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Let $X$ be a reflexive real Banach space, the complex case should be totally analogous. Define $$\tag{1} \mathcal{D}^\star(0, T; X)=\left\{ u\colon \mathcal{D}(0, T)\to X\ \text{linear and continuous}\right\} $$ where the topology on $\mathcal{D}(0, T)$, the space of real-valued test functions, is the usual one from distribution theory.

Now define $$ \tag{2} \left[\mathcal{D}(0, T; X^\star)\right]^\star = \left\{u \colon \mathcal{D}(0, T; X^\star)\to \mathbb{R}\ \text{linear and continuous}\right\}, $$ where $\mathcal{D}(0, T; X^\star)$ denotes the space of the smooth $f\colon (0, T)\to X^\star$ such that the support $\operatorname*{Supp}(f)$ is compact. We equip this vector space with the obvious analogue of the topology of $\mathcal{D}(0, T)$. Precisely, we consider the unique topology$^{[1]}$ such that, if $\phi_n, \phi\in \mathcal{D}(0, T; X^\star)$ then $\phi_n\to \phi$ is equivalent to $$ \begin{cases} \operatorname*{Supp}\phi_n \subset [a, b]\subset (0, T),\ \text{for fixed }a,b;\\ \left\lVert \frac{d^k \phi_n}{dx^k}-\frac{d^k\phi}{dx^k} \right\rVert_{\infty} \to 0,\quad\forall k\in \mathbb{N}. \end{cases} $$

Both definitions give rise to something which might be reasonably called "space of $X$-valued distributions".

Question. Are these two spaces isomorphic?

Example.

Let $X=\mathbb{R}^n$ and consider a continuous function $\boldsymbol{u}\colon (0, T)\to \mathbb{R}^n$. (The boldface font refers to vector valued functions). The two definitions above give rise to the following two representations of $\boldsymbol u$ as a vector valued distribution. Using definition (1) $$ \boldsymbol{u}\text{ acts on }\mathcal{D}(0, T)\text{ through the pairing }\langle \boldsymbol{u}, \phi\rangle = \int_0^T \boldsymbol{u}(t)\phi(t)\, dt,\text{ where }\phi\in \mathcal{D}(0, T).$$ Note that the test function $\phi$ is scalar-valued. On the other hand, using definition (2) $$ \boldsymbol{u}\text{ acts on }\mathcal{D}(0, T; \mathbb{R}^n)\text{ through the pairing }\langle \boldsymbol{u}, \boldsymbol{\psi}\rangle = \int_0^T \boldsymbol{u}(t)\cdot \boldsymbol \psi(t)\, dt,\text{ where }\boldsymbol\psi\in \mathcal{D}(0, T; \mathbb{R}^n).$$ Here the test function is vector-valued and the pairing uses the dot product of $\mathbb{R}^n$.

Note.

From some lecture notes which I found online it seems that Laurent Schwartz himself chose definition (1).


$^{[1]}$ Actually, I am cheating here. I know neither if such a topology exists nor if it is unique. I am just guessing that the usual construction which works for real valued test functions works here as well.

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  • 1
    $\begingroup$ This question on Hilbert-space valued distributions is strongly related. $\endgroup$ – Giuseppe Negro Aug 8 '18 at 10:21
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    $\begingroup$ Isn't this exactly Grothendieck's thesis problem? He defined tensor products in the category of topological vector spaces and proved that there are two "extreme" ones. A maximal one and a minimal one. The first is "projective" and the latter "injective". Their norms are dual of each other and they coincide once one of the tensor summands is "injective" (which spaces of smooth functions typically are) $\endgroup$ – Adrián González-Pérez Apr 18 '20 at 16:24
  • $\begingroup$ @AdriánGonzález-Pérez: this is very interesting, but I don't really understand what you are writing, I have no familiarity with abstract functional analysis, categories, all of that. If and when you have time and will, please elaborate in an answer. (I hope you are doing well). $\endgroup$ – Giuseppe Negro Apr 18 '20 at 16:41
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    $\begingroup$ I hope your doing well too! No problem. To be fair, this theory is no longer well known outside of the Banach space world since (i have the impression) more general locally convex spaces are out of fashion. Modern books like Ryan's or Diestel's cover the Banach space theory. So, perhaps the reference for the case of distributions is still Grothendieck's thesis. $\endgroup$ – Adrián González-Pérez Apr 19 '20 at 14:56
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Let me elaborate on what i meant.

