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The revised question

After more reflection on the problem, I might have found the answer by myself. Let $U$ be an open subset of $M$, irrespective of whether it has a boundary or not. Let

$$\mathcal D _\infty ' = \{ u \in \mathcal D ' (U) \mid \forall \varepsilon >0 \ \exists K _\varepsilon \ \text{compact such that} |\langle u, \varphi \rangle| < \varepsilon \\ \forall \varphi \in \mathcal D (U) \ \text{with supp} \ \varphi \cap K _\varepsilon = \emptyset \ \text{and} \ \sup \limits _\alpha \sup \limits _U |\partial _\alpha \varphi| \le 1 \} .$$

Note that if $u \in \mathcal D _\infty ' (M)$, then $\partial _\alpha u \in \mathcal D _\infty ' (M) \forall \alpha$ and that $\mathcal D _\infty ' (M)$ contains all the smooth functions $f$ such that $\partial _\alpha f$ vanishes at infinity $\forall \alpha$. It also contains all the compactly-supported distributions, which suggests that the given definition is a good one.

A new interesting question arises, though: does $\mathcal D _\infty ' (M)$ have a nice predual? In order for the question to make sense, I should specify the topology on $\mathcal D ' (M)$ (the weak* or the strong one). I do not know which one to choose, I guess the question could be reformulated as: does any of these two topologies (or any other one) give a nice predual?

The original question

If $U$ is an open subset in a Riemannian manifold $M$ (as a first step, $M = \Bbb R^n$ should suffice) with the boundary $\partial U$ a submanifold in $M$, does there exist something as "the space of distributions from $\mathcal D ' (U)$ that vanish on $\partial U$"?

The closest thing that comes to mind is the space of those distributions with the support (not necessarily compact) included in $U$, but this is needlessly restrictive for the problem (see rationale below).

The other thing that comes to mind is that this space should be the distributional-theoretic counterpart of the Sobolev spaces $W ^{k,p} _0 (U)$. If a "trace operator" $T : \mathcal D ' (U) \to \mathcal D ' (\partial U)$ existed, then this space could be defined as the kernel of $T$.

Rationale

Suppose that we want to show the uniqueness of the solution of the heat equation in the space of smooth functions that vanish on the boundary. The argument is a classic: under the assumption that the solution $u$ is real and $u(0, \cdot) = 0$, if $I(t) = \int u^2$ then $I'(t) = \int 2 \partial _t u \ u = \int 2 \Delta u \ u = -2 \int \| \nabla u \| ^2 \le 0$, so $I = I(0) = 0$, so $u = 0$. One also sees that $u$ cannot be complex since both its real and ist imaginary part would satisfy the equation and thus be $0$.

The core thing used above is the integration by parts in which $\int \Delta (u^2) = 0$ because $u$ vanishes on the boundary. I wonder whether this argument can be mimicked if instead of smooth functions one used "distributions vanishing on the boundary", provided that such a thing exist. Of course, I wouldn't know what to replace all the products of distributions that would show up with, but this is a different matter.

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    $\begingroup$ The usual thing to do is to take some space of compactly supported distributions and take its closure / completion in an appropriate topology. For instance, that's one way of defining the Sobolev spaces $W^{k,p}_0(U)$. $\endgroup$
    – Nate Eldredge
    Commented Sep 22, 2015 at 19:22
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    $\begingroup$ @NateEldredge: Plausible, but what would be "the appropriate topology"? And why take distributions with compact support? For Sobolev spaces, one starts with functions of compact support, not with generalized functions of compact (essential) support. Related to my question (should I ask it separately?), could one construct a theory of distributions, parallel to the usual one, but using the bounded smooth functions that vanish on the boundary (and drop the compact subsets from the seminorms) as test functions? $\endgroup$
    – Alex M.
    Commented Sep 22, 2015 at 19:45
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    $\begingroup$ About the modified definition: $\{\,u\in\mathcal D'(M):$ $\forall\varepsilon>0\, \exists K\,\text{compact such that}\,|\langle u,\varphi\rangle|<\varepsilon\ \forall\varphi\in\mathcal D(M)\text{ with supp}\,\varphi\cap K=\emptyset\,\} $ . If it is to be understood " $\forall\varepsilon\ \exists K\ \forall\varphi$ ..." with implicitly understanding ($*$) $K=K_\varepsilon\subset U\,$, then it is just the set of distributions having a compact support included in $U\,$. Without ($*$) it is just the set of distributions having a compact support. $\endgroup$
    – TaQ
    Commented Sep 26, 2015 at 9:09
  • $\begingroup$ (cnt.) Proof: Take $\varepsilon=1$ and deduce that $|\langle u, \varphi \rangle|=0\,$. $\endgroup$
    – TaQ
    Commented Sep 26, 2015 at 9:11
  • $\begingroup$ @TaQ: You are right, thank you. My definition couldn't have been useful (it was too rigid) because of its linearity in $\varphi$: if $| \langle u, \varphi \rangle | < \varepsilon$, then one cannot have $| \langle u, c \varphi \rangle | < \varepsilon \ \forall c$ unless $\text{supp} \ u$ is compact, as you note. Please note that I have altered the definition by including a boundedness condition on $\varphi$ and its derivatives that fixes this annoyance. $\endgroup$
    – Alex M.
    Commented Sep 26, 2015 at 11:20

