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?


3 Answers 3


Spaces like $\mathscr{S}'$ and $\mathscr{D}'$ 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 $\mathscr{S}$, $\mathscr{S}'$, $\mathscr{D}$, $\mathscr{D}'$, $\mathscr{E}$, $\mathscr{O}_M$, etc. are reflexive. In particular $\mathscr{S}$ with its familiar Fréchet topology can be recovered as $(\mathscr{S}')'$ 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".

Feb 2020 edit/addendum following the questions raised by LinearOperator32:

For the $\mathscr{S},\mathscr{S}'$ case (which is technically simpler than $\mathscr{D},\mathscr{D}'$), the Schwartz Kernel Theorem in its most powerful form (which is not in otherwise excellent books like the one by Friedlander and Joshi or Hörmander Vol. I) says the following.

The map

$$ \mathscr{S}'(\mathbb{R}^{m+n})\rightarrow {\rm Hom}(\mathscr{S}(\mathbb{R}^m),\mathscr{S}'(\mathbb{R}^n)) $$ $$ T\mapsto(f\mapsto(g\mapsto T(f\otimes g))) $$ is an isomorphism of topological vector spaces. Usually the versions found in the literature show the bijective property but do not mention the homeomorphism part.

To learn the proof of the above theorem, one basically has two options

1) Option 1 (the orthodox approach): Learn all the definitions in the diagram featuring in Kuperberg's MO answer


for instance, by

1.a) reading the 500 pages or so that precede the Kernel Theorem in the book "Topological Vector Spaces, Distributions and Kernels" by Trèves, or by

1.b) learning French and reading the book "Théorie des Distributions" by Schwartz as well as its sequel made of the two articles "Théorie des distributions à valeurs vectorielles. I" and "Théorie des distributions à valeurs vectorielles. II", or

1.c) by reading Vol. 2 of the book "Topological Vector Spaces and Distributions" by Horváth :)... (just kidding, the book does not exist, but it would have been awesome if it did, given the high quality of Vol. 1).

2) Option 2 (my approach): Use the isomorphism of $\mathscr{S}$ with a space of sequences to reduce the Kernel Theorem to a much more manageable discrete version with infinite column vectors and infinite matrices, where this becomes completely elementary (something I could give as 2h midterm exam in a first year graduate analysis course). This approach is sketched in

Understanding the proof of Schwartz Kernel Theorem

Together with the Kernel Theorem one can prove in this way its companion result, namely, Fubini's Theorem for distributions. With these tools one can derive effortlessly with a couple of algebraic manipulations some classical facts which usually are established via some cumbersome estimates. I explained that in


As for sequences used for defining continuity: This makes no sense in 2020.

When learning the notion of continuity, the definition is (correctly) given using the local version with neighborhoods and the global version with inverse images of open sets. Then, one learns as a useful proposition the sequential criterion for say metric spaces. Maybe later still, one learns about ultrafilters or nets as a substitute when spaces start being far from metrizable. Unfortunately, most expositions of the theory of distributions give the sequential criterion as the definition of continuity. Worse still, it does that for maps that are linear, i.e., not that complicated anyway. The reason is, these expositions dance around the topology yet carefully avoid saying what it is, as if this was too difficult. It is not! See the explanations I gave at

Doubt in understanding Space $D(\Omega)$

and also at


for a simple example.

The reason for the dance around the topology is mostly historical. As very well explained in the answer by Pedro at

Motivation for test function topologies

Laurent Schwartz developed his theory before figuring out the topology of $\mathscr{D}$, relying on sequential characterizations of continuity as a temporary solution in order to move forward quickly towards interesting applications. For some reason, this temporary fix, has stayed with us.

Of the most well known presentations, one can single out two which define the topology of $\mathscr{D}$ properly: the one by Rudin in the book "Functional Analysis", and the one by Tao in his blog entry on the subject


The presentations are similar, but Rudin's is a bit messy as discussed in

Rudin's Construction of Inductive Limit Topology: unnecessarily abstruse?

Tao's account is clean but has the following practical drawback, from the point of view of trying to get rid of sequential criteria for continuity. The set of defining seminorms is described by constraints: it is the set of seminorms satisfying some admissibility condition. In order to prove say continuity of a linear map with domain $\mathscr{D}$ without sequences one needs to show a statement of the form: for every continuous seminorm in the target, there exists a continuous seminorm in the domain, such that some inequality is satisfied. So one has to produce a seminorm for $\mathscr{D}$. This calls for a set of defining seminorms which is described parametrically. Fortunately, this was done by Horváth, as I discussed in the links above, but hardly anyone knows about this, as far as I can tell.

  • $\begingroup$ 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, 2018 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. $\endgroup$ Mar 6, 2018 at 14:44
  • $\begingroup$ @AbdelmalekAbdesselam, the sets of continuity are the same, see my answer. $\endgroup$ Feb 2, 2020 at 6:39
  • $\begingroup$ @LinearOperator32: well, I was trying to avoid sequences altogether. There is no need for them really. $\endgroup$ Feb 3, 2020 at 22:37
  • $\begingroup$ What proof did you have in mind that avoided sequences? An idea I have is: recall $\beta(\mathcal D', \mathcal D) = \mu(\mathcal D', \mathcal D)$, and recall $\sigma(\mathcal D', \mathcal D)$ and $\mu(\mathcal D', \mathcal D)$ have the same bounded sets as they are compatible lc topologies. $\mathcal D$ is bornological. These observations show the continuity sets are the same. However, the bornologicity of $\mathcal D$ relies on the bornologicity of metrizable lcs' (since it's an LF space), which is usually proven using sequences (as far as I know), so it's not totally "sequence-free". $\endgroup$ Feb 4, 2020 at 23:25

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.

  • $\begingroup$ 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, 2018 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, 2018 at 19:16

Let $\mathfrak S$ be a collection of bounded subsets of $\mathcal S$ (of $\mathcal D$) which covers $\mathcal S$ (which covers $\mathcal D$). Then, the Schwartz kernel theorem is true when equipping $\mathcal S'$ (equipping $\mathcal D'$) with the topology of uniform convergence on members of $\mathfrak S$, because $$\mathscr L(\mathcal S, \mathcal S'_{\sigma}) = \mathscr L(\mathcal S, \mathcal S'_{\mathfrak S}) = \mathscr L(\mathcal S, \mathcal S'_{\beta})~~ \text{and} ~~\mathscr L(\mathcal D, \mathcal D'_{\sigma}) = \mathscr L(\mathcal D, \mathcal D'_{\mathfrak S}) = \mathscr L(\mathcal D, \mathcal D'_{\beta}).$$

Let us focus on the second set of equalities (the proof is similar in the other case, anyways). First:

Lemma: $\sigma(\mathcal D', \mathcal D)$ and $\beta(\mathcal D', \mathcal D)$ convergent sequences are the same.

Proof: If $(u_n)_n$ is a sequence in $\mathcal D'$ converging in $\sigma(\mathcal D', \mathcal D)$ to some $u \in \mathcal D'$, then, as $\mathcal D$ is barreled, the convergence occurs uniformly on precompact subsets of $\mathcal D$. But bounded subsets of $\mathcal D$ are precompact. $\blacksquare$

Then one just needs to recall that for a linear map $T: \mathcal D \to E$, where $E$ is a TVS, $T$ is continuous if and only if it is sequentially continuous.


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