# $l_2(S)$is a hilbert space where S is a subset

Let S be a non-empty set and $$l_2(S)$$ be the set of all complex functions $$f$$ defined on $$S$$ with the following two properties:

$$(1) \{s:f(s)\ne0\}$$ is empty or countable. $$(2) \sum{|f(s)|}^2 < +\infty$$

Then show that :

(a) $$l_2(S)$$ forms a complex linear space with respect to pointwise addition and scalar multiplication.

(b) If norm and inner product is defined as, $$||f||=(\sum{|f(s)|}^2)^{\frac{1}{2}}$$ and $$=\sum f(s)\bar{g(s)}$$ respectively, then $$l_2(S)$$ is actually a Hilbert space.

My attempt :

(a) Easily done.

(b) Clearly a complex linear space and then assuming it to be Banach, considering for each $$f \in l_2(S)$$ , the form $$f(s)=f_1(s)+if_2(s)$$ i.e. considering its Real and Imaginary parts, computed and proved that the 'parallelogram law' holds for $$l_2(S)$$. But stuck on how to show that the space is actually complete.

You can reduce to the case that $$l_2$$ is complete:
Assume you have a cauchy sequence $$(f_n)$$ in $$l_2(S)$$. Set $$N=\bigcup_{n\in \mathbb{N} } \{s:f_n(s)\neq 0\}.$$ Then you identify $$l_2(N)$$ as a subspace of $$l_2(S)$$ and under this identification your sequence lives in $$l_2(N)$$. As $$N$$ is countable $$l_2(N)\cong l_2$$ or $$l_2(N)\cong \mathbb{C}^N$$ depending on whether $$N$$ is infinite or not, so your sequence converges.
A proof for the completeness of $$l_2$$ can be found here (just replace $$\mathbb R$$ by $$\mathbb C$$): Understanding the proof of $$l_2$$ being complete.