Show that $W^{s_2, 2}(\mathbb{T}) \hookrightarrow W^{s_1,2} (\mathbb{T})$ is a compact operator

Question

How can one prove that for $0 \le s_1 < s_2$, the embedding map $W^{s_2, 2}(\mathbb{T}) \hookrightarrow W^{s_1,2} (\mathbb{T})$ is a compact operator?

My proof

Let $f \in W^{s_2,2}(\mathbb{T})$, then $f \in L^2(\mathbb{T})$ and $\sum_{n \in \mathbb{Z}} (1+ n^2)^{s_2} |\widehat{f}(n)| < \infty$. Since $s_1 < s_2$, then we have

$$ \sum_{n \in \mathbb{Z}} (1+ n^2)^{s_1} |\widehat{f}(n)| \le \sum_{n \in \mathbb{Z}} (1+ n^2)^{s_2} |\widehat{f}(n)| < \infty.$$

Thus, $f \in W^{s_1,2}(\mathbb{T})$.

Consider the orthonormal basis $\{e_n \}_{n \in \mathbb{Z}}$ for $L^2(\mathbb{T})$. Let $f \in W^{s_1,2}(\mathbb{T})$. Since $f = \sum_{j \in \mathbb{Z}} \langle f, e_j \rangle e_j$, $\sum_{n \in \mathbb{Z}} (1+n^2)^{s_1} \left| \widehat{f}(n) \right|^2<\infty$, and $\widehat{f}(n) = \langle f, e_n \rangle$. Then we have

$\sum_{n \in \mathbb{Z}} (1+n^2)^{s_1} \left| \widehat{f}(n) \right|^2 = \sum_{n \in \mathbb{Z}} (1+n^2)^{s_1} \langle f, e_n \rangle^2<\infty.$

Given $\epsilon >0$, $\exists N >0$ such that $ \left| \sum_{|n| > N+1 } (1 + n^2)^{s_1} \langle f,e_n \rangle \right| < \epsilon$. Since $(1+n^2)^{s_1}$ is never zero, then $\underset{n \to \infty}{\lim}\langle f,e_n \rangle =0, \forall |n| > N+1$. Then for every $f \in W^{s_1,2}(\mathbb{T})$, $f$ can be expressed as a finite dimensional orthonormal bases $\{ e_n\}_{|n| \le N}$. Namely

$f = \sum_{|n| \le \mathbb{Z}} \langle f, e_n \rangle e_n.$

Thus the embedding operator is compact.

Added

2- If $m \in \mathbb{N}_0$ and $s > m + \frac{1}{2}$, then $W^{s,2}(T) \subset C^n (T) \subset W^{n,2}(T)$. Where $T=\mathbb{R}/2 \pi \mathbb{Z}$.

Is my proof to (1) is ok? could help me with the point (2).

Edited

I am thinking of using this theorem:

Let $T \in B(H)$ be a diagonal operator, i.e; $\exists \{ e_i \}_{i \in I}$ orthonormal basis of $H$ such that $Te_i = \lambda_i e_i$, $\lambda_i \in \mathbb{C}$. Then the following conditions are equivalent:

1- $T \in K(H)$

2- $\lim_{i \to \infty} \lambda_i =0$, i.e. $\forall \epsilon >0, \exists F \subset I$ finite subset such that $\forall i \in I \subset F$, $|\lambda_i|<\epsilon$

Answer 1

On your effort

Your efforts are in the right direction, but what you've got wrong is the jump from the step that $\langle f,e_n \rangle = 0$ for large enough $n$. We don't actually know this : and $\lim_{n \to \infty} \langle f,e_n\rangle = 0$ clearly doesn't imply this.

That's the mistake : but otherwise I'm very happy. You saved me the trouble of proving the inclusion is well defined, and continuity of the inclusion is fairly clear as well from the inequality used to establish the well-defined nature of the inclusion. (Because it makes it clear that the operator is bounded hence continuous).

On the theorem to be used

Let's see : The theorem provides iff conditions for a diagonal operator to be compact. However, the problem is quite simple : $T \in B(H)$ is not true, because we have here, two different Hilbert spaces $W^{s_1,2}$ and $W^{s_2,2}$ with their own norms, and while one can possibly think of generalizing the given theorem to this scenario, it clearly can't be used as is, because we don't have an operator that's going from one space back to the same space.

Attempting something with the direct sum , in order to bring the spaces under the same umbrella do not seem to be useful either , particularly because doing this makes the operator lose diagonalizability.

So, what you'll have to do if you were to use the theorem , is do what Alex suggested : interpret $W^{s_1,2}, W^{s_2,2}$ as subspaces of $L^2(\mathbb T)$ using appropriate maps, so you can try to create the inclusion using a self-mapping on $L^2(\mathbb T)$, and you are in good shape. Making sure the appropriate map is bounded will be sufficient, since the composition of a compact map with any number of bounded maps on either side remains compact.

This is a very important point : to use a theorem that is known for an operator on a space for an operator between two different spaces, we try to find a space where those two are embedded in, and use the theorem for an appropriate operator on that space. A technique to keep in mind.

The details

$W^{s_2,2}$ is the set of all functions in $L^2(\mathbb T)$ endowed with the norm $\|f\|^2_{W^{s_2,2}} = \sum_{n \in \mathbb Z} (1+n^2)^{s_2} |\hat{f}(n)|^2$ (so that only functions with that particular norm finite are considered). Similarly for $W^{s_1,2}$.

