Answer to Question #259380 in Functional Analysis for Jayanng

Question

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

Answer 1

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

Axioms of addition

$\overline{x}+\overline{y}=\overline{y}+\overline{x};\\ \overline{x}+(\overline{y}+\overline{z})=(\overline{x}+\overline{y})+\overline{z}$

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

$\overline 0=(0,0,0,..)$ or vector with zeros only.

Next property takes place

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

because it is true for each component.

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

$(c\cdot \overline x)_i=c\cdot x_i,i\in N,\space c\in R,\space \overline{x}\in L(R)$

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

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

$c\cdot (\overline x+\overline y)=c\cdot \overline{x}+c\cdot \overline{y};\\ c\cdot (d\cdot \overline x)=(c\cdot d)\cdot \overline {x},\space c, d\in R$

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

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

We have:

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

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