Answer to Question #259380 in Functional Analysis for Jayanng

Question #259380

Show that the space L(R) of all bounded real sequence is a linear space over a vector field R


1
Expert's answer
2021-11-01T13:31:24-0400

Let x=(x1,x2,...), y=(y1,y2,...)L(R)\overline{x}=(x_1,x_2,...),\space \overline{y}=(y_1,y_2,...)\in L(R) be two elements from the set L(R), We define their sum as x+y=(x1+y1,x2+y2,...,xi+yi,..),iN\overline{x}+\overline{y}=(x_1+y_1,x_2+y_2,...,x_i+y_i,..),i\in N or as sum for each component. We should ptove, that x+yL(R)\overline x+\overline {y}\in L(R) but it is easy because iNxi+yixi+yixL(R)+yL(R)x+yL(R)xL(R)+yL(R)<\forall i\in N |x_i+y_i|\le |x_i|+|y_i|\le ||\overline{x}||_{L(R)}+||\overline{y}||_{L(R)}\rarr\\ ||\overline{x}+\overline{y}||_{L(R)}\le ||\overline{x}||_{L(R)}+||\overline{y}||_{L(R)}<\infty

Axioms of addition

x+y=y+x;x+(y+z)=(x+y)+z\overline{x}+\overline{y}=\overline{y}+\overline{x};\\ \overline{x}+(\overline{y}+\overline{z})=(\overline{x}+\overline{y})+\overline{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,..)\overline 0=(0,0,0,..) or vector with zeros only.

Next property takes place

xL(R) x+0=x\forall \overline{x} \in L(R)\space \overline{x}+\overline 0=\overline x

because it is true for each component.

Operation of multiplication elements of L(R) by scalar from R is defined by the formula

(cx)i=cxi,iN, cR, xL(R)(c\cdot \overline x)_i=c\cdot x_i,i\in N,\space c\in R,\space \overline{x}\in L(R)

We have cxL(R)c\cdot \overline x\in L(R) and cxL(R)cxL(R)||c\cdot \overline {x}||_{L(R)}\le |c|\cdot ||\overline{x}||_{L(R)}

Operation of multiplication is defined through components , therefore next properties take place:

c(x+y)=cx+cy;c(dx)=(cd)x, c,dRc\cdot (\overline x+\overline y)=c\cdot \overline{x}+c\cdot \overline{y};\\ c\cdot (d\cdot \overline x)=(c\cdot d)\cdot \overline {x},\space c, d\in R

1(x)=x1\cdot \overline(x)=\overline{x}

The opposite to x\overline{x} element x-\overline{x} we define as (x)i=xi,iN(-\overline{x})_i=-x_i,i\in N

We have:

(x)+x=0,x=(1)x(-\overline{x})+\overline {x}=\overline {0},\\ -\overline{x}=(-1)\cdot \overline{x}

Thus all axioms of linear space for L(R) have verified by us.


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