So if $e_n = \frac{e^{inx}}{\sqrt 2 \pi}$ then $e_n$ is an orthonormal basis of $L^2(\mathbb T)$. Under the $W^{s_2,2}$ norm and inner product resulting from it, we know that the $e_n$ continue to belong to, and be orthogonal in $W^{s_2,2}$ (recall that $\hat{f}(n) = \langle f,e_n\rangle$ so this makes this fact obvious). To make the $e_i$ orthonormal, observe that $\langle e_i,e_i\rangle_{W^{s_2,2}} = (1+i^2)^{s_2}$. Therefore, it follows that $d_i = \frac{e_i}{(1+i^2)^{\frac {s_2} 2}}$ is an orthonormal basis of $W^{s_2,2}$.

This fact is useful, because now we define the following map from $W^{s_2,2}$ to $\mathbb L^2(\mathbb T)$ : it is given by $T(e^{inx}) = (1+i^2)^{\frac{s_2}{2}}e^{inx}$ (scaled to the orthonormal basis $d_i$, and extended by linearity to the rest of the space). It's fairly clear to see that this is an isometric embedding.

Similarly , define $T': L^2(\mathbb T) \to W^{s_1,2}$ given by $e^{inx} \to (1+n^2)^{-\frac {s_1}2}e^{inx}$. Note that this is well-defined and is an isometry as well.

Now, consider the map $S : \mathbb L^2(T) \to \mathbb L^2(T)$ given by $S(e^{inx}) = (1+n^2)^{\frac{s_1-s_2}{2}} e^{inx}$. This is a diagonal (continuous) map, and is compact because of the theorem!

You just need to verify that $T'ST$ is the inclusion map from $W^{s_2,2}$ to $W^{s_1,2}$ and you are done, because $S$ is compact and the other maps are bounded.

The second part

For the second part, you seem to only be demanding inclusion, rather than compact inclusion.

For this, the point is quite simple : if $f$ is (classically/weakly) differentiable, then the Fourier coefficients of $f'$ have a nice relation with that of $f$ : indeed, roughly speaking, an integration by parts can be used to prove that $\hat{f'}(n) = --in \hat{f}(n)$. It's instantly obvious that an inductive argument can be used for proving that $W^{s,2}(\mathbb T) \subset C^n(\mathbb T)$ for $s > n + \frac 12$, since going down the derivative chain leaves us to prove that $W^{s,2}(\mathbb T) \subset C^0(T)$ for $s>\frac 12$, which is clear from making such a function a uniform limit of trigonometric polynomials and using the Weierstrass M-test. (I leave you to fill in the details, they are fairly standard and can be found in e.g. Rudin's book, or in Stein-Shakarchi).

That $C^n(\mathbb T) \subset W^{n,2}(\mathbb T)$ works out from a continuous function attaining its supremum/infimum on a compact set, and then stuff like integrability etc. is obvious from this point.

The generalized Rellich-Kondrasov theorem

The generalized Rellich-Kondrasov theorem (cited as Theorem 6.2, Chapter 6, Adams, Sobolev spaces) seeks to generalize the embedding of Sobolev spaces. While there are many, many conditions that can be of concern to us, the one that matters is part I and II of the conditions : If the domain $\Omega$ satisfies the cone conditions, then the map $W^{j+m,p}(\Omega) \to W^{j,q}(\Omega)$ is compact if $mp > n$ (where $\Omega \subset \mathbb R^n$) and $1 \leq q \leq \infty$, and is also compact if $mp \leq n$ and $q \leq \frac{np}{n-mp}$. (where if $n=mp$ then the RHS is infinite)

The cone condition is a technical condition that is quite easy to verify for the torus, so skipping that we have $j=s_1,m = s_2-s_1,p=q=2$ and $n=1$. If $s_2-s_1 >\frac 12$ we are done by the first condition, and if $s_2-s_1 \leq \frac 12$ then we are done by (doing some algebra and verifying) the second condition.

Of course, we barely used the full strength of the RK theorem, but this still shows its power in the face of general embedding theorems.

Answer 2

As pointed out in the comments, you're saying that $\widehat{f}(n) \to 0$ implies $\widehat{f}(n) = 0$ for large $|n|$. This is false.

However, given the setup, you're not very far from the proof. I suggest you try a direct path: take a bounded sequence $f_k \in W^{s_2,2}$ (i.e., $\| f_k \|_{W^{s_2,2}} \le 1$) and choose a subsequence for which $f_{k_j} \to f$ in $W^{s_1,2}$. Consider the following steps:

• Take a subsequence (here still denoted by $f_k$, for brevity) for which $f_{k} \to f$ weakly in $L^2$. (note that this fact hides a diagonal argument inside)
• Note that $\widehat{f_{k}}(n) \to \widehat{f}(n)$ for each $n$.
• All you need now is to show that $f_k$ is a Cauchy sequence in $W^{s_1,2}$. In this direction, you already observed that for each $\varepsilon > 0$ and $k \in \mathbb{N}$ there is $N$ such that $$ \sum_{|n| > N} (1 + n^2)^{s_1} |\widehat{f_k}(n)|^2 < \varepsilon $$ Using the fact that $\sum_{|n| > N} (1 + n^2)^{s_2} |\widehat{f_k}(n)|^2 \le 1$ for all $k$, you should be able to pick common $N$ for all $k$'s. Together with $\widehat{f_{k}}(n) \to \widehat{f}(n)$ (used for $|n| \le N$), it implies the Cauchy criterion.