3 Answers 3

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Here's a suggestion. For simplicity, assume that $\partial U$ is the zero level set of a scalar function $f$, for which $0$ is a regular value. Let $C^\infty_b(U)$ be the space of smooth functions in $U$ whose derivatives of all orders extend continuously to $\overline{U}$. Make it into a Fréchet space by using sup-seminorms over compacts in $\overline{U}$. Let $C^\infty_f(U) \subset C^\infty(U)$ be the pre-image of $C^\infty_b(U)$ with respect to the multiplication-by-$f$ map, $C^\infty(U) \stackrel{f\cdot}{\longrightarrow} C^\infty(U) \supset C^\infty_b(U)$. Use the initial topology with respect to the map $C^\infty_f(U) \stackrel{f\cdot}{\longrightarrow} C^\infty_b(U)$. Let $\mathcal{D}_f(U) = \bigcup_{k=1,2,\ldots} C^\infty_{f^k}(U)$ and give it the inductive limit topology.

You can interpret $\mathcal{D}_f(U)$ as the set of smooth functions on $U$ that grow at most polynomially near $\partial U$. Then, I think $\mathcal{D}'_f(U)$ will be your desired space of distributions vanishing (to arbitrary order) on $\partial U$.

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    $\begingroup$ I see what you are doing: you are mimicking the construction of the space of tempered distributions, replacing the monomials $x^\alpha$ ($\alpha$ multiindex) by $f^k$. Ingenious! Please note, though, that I have edited my post adding my own attempt at answering it and a completely new question. Would you please take a look at the new version? Thank you. $\endgroup$
    – Alex M.
    Commented Sep 25, 2015 at 18:10
  • $\begingroup$ @AlexM., I'm afraid that after five minutes of staring at your definition, I still can't parse it. Sorry. $\endgroup$
    – Igor Khavkine
    Commented Sep 25, 2015 at 21:53
  • $\begingroup$ I suppose the fact that you chose $\mathcal{D}_f(U)$ instead of $\mathcal{D}(\partial U)$ as notation means that this space depends on the chosen function $f$ ? Can we have some notion of vanishing at the boundary that is independent of such arbitrary choices? $\endgroup$
    – Johannes Hahn
    Commented Oct 1, 2015 at 17:59
  • $\begingroup$ @JohannesHahn, the dependence on $f$ is rather loose. Any function that vanishes only on the boundary and has $0$ as a regular value would give the same definition. However, replacing $f$ by say $f^{1/2}$ or $e^{-1/f^2}$ would give different definitions, I think. $\endgroup$
    – Igor Khavkine
    Commented Oct 1, 2015 at 20:06
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If $P$ is a linear partial differential operator in $\overline{U}$ and if the boundary $\partial U$ is non-characteristic for $P$, then every extendible distribution solution $u$ of $Pu=0$ in $U$ belongs, in a boundary collar $0\leq x<\varepsilon$, to $C_x^\infty([0,\varepsilon[,\mathcal{D}'(\partial U))$. This implies that boundary traces are defined by evaluating at $x=0$. (Here $x$ is a defining function; the same as $f$ in Igor Khavkine's answer.) This result is due to Peetre, and it can be found under the label partial hypoellipticity at the boundary in Hörmander's 1963 book. Melrose's microlocal theory of boundary problems (Acta Math. (1981)) gives a class of distributions on a manifold with boundary such that a canonical restriction (or boundary trace) map is defined; see Proposition 18.3.21 in Hörmander's volume III. Peetre's theorem is given an invariant generalization by Melrose.

