  ∣ƒ(x)—L<ε∣
x evaluates in its domain as
               d(x,a) < δ

If the limit exists, it is unique. 
            If a is in the interior of the domain: 
            The limit exists if and only if the function is continuous at a
            For our instance
ƒ(a)=lim┬(x→a)  ƒ(x)
            
            
*When a is in the boundary of the domain of f

f has a limit at a
               the latter formula allows the extension of continuity 

which essentially is the domain of  

f to a 

So since the seminorm C[0,1] can accommodate a series of functions, such as;
C(x,y)=C(x,y)=C(x₁,…,xᵤ,y₁,...yᵤ)= √((x₁-y₁)²+…+(xᵤ-yᵤ)²) 


On another note, I believe that Functional Composition can restate:
ξ : X→Ξ
ζ : Ξ→ℝ













