# Completeness of a specific Hilbert space

I am reading Zehnder's book "lectures on dynamical systems". In the chapter 7 he defines a space and states that it is a Hilbert space. I am struggling to show that the space is indeed complete. Here's the construction.

Let $$\Omega := C^{\infty}(S^1,\mathbb{R}^{2n} )$$. We represent an arbitrary element $$x \in \Omega$$ with its Fourier series, i.e.

$$x(t)= \sum_{j \in \mathbb{Z} } e^{j 2 \pi t J}x_j, \quad x_j \in \mathbb{R}^{2n}.$$

Now let $$H^s := \{ x \in L^2(S^1,\mathbb{R}^{2n} ) \mid \sum_{j \in \mathbb{Z} \setminus \{0\} } \mid j \mid^{2s} \mid x_j \mid^2 < \infty \}$$.

Equipped with the scalar product given by

$$ = _{\mathbb{R}^{2n}} + 2 \pi \sum_{j \in \mathbb{Z} \setminus \{0 \} } \mid j \mid^{2s} _{\mathbb{R}^{2n}}$$

My question is : is it sufficient to say that $$H^s \subset L^2$$ and since $$L^2$$ is an Hilbert space, then $$H^s$$ also is ? I guess that the answer is no, but I am not able to explain why.

I also tried to a direct computation, but I can't conclude. Let $$(x_n)$$ be a Cauchy sequence in $$H^s$$, then we have $$\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n,m \in \mathbb{N} : n,m \geq N$$ $$\parallel x_n - x_m \parallel \leq \epsilon.$$

My attempt : let $$\{ e_k \}$$ be the sequences with $$e_k(i)= \delta_{i,k}$$, where $$e_k(i)$$ denotes the $$i$$-th term of the sequence. Then, from the definition of the scalar product I get that $$=x_n(0)$$ and $$\frac{1}{2 \pi \mid k \mid^{2s}}=x_n(k)$$ if $$k \neq 0$$. Assume that $$k \neq 0$$, then $$\forall n,m \in \mathbb{N} : n,m \geq N$$

$$\mid x_n(k)- x_m(k) \mid = \frac{1}{2 \pi \mid k \mid^{2s}} \mid \mid \leq \parallel x_n - x_m \parallel \parallel e_k \parallel = \parallel x_n - x_m \parallel \leq \epsilon.$$

Hence $$\{ x_n(k) \}_{n \in \mathbb{N}}$$ is a convergent sequence of real numbers. Call the limit $$\widehat{x}(k)$$.

I can't find a proper argument to show that $$\widehat{x} \in H^{s}$$. Is it really clear that $$\widehat{x} \in L^2(S^2,\mathbb{R}^{2n})$$ ? I guess yes, since the sequence was taken in $$H^s$$ and $$L^2$$ is a Hilbert space.

This is not a complete answer, but what I have to say is too long to fit in a comment: To answer your first question it is indeed true that just because $$H^s\subset L^2$$ does not mean that $$H^s$$ has to be complete. Just consider $$\mathbb Q\subset \mathbb R$$ as a counterexample. What is true though, is that if $$Y$$ is a closed subset of a complete metric space, then $$Y$$ is complete. However, what is important to note is that completeness is a property of the metric, and because $$H^s$$ has a different inner-product than $$L^2$$ we cannot use this method (unless you prove the norms produced by the inner-product are equivalent). Thus we have to use an alternative approach.
Your idea of a direct proof is a good idea, but I am not sure what you are trying to do with your $$\hat x$$. This is only a function mapping into $$\mathbb R$$ as far as I can tell, and so definitely is not in $$L^2(S^1,\mathbb R^{2n})$$. It is also not clear to me what this sequence $$\{e_k\}$$ is. You are referring to them as sequences, but we are dealing with a function space $$L^2(S^1,\mathbb R^{2n})$$ so I am not certain how you are interpreting them as elements of this space.
• Thank you for your answer. Indeed my proof doesn't work out. I think that I need to find a suitable expression for $\parallel x_n - x_m \parallel$, but unfortunately it's not an easy task with this scalar product. The author just states that this space is complete, maybe there's another way of solving this problem but I just feel lost – Alain Mar 18 at 20:30