Weak convergence in the space of tempered distributions and weighted Sobolev spaces

Question

It is well known that, at least as sets, $$\mathcal{S}'(\mathbb{R}^n)=\bigcup_{m\in\mathbb{N}}(1+|x|^2)^mH^{-m}(\mathbb{R}^n)$$ where $\mathcal{S}'(\mathbb{R}^n)$ is the space of tempered distribution (i.e. the dual of Schwartz space), $H^{-m}(\mathbb{R}^n)$ denotes the $L^2$-Sobolev space of order $-m$ with norm $\Vert u\Vert^2=\int_{\mathbb{R}^n}(1+|\xi|^2)^{-m}|\hat{u}(\xi)|^2\,d\xi$ and $(1+|x|^2)^mH^{-m}(\mathbb{R}^n)$ is the weighted Sobolev space $$(1+|x|^2)^mH^{-m}(\mathbb{R}^n)=\{u\in\mathcal{S}'(\mathbb{R}^n): (1+|x|^2)^{-m}u\in H^{-m}(\mathbb{R}^n)\}$$

I also read somewhere that the strong topology on $\mathcal{S}'$ is the strict inductive limit topology induced by the RHS, however, I have not been able to find a proof of this fact.

I would be content with a weaker sequential statement: I would like to prove that $\{u_n\}\subset\mathcal{S}'(\mathbb{R}^n)$ converges weakly to $u$ in $\mathcal{S}'(\mathbb{R}^n)$ (i.e. $u_n(\phi)\rightarrow u(\phi)$ in $\mathbb{C}$ as $n\rightarrow\infty$ for all $\phi\in\mathcal{S}(\mathbb{R}^n)$) if and only if there is some $m\in\mathbb{N}$ for which $u_m\rightarrow u$ in the topology of $(1+|x|^2)^mH^{-m}(\mathbb{R}^n)$.

Does anyone know a proof of this fact?

Answer 1

Taking the real part of everything we can restrict to real-valued Schwartz functions and real valued distributions.

Let the order of a tempered distribution be the least positive integer $m$ such that it extends to a continuous functional on the Hilbert space $B_m=(1+|x|)^{-m} H^m$

Let a sequence $(f_k ) \in S'$ such that $\sum_k f_k$ converges weakly to some distribution in $S'$ but such that $order(f_k)$ is unbounded, wlog. we can assume $order(f_k)=k$.

Let $c_k = \sup_{m\le k}\|f_m\|_{\textstyle (B_m)'}$

Construct a sequence $\phi_k\in S$ such that $$\|\phi_k\|_{B_{k-1}} \le \frac1{c_k} 2^{-k}\ \ and \ \ \langle f_k,\phi_k\rangle > e^k+\sum_{m<k} |\langle f_k,\phi_m\rangle|$$ (by definition of the order this is always possible)

$$\Phi= \sum_k \phi_k \ converges \ in \ all \ B_m \ thus \ in\ S$$

Also for $m>k$, $$|\langle f_k,\phi_m \rangle|\le \|f_k\|_{(B_k)'}\|\phi_k\|_{B_k}\le c_m\|\phi_m\|_{B_{m-1}}\le 2^{-m}$$

Thus

$$\langle f_k,\Phi \rangle = \sum_{m<k} \langle f_k,\phi_m\rangle+\langle f_k,\phi_k\rangle+ \sum_{m>k} \langle f_k,\phi_m\rangle$$ $$ > e^k-\sum_{m>k} 2^{-m}$$ And hence $$\sum_k \langle f_k,\Phi\rangle = \infty$$ which is a contradiction.

Answer 2

Indeed, the strong topology in $\mathscr{S}'(\mathbb{R}^n)$ is equal to the inductive limit of your sequence.

It is simplier to consider family of weighted Sobolev spaces with two indices: $$ H^s_l(\mathbb{R}^n) = \{f \in \mathscr{S}': (1 + |x|^2)^{l/2} \in H^s\}. $$ Theese are Hilbert spaces and if $s \ge s'$ and $l \ge l'$ then we have the continuous inclusion $H^s_l \subset H^{s'}_{l'}$. It is obvious that $\mathscr{S} \subset H^s_l \subset \mathscr{S}'$ for all real $s,l$.

