I am convinced that B(R^N) is a sub-algebra of R^N. B(R^N) is closed under the operations of addition, scalar multiplication, and multiplication. This makes sense because B(R^N) represents the set of all bounded real-valued sequences. It seems that you can treat these sets of sequences as vectors and therefore addition, scalar multiplication and multiplication would all be possible.
- <s, s> > 0 for all s not equal to the empty set in the set of all bounded real-valued sequences. The inner product of two bounded sequences s, t: N → R is equal to <s, t> = the summation of s(n)t(n)/2n while n is greater than or equal to 1. This would mean that <s, s> = the summation of s(n)s(n)/2n = s2(n)/2n while n is greater than or equal to 1. The multiplication between two bounded sets is possible, therefore s(n)s(n) = s2(n)
- Similar to how the dot product between two vectors is the same whether it is a dot b ot b dot a, <s, t>=<t, s> for all s,t in the set of B(R^N). <s, t> := the summation of s(n)*t(n)/2^n for n is greater than or equal to 1. <t, s>:= the summation of t(n)*s(n)/2^n for n is greater than or equal to 1. Because of the commutative property of multiplication, s(n)*t(n) = t(n)*s(n), therefore <s, t> = <t, s> for all s, t in the set of B(R^N).
- This statement, “<s_1+s_2, t> is greater than or equal to <s_1, t> + <s_2, t> for all s_1, s_2, t in the set of B(R^N)” is true because <s_1+s_2, t> := the summation of (s_1(n)+s_2(n))*t(n)/2^n for n is greater than or equal to 1 and <s_1, t> + <s_2, t> := the summation of s_1(n)*t(n)/2^n + s_2(n)*t(n)/2^n which equals (s_1(n)*t(n) + s_2(n)*t(n))/2^n which equals (s_1(n)+s_2(n))*t(n)/2^n because of the distributive property.
- <λs, t> = λ<s, t> for all s, t in the set of B(R^N) and all λ in the set of real numbers because <λs, t>:= the summation of λ*s(n)*t(n)/2^n for n is greater than or equal to 1. λ<s, t> := λ(s(n)*t(n)/2^n) hich equals λ*s(n)*t(n)/2^n.