17
$\begingroup$

Let $X$ be a reflexive Banach space. 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 the space of test functions $\mathcal{D}(0, T)$ 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 we have let $\mathcal{D}(0, T; X^\star)$ denote the space of smooth functions compactly supported in $(0, T)$ and taking values in $X^\star$. We equip this vector space with the obvious analogue of the topology of $\mathcal{D}(0, T)$: namely, 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 $$ \begin{cases} \mathrm{Spt}\phi_n \subset [a, b]\subset (0, T)\\ \left\lVert \frac{d^k \phi_n}{dx^k}-\frac{d^k\phi}{dx^k} \right\rVert_{\infty} \to 0,\quad\forall k\in \mathbb{N}. \end{cases} $$

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

Question. Are those 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.

$\endgroup$
  • 1
    $\begingroup$ This question on Hilbert-space valued distributions is strongly related. $\endgroup$ – Giuseppe Negro Aug 8 '18 at 10:21

Your Answer

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

Browse other questions tagged or ask your own question.