Set of the all sequences of real numbers $x = (x_1,x_2,...,x_n,...)$ with $\sum_{n=1}^{\infty}{x_i}^2 < \infty$

Question

Is the set of the all sequences $(x_n)_n$ over $\Bbb R$ such that $(x_n)_n = (x_1,x_2,...,x_n,...)$ with $$ \sum_{n=1}^{\infty}{x_i}^2 < \infty $$ a vector space over $\Bbb R$ with co-ordinate wise addition and scalar multiplication ?

I don't know how to proceed this type of problems.

Answer 1

Closure:

We have $\sum x_i^2<\infty, \sum y_i^2<\infty$

Then $\sum (x_i+y_i)^2=\sum x_i^2+\sum y_i^2+2\sum x_iy_i$.

$\sum x_iy_i\le (\sum x_i^2)^{\frac{1}{2}}\times (\sum y_i^2)^{\frac{1}{2}}$(Cauchy-Schwarz Inequality).

Hence $\sum (x_i+y_i)^2<\infty$

For any $\lambda$,$\sum \lambda^2x_i^2=\lambda^2\sum x_i^2<\infty$

Answer 2

Hint:

Here's a start: you have show that if $s$ is such a sequence, so is $cs$ for any real number $c$, i.e., you must show that if $$ \sum_i (s_i)^2 < \infty $$ then $$ \sum_i (cs_i)^2 < \infty $$ as well. Can you do that?

That shows that the operation $M$ of scalar multiplication, which is defined on the set $S$ of ALL sequences, has the property that when it's restricted to those with finite sum-of-squares --- let's call that set $V$ --- turns out to satisfy

$$ M: \Bbb R \times V \to V \subset S $$ i.e., that $V$ is "closed under scalar multiplication".

Answer 3

In order to show that the set $V$ of all sequences $$x = (x_1,x_2,...,x_n,...)$$ such that $$\sum_{n=1}^{\infty}{x_i}^2 < \infty$$ is a vector space, you need to verify all requirements of a vector space are satisfied.

Note that $V$ is non-empty because the sequence of $ (0,0,0,.....)\in V $

$V$ is closed under scalar multiplication because if $$\sum_{n=1}^{\infty}{x_i}^2 < \infty$$ Then $$\sum_{n=1}^{\infty}{(\lambda x_i)}^2< \infty$$ for any scalar $\lambda .

The closure of addition is a challenge but it is the result of the Cauchy-Schwarz inequality. $$\sum_{n=1}^{\infty}{(x_iy_i)}^2< \sum_{n=1}^{\infty}{x_i}^2 \sum_{n=1}^{\infty}{y_i}^2 $$

Other requirements such as commutative and distributive property of scalar multiplication over vector addition are inherited from the same properties of sequences of real numbers.

Thus all the requirement are met and $V$ is a vector space.