$\mathit{Proposition\; 1}$. $\mathscr{S} = \bigcap\limits_{s,l \in \mathbb{R}}H^s_l$ and topology of $\mathscr{S}$ coincides with projective limit topology.

$\mathit{Proof}$. Let's introduce the banach space $C^s_l$ ($s \in \mathbb{N}$) of functions $f \in C^s(\mathbb{R}^n)$ s.t. $$||f||^{(s)}_l = \sup\limits_{x \in \mathbb{R}^n, |\alpha| \le s}(1 + |x|^2)^{l/2} |D^\alpha f(x)| < \infty$$ By definition $\mathscr{S}$ is equal to the intersection $\bigcap\limits_{s,l \in \mathbb{N}}C^s_l$ with projective limit topology. All that remains is to prove the coincidence of two pretty similar projective limits. The idea is to use the analogue of Sobolev embedding theorem. In our situation there are continuous inclusions $C^s_{l + \chi} \subset H^s_l$ and $H^{s + \chi}_l \subset C^s_l$ for arbitrary $\chi > n/2$. First inclusion is pretty obvious. Second one can be proved by Fourier transform (in the fasion of Sobolev embedding theorem). Theese inclusions obviously lead us to algebraic and topological coincidence of $\mathscr{S}$ and $\bigcap\limits_{s,l \in \mathbb{R}}H^s_l$. $\blacksquare$

Now we can say a bit more about $\mathscr{S}'$. We already by definition have an inclusion $\bigcup\limits_{s,l \in \mathbb{R}} H^s_l \subset \mathscr{S}'$.

$\mathit{Theorem\; 1}$. This inclusion turns out to be equatily and strong topology on $\mathscr{S}'$ is equal to inductive limit topology on the union.

$\mathit{Proof}$. This is a direct application of duality theorem for projective limits. It is known that Mackey dual of a projective limit that satisfies an additional density condition (that canonical image of projective limit is dense in all terms) is equal to inductive limit of Mackey duals (theorem 4.4 Schaefer TVS). All you need is to observe that Mackey topology on $\mathscr{S}'$ is equal to strong topology and that $(H^s_l)' = H^{-s}_{-l}$ (also with coincidence of Mackey topology on lhs and usual topology on rhs). $\blacksquare$

And finally your sequence of spaces $(1+|x|^2)^m H^{-m}$ is equal to a sequence $H^{-m}_{-2m}$ which is cofinal in directed set of weighted Sobolev spaces $H^s_l$ and therefore it gives the same inductive limit (note that inductive limit goes in the direction of $-\infty$).

$\mathbf{Edit.}$ Convergence of sequences.

Consider four types of convergence for $\phi_n \in \mathscr{S}'$.

1. $\phi_n \overset{1}{\rightarrow} \phi \in \mathscr{S}'$ wrt $*$-weak topology

2. $\phi_n \overset{2}{\rightarrow} \phi \in \mathscr{S}'$ wrt strong topology

3. $\phi_n \overset{3}{\rightarrow} \phi \in \mathscr{S}'$ iff there exist some $s,l \in \mathbb{R}$ s.t. $\phi_n,\phi \in H^s_l$ and convergence holds there (in the sence of Sobolev space norm)

4. $\phi_n \overset{4}{\rightarrow} \phi \in \mathscr{S}'$ iff there exist some $s,l \in \mathbb{R}$ s.t. $\phi_n,\phi \in (C^s_l)'$ and convergence holds there (in the norm of dual space)

$\mathit{Theorem\; 2}.$ All four types of convergence are equivalent.

It is already obvious that $3 \Rightarrow 2 \Rightarrow 1$. First implication is due to first theorem and second implication holds since strong topology is stronger than $*$-weak. Also from inclusions that were mentioned in first proposition follows equivalence $3 \Leftrightarrow 4$. Therefore we only have to prove that $1 \Rightarrow 3$. To prove this we need some more information about inductive limits.

For propositions 2,3,4 we fix the following data. Let $E = \bigcup\limits_{n \in \mathbb{N}} E_n$ be the inductive limit of a sequence $\{E_n\}_{n \in \mathbb{N}}$, $E_n \subset E_m$ for $n \le m$ and inclusion is continuous ($E_n$ are locally convex spaces).

