2
$\begingroup$

We define the space of linear and continuous operators from $\mathcal{S}(\mathbb{R}^d)$ to $\mathcal{S}'(\mathbb{R}^d)$ as $\mathcal{L}(\mathcal{S}(\mathbb{R}^d), \mathcal{S}'(\mathbb{R}^d))$.

The Schwartz kernel theorem says that there is an isomorphism between $\mathcal{L}(\mathcal{S}(\mathbb{R}^d), \mathcal{S}'(\mathbb{R}^d))$ and $\mathcal{S}'(\mathbb{R^d}\times \mathbb{R}^d)$, thanks to the relation that associated to a "kernel" $ K \in \mathcal{S}'(\mathbb{R^d}\times \mathbb{R}^d)$ the operator $\mathcal{K}$ defined by $$\langle \mathcal{K}\{\varphi_1\} , \varphi_2 \rangle = \langle \varphi_1\otimes \varphi_2 , K \rangle $$ for any $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R}^d)$.

It is often only implicit that the adequate topology on the dual is the weak*-topology. Is there a good reason to consider this topology for the dual?

Is the result false if one endows $\mathcal{S}'(\mathbb{R}^d)$ with the strong topology?

$\endgroup$
1
$\begingroup$

I try to give an answer, but I think the question is "wide". We can also reason in terms of the space of distributions, then, essentially, these results are valid for restriction on the space of tempered distributions. First, there is a result that demonstrates the equivalence of the following properties

If $\Omega \subseteq \mathbb{R}^n$ is open and if $u : \mathcal{D}(\Omega) \longrightarrow \mathbb{C}$ is a linear map, then the following properties are equivalent

  1. $u: \mathcal{D}(\Omega) \longrightarrow \mathbb{C}$ is continuous

  2. $u: \mathcal{D}(\Omega) \longrightarrow \mathbb{C}$ is limited

  3. If $\lbrace \varphi_k\rbrace \subset \mathcal{D}(\Omega)$ converges to $\varphi \in \mathcal{D}(\Omega)$ respect the topology of the test functions $\mathcal{T}$, then $u(\varphi_k) \rightarrow u(\varphi)$.

  4. The restriction of $u: \mathcal{D}(\Omega) \longrightarrow \mathbb{C}$ to $\mathcal{D}_K(\Omega)$ is continuous for all compact $K \subset \Omega$

  5. For all compact set $K \subset \Omega$ there are $N\in \mathbb{N}$ and $C_K > 0$ such that $|u(\varphi)| \leq C_K \left \| D^\alpha \varphi \right \|_{K_{N}}$, where $\left \| D^\alpha \varphi \right \|_{K_{N}} := \sup_{x \in K} |D^\alpha \varphi| , \forall \varphi \in \mathcal{D}_K(\Omega)$

From this fact follows a good definition of distribution. The couple $(\mathcal{D}(\Omega),\mathcal{D}'(\Omega))$ is a dual couple in the sense that $u(\varphi)=0$ $\forall u \in \mathcal{D}'(\Omega)$ $\Longrightarrow$ $\varphi=0$, ($L^1_{loc}(\Omega) \subset \mathcal{D}'(\Omega))$ then $\int_{\Omega} |\varphi|^2 dx = \langle\overline{\varphi}, \varphi \rangle=0$ and generalizing this situation if $u$ is a distribution $\varphi$ a test function, use this notation $\langle \varphi, u \rangle$=$\langle u, \varphi \rangle$= $u(\varphi)$. Furthermore in $\mathcal{D}'(\Omega)$ there is weak*-toplogy $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ defined through the separable family of seminorm $\mathcal{F}:=\lbrace p_\varphi(u)=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace$ and $\mathcal{D}'(\Omega)$ is locally convex spacea. The sequential convergence $\langle \varphi, u_k \rangle \rightarrow \langle \varphi , u \rangle$ $\forall \varphi \in \mathcal{D}(\Omega)$ is the weak* convergence, i.e. the convergence with respect to weak*-topology. These last facts are also valid in the space of temperate distributions, it is the same.