So, regarding your original question, there is a natural space of distributions vanishing (to first order) at the boundary, the null space of Melrose's restriction map. As for your rationale: the space-time boundary is non-characteristic for the heat operator. So boundary values (and their vanishing or not) are well-defined. As for the boundary terms in integration by parts it may be useful to look at the theory of the Calderon projector for boundary problems.

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  • $\begingroup$ I somehow got the impression that OP's main motivation for getting a "suitable space" of distributions was to be able to prove some kind of uniqueness result in that space for an initial value problem for the heat equation or possibly "something related". Are you suggesting for this purpose some space whose elements locally near the boundary are like those in $C^\infty([0,1[,\mathscr D'(\mathbb R^N))\,\,$, or in $\mathscr H^{loc}_{(s,t)}(\overline{\mathbb R}{}^n_+)$ which appear in Hörmander's book? $\endgroup$
    – TaQ
    Commented Oct 4, 2015 at 8:46
  • $\begingroup$ Hörmander's book, the link is too long for one comment! By Tikhonov's result neither of these seem to be suitable for this purpose. $\endgroup$
    – TaQ
    Commented Oct 4, 2015 at 8:48
  • $\begingroup$ I understand the OP as being about justification of partial integration and the vanishing of boundary terms. This issue arises in the uniqueness argument sketched. Of course, by Tikhonov's theorem there is no uniqueness without some growth condition, but I think the OP was not about this. $\endgroup$
    – user80744
    Commented Oct 4, 2015 at 13:17
  • $\begingroup$ So, the basic (interpretation) problem is the matter of which one of the following two being the OP's main concern: 1) Suppose that we want to show the uniqueness of the solution ..., or 2) I wonder whether this argument can be mimicked if ... . You seem to think it is the latter, and to me it seems to be the former. $\endgroup$
    – TaQ
    Commented Oct 4, 2015 at 13:55
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$\def\bbR{\mathbb R}\def\bbN{\mathbb N}\def\sp{\kern.4mm}$Working locally in a manifold, one can transport the problem to $\bbR\times\bbR^N$ for some $N\in\bbN$ if $n\ge 2\,$. Then a distribution on $\Omega=\bbR^+\kern-.9mm\times\bbR^N$ having boundary values zero on $\{\sp 0\sp\}\times\bbR^N$ can be defined to be any $T$ in $\mathscr D'(\Omega)$ with the following property: There is a function $c:\bbR^+\to\mathscr D_\sigma'(\bbR^n)$ where the latter space is equipped with the weak$^*$ topology, with limit zero at $0\sp$, and such that $\bbR^+\owns t\mapsto c(t)\,(\varphi(\sp t\sp,\cdot\sp))$ being $L ^1(\bbR^+)$ with $T(\varphi)=\int_{\,\bbR^+}c(t)\,(\varphi(\sp t\sp,\cdot\sp))\,\mathrm d\sp t$ hold for all $\varphi$ in $\mathscr D(\Omega)\sp$. This, however, is only one possibility, and whether it is usefull or not depends on the context in which one is working.

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  • $\begingroup$ How is your suggestion related to the one of Igor Khavkine's? $\endgroup$
    – Jochen Wengenroth
    Commented Sep 24, 2015 at 6:18
  • $\begingroup$ What do you mean by the notation $\mathcal D ' _\sigma$? $\endgroup$
    – Alex M.
    Commented Sep 24, 2015 at 7:12
  • $\begingroup$ @Jochen Wengenroth: Though not having checked the details, it seems to me that neither of them implies the other but it would not be hard to give a third example which is a weaker concept than both of those, i.e. is implied by both ones. To Alex M.: $\mathscr D_\sigma'$ denotes the weak$^*$ topological dual of $\mathscr D\sp$, i.e. has the topology of pointwise convergence. BTW instead of $\mathscr D_\sigma'(\bbR^n)$ there should be $\mathscr D_\sigma'(\bbR^N)\sp$. I'll fix this later making all the required edits at the same time in case there appears to be need for others. $\endgroup$
    – TaQ
    Commented Sep 24, 2015 at 17:47

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