Completion of a Inner Product Space

Question

I was reading Introductory functional analysis with applications by Kreyszig and I'm stuck trying to understand the proof that every inner product space can be completed.

I have many doubts when it comes to using equivalence classes because it is used in the proof. Also, it mentions 'By Lemma 3.2.2,' the lemma being that the inner product is continuous. The cited Theorem 2.3.2 is the one that guarantees every metric space can be completed.

In short, could someone help me by providing a more detailed explanation of this proof?

I believe the use of equivalence classes is because, when we want to complete a space, we want to add to this space only the points that are limits of some Cauchy sequence and are not in this space. So, in my understanding, the equivalence class plays the role of dealing only with the points that we need to add.

Answer 1

It is possible to provide an alternative proof. Let $X$ be the completion of the normed space $W.$ Then $W$ is dense in $X.$ Assume $x_n,y_n\in W$ and $x_n\to x,$ $y_n\to y.$ Then $\langle x_n,y_n\rangle$ is convergent. Indeed, we have $$ \langle x_n,y_n\rangle ={1\over 4}\sum_{k=1}^4i^k\|x_n+i^ky_n\|^2\quad (*)$$ The right hand side is convergent so the limit of $\langle x_n,y_n\rangle $ exists and does not depend on the sequences $x_n,y_n$ but on $x,y.$ Thus we can define the function on $X\times X$ by $$\langle x,y\rangle=\lim _n\langle x_n,y_n\rangle $$ This function clearly satisfies the properties of the inner product and corresponds to the norm on $X.$

Remark The equivalence classes of Cauchy sequences are used in constructing the norm completion of $W,$ but we do not need to discuss that taking for granted the completion exists.

Answer 2

It is not correct that the use of equivalence classes is to deal with only the points we need to add. The equivalence classes also deal with the already existing points. If $x\in X$ then the sequence $(x_n)$ with $x_n=x$ for all $n$ is a Cauchy sequence and its equivalence class is found in $W.$

The use of Cauchy sequences is because completeness is defined in terms of such. A converging sequence is always Cauchy, but a Cauchy sequence does not always converge. A space where a Cauchy sequence always converges is called complete.

Basically we define a new space where each point is the virtual limit (or "the should be limit") of a Cauchy sequence.

The use of equivalence classes is because there are many Cauchy sequences for every limit. For example, here are some Cauchy sequences in $\mathbb R$ that have limit $0$: $(1/n)$, $((-1)^n/n^2)$, $(e^{-n})$.

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I don't have that book so I don't know the exact steps of the proof, but it should be something like the following (not necessarily in this order and I might have missed some steps):

1. Let $H$ be the set of equivalence classes of Cauchy sequences on $X.$
2. Explain how $X$ is embedded as a subset $W$ of $H.$
3. Give $H$ a linear structure derived from $X.$
4. Give $H$ an inner product derived from the inner product on $X.$
5. Show that $H$ is complete.
6. Show that $X$ is isomorphic to $W.$