$\mathit{Proposition\; 2}$. Assume that there is a sequence $V_n \subset E_n$ of neighbourhoods of zero s.t. $V_n \subset V_m$ for $n \le m$ and $V_n$ is closed in $E$ for all $n$. Then for arbitrary bounded $S \subset E$ exist $n$ and $\lambda > 0$ s.t. $S \subset \lambda V_n$.

$\mathit{Proof}.$ Assume that the statement is false. Then there exists a sequence $x_n \in S$ s.t. $x_n \notin n V_n$. Then sequence $y_n = \frac{1}{n}x_n$ converges to $0$ (since $x_n$ is bounded) and $y_n \notin V_n$. Now we are able to construct a sequence $\tilde{V}_n$ of neigbourhoods of zero in $E_n$ s.t. $\tilde{V}_n \subset \tilde{V}_m$ for $n \le m$ and $y_m \notin \tilde{V}_n$ for all $m,n$. Assume that we have succeded in this construction. Then $V = \bigcup\limits_{n \in \mathbb{N}} \tilde{V}_n$ is a neighbourhood of zero in $E$ s.t. $y_m \notin V$ for all $m$ which cotradicts with convergence of $y_n$.

Now we construct a sequence $\tilde{V}_n$ by induction. At first step consider neighbourhood of zero $U \subset E$ s.t. $y_1 \notin \overline{U + V_1}$ (it is possible since $V_1$ is closed and $y_1 \notin V_1$). Then let $V^1_n = V_n \bigcap \overline{U + V_1}$. Then $V^1_n$ is a closed neighbourhood of zero in $E_n$, $V^1_n \subset V^1_m$ for $n \le m$ and $y_1,y_n \notin V^1_n$. Now assume that we have already constructed a sequence of closed neighbourhoods of zero (in $E_n$) $V^k_n$ s.t. $y_1, y_2, \dots, y_k, y_n \notin V^k_n$, $V^k_n \subset V^k_m$ for $n \le m$. Then take neighbourhood of zero $U \subset E$ s.t. $y_{k+1} \notin \overline{U + V^k_{k+1}}$ and let $V^{k+1}_n = V^k_n \bigcap \overline{U + V^k_{k+1}}$.

Now let $\tilde{V}_n = \bigcap\limits_{k \in \mathbb{N}} V^k_n$. It is obvious that $y_m \notin \tilde{V}_n$ for all $m,n$ and that $\tilde{V}_n \subset \tilde{V}_m$ for $n \le m$. All that remains is to prove that $\tilde{V}_n$ is a neighbourhood of zero in $E_n$. To see this note that in construction of sequence $V^k_n$ on step $k$ terms $V^k_1, V^k_2, \dots, V^k_k$ don't change. Therefore sequence $V^k_n$ stabilizes for $k \ge n$ and therefore $\tilde{V}_n$ is a $\mathit{finite}$ intersection of neighborhoods of zero. $\blacksquare$

$\mathit{Proposition\; 3}$. Now assume that all $E_n$ are normed spaces and $B_n$ is the closed unit ball in $E_n$. Assume also that $B_n$ is closed in $E$ for all $n$. Then if $S \subset E$ is bounded then there exists $n \in \mathbb{N}$ s.t. $S \subset E_n$ and $S$ is bounded there.

$\mathit{Proof}$. Let's construct a sequence $\lambda_n > 0$ s.t. $B_n \subset \lambda_n B_{n+1}$ (it is possible since inclusion of $E_n$ into $E_{n+1}$ is continuous). Then let $V_n = \lambda_1 \dots \lambda_{n-1}B_{n-1}$. We know that $V_n \subset V_{n+1}$ and $V_n$ is closed in $E$. Therefore by previous proposition there exist $n$ and $\lambda > 0$ s.t. $S \subset \lambda V_n$. Then $S \subset E_n$ and it is bounded there. $\blacksquare$

$\mathit{Proposition\; 4}$. Now let $E_n$ be reflexive Banach spaces and $B_n$ is still a unit ball in $E_n$. Then $B_n$ is closed in $E$.

$\mathit{Proof}.$ Since $E_n$ are reflexive $B_n$ is compact in weak topology. Since every continuous linear map from one LCS to another is also continuous wrt weak topologies it follows that $B_n$ is compact in all $E_m$ for $m \ge n$ and $E$ wrt weak topology. All that remains is to prove that $E$ is Hausdorff (then weak topology in $E$ is also Hausdorff and therefore $B_n$ is closed in weak topology of $E$ and consequently it is closed in $E$).

