2

Let $\Omega \subseteq \mathbb R^n$ be an open convex set and $T \in D'(\Omega)$. Is it possible to extend $T$ to a distribution in $D'(\mathbb R^n)$?

I've been looking in some texts (Rudin and this notes of Thierry Ramond) but I haven't found anything.

Gonzalo Benavides
  • 981

1 Answers1

3

(Edited:) Yes and no. For example, $u=\sum_{n=1}^\infty \delta_{1/n}$ is a perfectly fine distribution on the open interval $(0,1)$, but does not extend _by_that_formula_. It does have an extension, however! So this may raise the issue of whether we really do or don't want some further auxiliary features of an extension...

EDIT, here I mean $\delta_a(\varphi)=\varphi(a)$, Dirac delta.

EDIT-EDIT: there exist test functions $\varphi$ on $\mathbb R$ for which that sum does not converge: take any $\varphi$ with $\varphi(0)\not=0$, although for $\varphi$ a test function on $(0,1)$ its support is necessarily compact not including $0$, so the indicated sum is finite (and, further, we can prove continuity in the relevant topology).

EDIT-EDIT-EDIT: and, as @Bob noted in a comment, the mere fact that the indicated formula gives something divergent is not complete proof that there is no extension whatsoever. In fact, my purported example fails in this regard, since $\lim_N \sum_{n=1}^N (\delta_{1/n}-\delta_{-1/n}$ does extend the given distribution, for the usual reasons that Cauchy principal value integrals $\lim_N \int_{|x|>1/N} {f(x)\over x}\;dx$ do converge.s For that matter, the distributions $1/|x|^s$ and ${\mathrm sgn}(x)/|x|^s$ are not literally $L^1$ for $\Re(s)>> 1$, but there are meromorphic continuations, nevertheless.

Ok, so, still, the (unsurprising) answer is that, no, not everything extends in a reasonable fashion, such as degrees of homogeneity.

But, yes, a continuous linear functional on some subspace $W$ of the space $V$ of all test functions (with the subspace topology), extends uniquely, by continuity to the closure $\overline{W}$ of $W$. Then use Hahn-Banach to extend (randomly...) to the whole space of test functions.

So, again, without any further conditions, everything extends, but not in a necessarily useful or satisfying fashion.

If, for example, we ask for an extension of $u(\varphi)=\int_{\mathbb R} {\varphi(x)\over x}\;dx$ that ALSO PRESERVES HOMOGENEITY, then we find that there is no such (either via the snake lemma or an ad-hoc recap of the argument for it...), because of the existence of Dirac $\delta$, and the long exact sequence.

paul garrett
  • 52,465
  • Do you mean $\langle u,\varphi \rangle= \sum_{n=1}^\infty \varphi(\frac{1}{n})$? – Gonzalo Benavides Jul 01 '18 at 21:49
  • Yes. Edited to reflect the intent of my notation. – paul garrett Jul 01 '18 at 21:51
  • Isn't $v=\sum_{m \in \mathbb Z} \sum_{n=1}^\infty \delta_{m+1/n} $ an extension to $D'(\mathbb R)?$ – Gonzalo Benavides Jul 01 '18 at 21:52
  • 1
    Your $\nu$ would be an extension, if the sum were guaranteed to converge, to begin with, but it does not, etc. – paul garrett Jul 01 '18 at 21:55
  • Can you please expand why, if such an extension of $T$ exists, it should act on the generic test function of $\mathbb{R}$ with the same formula that defines it on $(0,1)$? Probably it's obvious, but here is midnight, so... :) – Bob Jul 01 '18 at 22:05
  • 1
    @Bob, you are indeed correct that there is something more to show, as you say, that just because the indicated formula fails does not imply that there is no extension. I will add something... – paul garrett Jul 01 '18 at 22:12