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

$$ <x,y> = <x_0,y_0>_{\mathbb{R}^{2n}} + 2 \pi \sum_{j \in \mathbb{Z} \setminus \{0 \} } \mid j \mid^{2s} <x_j,y_j>_{\mathbb{R}^{2n}}$$

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 $<x_n,e_0>=x_n(0)$ and $\frac{1}{2 \pi \mid k \mid^{2s}}<x_n,e_k>=x_n(k)$ if $k \neq 0$. Assume that $k \neq 0$, then $\forall n,m \in \mathbb{N} : n,m \geq N$

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

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.

Thank you for your help.


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.

  • $\begingroup$ 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 $\endgroup$ – Alain Mar 18 at 20:30

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