Preliminaries. Let $X$ and $Y$ be two vector spaces, their (algebraic) tensor product $X \otimes Y$ is the span of all formal combinations $x \otimes y$, with $x \in X$ and $y \in Y$, where the operator "$\otimes$" is assumed to be bilinear. Alternatively, you can think of $X \otimes Y$ in terms of bases as the space spanned by the product of the bases of $X$ and $Y$.

The theory of topological tensor products deals with the following problem: if $X$ and $Y$ are Banach/Frechet spaces, is there a norm/collections of norms on $X \otimes Y$ that is "compatible" with the topologies of $X$ and $Y$? In that case the "completion" (in the Banach case or in Frechet spaces with countable norms) of $X \otimes Y$ with respect to a tensor norm $\alpha$ is denoted by $X \otimes_\alpha Y$. Of those crossed product topologies the coarsest one is denoted by $\epsilon$ (and is "injective") and the finest one is denoted by $\pi$ (and is "projective"). When a Frechet space $E$ satisfies that its injective and projective tensor products coincide it is said to be nuclear. The space $D(\mathbb{R}^n)$ and its dual are both nuclear spaces.

The question: Ok, enought generalities. Given spaces $X$, $Y$ the injective tensor product satisfies that the natural embedding: $$ Y^\ast \otimes_\epsilon X \to \mathcal{B}(Y,X), $$ given by sending $y^\ast \otimes x$ to the operator $z \mapsto \langle y^\ast, z\rangle x$ is an isometry (or a linear homeomorphism in his range in the non-Banach case). The image of the "isometry" above is given by the maps spanned by finite tensors and contained inside the compact operators $\mathcal{B}_0(Y,X)$. The image is exactly the compact operators if either $X$ or $Y$ as the approximation property. Nuclear spaces have the AP.

Now the question-specific part. Approximate the space $D(\mathbb{R}^n;X^\ast)$ using partitions of unity, by simple tensors of the form $$ \sum_{k = 1}^N \psi_k(x) \, x_k^\ast = \sum_{k = 1}^N \psi_k(x) \otimes x_k^\ast, $$ where $x_k^\ast \in X^\ast$ and $\psi_k \in D(\mathbb{R}^n)$. Then, $D(\mathbb{R}^n;X^\ast) = D(\mathbb{R}^n) \otimes_\epsilon X^\ast$. Its dual, by nuclearity of $D$ and reflexivity of $X$, its given by $$ D(\mathbb{R}^n;X^\ast)^\ast = \big( D(\mathbb{R}^n) \otimes_\epsilon X^\ast \big)^\ast = D'(\mathbb{R}^n) \otimes_\pi X^{\ast \ast} = D'(\mathbb{R}^n) \otimes_\epsilon X. $$ But, as we have seen before $D'(\mathbb{R}^n) \otimes_\epsilon X$ is just a subset of $\mathcal{B}(D(\mathbb{R}^n), X)$, the compacts. So, your first definition gives you a larger space and if you want them to be equal you need to change:

$$\tag{1} \mathcal{D}^\star(0, T; X)=\left\{ u\colon \mathcal{D}(0,T)\to X\ \text{linear and } \color{#c00}{\mathrm{continuous}} \right\} $$

by

$$\tag{1} \mathcal{D}^\star(0, T; X)=\left\{ u\colon \mathcal{D}(0,T)\to X\ \text{linear and } \color{#c00}{\mathrm{compact}} \right\} $$

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  • $\begingroup$ Btw, that probably explains why Schwartz choose the definition (1). Definition (2) is more restrictive having an implicit "compactness" assumption. $\endgroup$ – Adrián González-Pérez Apr 20 '20 at 14:32
  • $\begingroup$ This is a fantastic answer. Thanks! $\endgroup$ – Giuseppe Negro Apr 24 '20 at 13:45

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