A basic qn on the inner product of a vector with an infinite sum of vectors

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An infinite sum of vectors can be a vector in the same vector space, but this depends on the definition of the sum and the nature of the vector space. While vector addition is defined for two vectors and can be extended to finite sums, the sum of an infinite number of vectors requires a limit process, which necessitates a metric that not all vector spaces possess. In finite-dimensional spaces, if the limit of the nth partial sum exists, the sum will belong to the vector space. However, in infinite-dimensional spaces, convergent sequences can lead to limits that fall outside the subspace. An example is the infinite sum of polynomials resulting in a function that is not a polynomial, illustrating the complexities involved.
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A basic qn:An infinite sum of vectors will also be a vector in the same vector space?

By definition, the sum of any two vectors of a vector space will be a vector in the same vector space. But does this mean the sum of an uncountable or countable number of vectors will also be a vector in the same vector space?
 
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unfortunately, your question does not make sense until you have defined what you mean by a sum of a countable or uncountable number of vectors. Vector addition is defined for two vectors and can then be extended, by induction, to the sum of any finite number of vectors but the sum of an infinite number of vectors, whether countable or uncountable, is undefined.

To define the sum of an countable number of vectors, as we do for infinite series, would require a limit process which would, in turn, require a metric which vector space, in general, do not have.
 
That metric is very often defined from an inner product

d(x,y)=\|x-y\|=\sqrt{\langle x-y,x-y\rangle}

This allows you to define limits of sequences (of which infinite sums is a special case). If the sum (i.e. the limit of the nth partial sum as n goes to infinity) exists at all, and we're dealing with a finite-dimensional vector space, then the sum will certainly be a member of the vector space. It we're dealing with an infinite-dimensional vector space, there can exist convergent sequences of members of a subspace, that converge to a vector that doesn't belong to the subspace. This is why books on functional analysis talk about "closed subspaces" sometimes. A subspace is closed if the limit of every convergent sequence of members of the subspace belongs to the subspace.
 


seeker101 said:
By definition, the sum of any two vectors of a vector space will be a vector in the same vector space. But does this mean the sum of an uncountable or countable number of vectors will also be a vector in the same vector space?

An easy counter-example is the vector space of polynomials. 1/(1-x) is an infinite sum of polynomials but not a polynomial.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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