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 $ <f,g>=\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.

Thanks in advance for help.


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.


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