Now, if $\Omega_1, \Omega_2 \subseteq \mathbb{R}^n$ are open sets. If $T:\mathcal{D}(\Omega_1) \longrightarrow \mathcal{D}(\Omega_2)$ is linear map such that $T'u=u \circ T \in \mathcal{D}'(\Omega_1)$ $\forall u \in \mathcal{D}'(\Omega_{2})$, then $T':\mathcal{D}'(\Omega_2) \longrightarrow \mathcal{D}'(\Omega_1)$ is dual or transponse operator of $T$, and it is characterized by the following relation \begin{align*} \displaystyle \langle \varphi, T'u \rangle = \langle T \varphi, u \rangle \end{align*} it occurs that $T$ and $T'$ are respectively weak and weak* continuous, and the following result exists

  1. If $T: \mathcal{D}(\Omega_1) \longrightarrow \mathcal{D}(\Omega_2)$ and $T^*:\mathcal{D}(\Omega_2) \longrightarrow \mathcal{D}(\Omega_1)$ are continuous linear operator, and such that $\langle \varphi, T^* \psi \rangle = \langle T \varphi , \psi \rangle$. Then $T^*$ extends to a weak* continuous linear operator $T':\mathcal{D}'(\Omega_2) \longrightarrow \mathcal{D}'(\Omega_1)$.

The result in 6. is widely used to define new distributions by duality, also to define the transpose kernel of a Schwartz kernel, and from estimates with seminorm on $\mathcal{D}$ or $\mathcal{S}$ in your case. Essentially the Schwartz kernel theorem says that

  1. Let $X, Y \subset \mathbb{R}^n$ be open sets. A linear map $\mu : \mathcal{D}(Y) \rightarrow \mathcal{D}'(X)$ is sequentially continuous if and only if it is generated by a Schwartz kernel $k \in \mathcal{D}'(X \times Y)$.

it's just to have the sequentially continuous map definition $\mu : \mathcal{D}(Y) \rightarrow \mathcal{D}'(X)$ is necessary to have a convergence with respect to weak*-topology. I believe that all these are good regions not to consider other types of topologies, on the other hand the whole theory of distributions is based on particular topologies in the spaces of the test functions that allow an extension by duality and considering the weak* topology.

$\endgroup$
  • $\begingroup$ This is very interesting. Following your explanations, I understand better why the weak*-topology is natural when considering operators. My question was also motivated by function space theory on Banach spaces. In some context, one has good reason to consider the norm topology on the Banach space $X'$, with $\lVert u \rVert_{X'} = \sup \{ \langle u , f \rangle, \lVert f \rVert_X = 1\}$ (even if the weak* topology also appear very often, e.g. with the Banach-Alaoglu theorem). With this in mind, I was wondering about the role of the strong topology for $\mathcal{D}'$ or $\mathcal{S'}$. $\endgroup$ – Goulifet Feb 3 '18 at 18:19
  • $\begingroup$ I understand what you mean, but the discourse that topologies in test spaces are locally convex (a lower level of topologies induced by metrics), and by duality it is more natural to consider weak* topologies. $\endgroup$ – user288972 Feb 3 '18 at 19:16
1
$\begingroup$

Spaces like $S'$ and $D'$ have a canonical topology which is the strong one. It is slightly more involved than the weak-$\ast$ topology, and I think that's the only reason people prefer the latter but that is misguided. If for a locally convex space $V$ you always equip $V'$ with the strong topology you get the following benefits.

1) If $V$ is a Banach space, $V'$ is the usual dual with the operator norm topology.

2) Spaces like $S$, $S'$, $D$, $D'$, $\mathcal{E}$, $\mathcal{O}_M$, etc. are reflexive. In particular $S$ with its familiar Frechet topology can be recovered as $(S')'$ provided you use the strong rather than weak-$\ast$ topologies when you take duals.

3) The tensor product of distributions is continuous.

4) Moments of probability measures are continuous as soon as they are well defined (integrability condition). This is a consequence of 3) and the closed graph theorem.

I'm sure one can add more items to this list. The bottom line is that these spaces of test functions or distributions with the correct topology are essentially finite-dimensional.

$\endgroup$
  • $\begingroup$ I admit that these are very good reasons in favor of the strong topology as "natural". But do you agree that Schwartz kernel theorem is not valid for the strong topology (there are more operators continuous from $S$ to $(S',weak*)$ than from $S$ to $(S',strong)$? $\endgroup$ – Goulifet Mar 6 '18 at 2:03
  • 1
    $\begingroup$ The kernel theorem is valid for continuous maps from $S$ to strong $S'$. A priori there should be more continuous maps to weak-star $S'$ but I am not sure this is the case. Sometimes Baire type theorems come into play in this kind of situation. Although I have to think more to see if the sets of continuous maps are different or not. In any case this is best studied by replacing S with sequences of fast decay and S' by sequences of at most polynomial growth. This is how I proved the kernel theorem in the course I taught recently. $\endgroup$ – Abdelmalek Abdesselam Mar 6 '18 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.