Two possible definitions of "vector-valued distribution"

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.

Solution 1:

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

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