Let x=(x1,x2,...), y=(y1,y2,...)∈L(R) be two elements from the set L(R), We define their sum as x+y=(x1+y1,x2+y2,...,xi+yi,..),i∈N or as sum for each component. We should ptove, that x+y∈L(R) but it is easy because ∀i∈N∣xi+yi∣≤∣xi∣+∣yi∣≤∣∣x∣∣L(R)+∣∣y∣∣L(R)→∣∣x+y∣∣L(R)≤∣∣x∣∣L(R)+∣∣y∣∣L(R)<∞
Axioms of addition
x+y=y+x;x+(y+z)=(x+y)+z
are valid because it is true in R for each component of vectors.
In linear space must be zero element and we define it by the formula
0=(0,0,0,..) or vector with zeros only.
Next property takes place
∀x∈L(R) x+0=x
because it is true for each component.
Operation of multiplication elements of L(R) by scalar from R is defined by the formula
(c⋅x)i=c⋅xi,i∈N, c∈R, x∈L(R)
We have c⋅x∈L(R) and ∣∣c⋅x∣∣L(R)≤∣c∣⋅∣∣x∣∣L(R)
Operation of multiplication is defined through components , therefore next properties take place:
c⋅(x+y)=c⋅x+c⋅y;c⋅(d⋅x)=(c⋅d)⋅x, c,d∈R
1⋅(x)=x
The opposite to x element −x we define as (−x)i=−xi,i∈N
We have:
(−x)+x=0,−x=(−1)⋅x
Thus all axioms of linear space for L(R) have verified by us.
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