We have already proved that $B_n$ is closed in $E_m$ for all $m \ge n$. Let $x \in E$ be a non-zero element. Then $x \in E_n$ for some $n$. Let's construct a sequence $\lambda_n,\lambda_{n+1},\dots$ of positive numbers s.t. $x \notin \lambda_n B_n + \lambda_{n+1}B_{n+1} + \dots + \lambda_m B_m = V_m$ for all $m \ge n$. Assume that we succeded in this construction. Then $x \notin V = \bigcup\limits_{n \in \mathbb{N}} V_n$ where $V_k = E_k \bigcap V_n$ for $k < n$. $V$ is a neighbourhood of zero in $E$ and therefore $E$ is Hausdorff.

Now we construct sequence $\lambda_m$, $m \ge n$ by induction. First step is trivial. Assume that we have already constructed $\lambda_n, \dots, \lambda_m$ s.t. $x \notin V_m$. Then $V_m$ is a sum of sets that are weakly compact in $E_{m+1}$. Therefore $V_m$ is closed in $E_{m+1}$ and we can continue the construction. $\blacksquare$.

Now it is trivial to prove an important statement about $\mathscr{S}'$ since $H^s_l$ are reflexive (Hilbert) spaces.

$\mathit{Corollary}.$ Let $S \subset \mathscr{S}'$ be a strongly bounded set. Then there exist $s,t \in \mathbb{R}$ s.t. $S \subset H^s_l$ and is bounded there. $\blacksquare$

Also before we prove theorem 2 let's consider another fact.

$\mathit{Lemma}$. Let $F$ be a Banach space and assume that $\alpha_n \in F'$ is a bounded sequence, s.t. $\alpha_n(x)$ converges for all $x \in S \subset F$ where $S$ is dense. Then $\alpha_n$ $*$-weakly converges (to some element $\alpha$).

$\mathit{Proof}$. This is relatively simple. Since $\alpha_n$ is bounded and it converges pointwise on some dense set $S$ it is simple to check that it converges pointwise on all $F$. Then Banach-Steinhaus theorem finishes the proof. $\blacksquare$

$\mathit{Proof\; of\; theorem\; 2}$. Let $\phi_n \in \mathscr{S}'$ be a $*$-weakly convergent sequence. Then it is $*$-weakly bounded and therefore it is strongly bounded since $\mathscr{S}$ is reflexive. Then by propositions 2,3,4 (see corollary) we know that this sequence belongs to some $H^s_l$ and is bounded there. Then we can use lemma since $(\phi_n, f)$ converges for all $f \in \mathscr{S}$ and it is dense in $H^s_l$. So $\phi_n$ weakly converges in $H^s_l$ (weak and $*$-weak convergence on $H^s_l$ are the same). Then by compactness of embeddings of Sobolev spaces it follows that $\phi_n$ converges wrt norm in $H^{s'}_l$ for arbitrary $s' < s$. $\blacksquare$.

$\mathbf{Some\; additional\; notes}.$

1. In all theese proofs it is very important that spaces $\mathscr{S}$ and $H^s_l$ are reflexive. For example it is very hard to prove that fourth type of convergence is equivalent to first or second without going through third one since $C^s_l$ are not reflexive. Reflexivity is important in first theorem since it gives equality of strong and Mackey topology and it is freely used in proposition 4 and theorem 2.

2. Strong and weak topology on $\mathscr{S}'$ don't coincide. Howewer they induce equal convergence on sequences (but not equal convergence of nets).

3. It is also important that $\mathscr{S}$ is a Frechet-Schwartz space, since we have used compactness of embeddings for Sobolev spaces

4. It is possible to prove that in the same assumptions as in proposition 4 the inductive limit is reflexive. Therefore I suppose that we can generalize the results about convergence to a following situation: $E$ an projective limit of a sequence of reflexive Banach spaces with compact inclusions, s.t. $E$ is dense in $E_n$ for all $n$. Then $E'$ with strong topology is an inductive limit of $E_n'$ with strong topology and three types of convergence in $E'$ coincide: $*$-weak, strong, convergence in $E'_n$ for some $n$.