Proving continuity on spaces of distributions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:55:40Z http://mathoverflow.net/feeds/question/111162 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions Proving continuity on spaces of distributions? goci 2012-11-01T15:19:37Z 2012-11-07T08:42:53Z <p>Let $\mathcal{D}'(\Omega)$ be the space of distributions on an open set $\Omega$, and $\mathcal{E}'(\Omega)$ the compactly supported ones.</p> <p>When you have a linear operator $T:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ or $T:\mathcal{E}'(\Omega)\rightarrow\mathcal{E}'(\Omega)$, I often find it easy to prove that convergent sequences get mapped to convergent sequences (in the weak topology), but proving continuity with respect to the weak topology is much harder. Since these spaces are not first-countable (I think), one must work with a local basis around 0.</p> <p>For example, if you have a pseudodifferential operator $P:\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ which is properly supported, then it continuously extends to a (necessarily unique) operator $P:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ and maps $\mathcal{E}'(\Omega)$ continuously into $\mathcal{E}'(\Omega)$. I was able to prove the first statement, but I was able to prove the second statement only in terms of sequential continuity. Note that the weak topology of $\mathcal{E}'(\Omega)$ is not the subspace topology that it inherits from the weak topology of $\mathcal{D}'(\Omega)$ (I think), so this is not trivial.</p> <p>In many textbooks involving distributions, authors don't seem to be careful about this. They give an argument proving sequential continuity and claim that it is continuous in the more general sense. Is there a general scheme for converting sequential continuity arguments to actual continuity arguments for these spaces? Or can you point me to a detailed proof of the aforementioned fact about proper PDOs?</p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111167#111167 Answer by Peter Michor for Proving continuity on spaces of distributions? Peter Michor 2012-11-01T16:01:00Z 2012-11-05T10:30:52Z <p>$\mathcal D(\Omega) = \varinjlim_K \mathcal D(K\subset\Omega)$; it is the direct limit of the space of distributions with support in a fixed compact subset $K\subset \Omega$. Each of these spaces is Frechet. Thus a linear mapping $T:\mathcal D'(K\subset\Omega) \to E$ into any locally convex space $E$ is continuous if it maps convergent sequences to convergent sequences. By the universal property of the direct limit $T$ is then continuous on $\mathcal D'(\Omega)$ also.</p> <p>$\mathcal E(\Omega)$ is itself a Frechet space. </p> <p>Edit: Sorry, I mixed the spaces with their duals, not really awake. Commenters are right. I changed it now, and it is not an answer to the question. </p> <p>SECOND EDIT: Let me try again (after my blunder before) with an explanation, along the lines: The operator is important, locally convex topologies are auxiliary.</p> <p>All spaces in the question are reflexive (for the bornological topology) complete locally convex spaces. For a convenient locally convex space $E$ with (a point separating subspace of the) dual space $E'$ we can consider locally convex topologies on $E$ which are compatible with the duality (all elements of $E'$ are continuous): There is the weakest one $\sigma(E,E')$, and among all of those with the same system of bounded sets (namely the system of weakly bounded sets) there is the strongest one (the bornological one: each bornivorous set is a 0-neighbourhood). If a linear mapping $T:E\to F$ is bounded then it is continuous for the bornological topologies. </p> <p>And now it comes: A linear mapping is bounded if and only if it is bounded on each sequence which is Mackey convergent to 0. (See 5.4 of the ref. below)</p> <p>A sequence $x_n$ is Mackey convergent to 0 in $E$: There exists a sequence $0&lt; t_n \to \infty$ in $\mathbb R$ such that $t_n x_n$ is bounded in $E$.</p> <p>(Added in edit) Proof: Suppose that $T(B)$ is not bounded in $F$ for some bounded $B\subset E$. So there are a semi norm $p$ on $F$ and $x_n\in B$ with $t_n^2 := p(T(x_n)) \to \infty$. Then $p(T(x_n/t_n)) = t_n \to \infty$, but $x_n/t_n$ is Mackey convergent to 0 in $E$. QED.</p> <p>In this sense the question can now be answered: If $T:E\to F$ is sequentially continuous, then it is continuous for the bornological topologies.</p> <p>Reference: Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997 <a href="http://www.mat.univie.ac.at/~michor/apbookh-ams.pdf" rel="nofollow">(pdf)</a> </p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111171#111171 Answer by goci for Proving continuity on spaces of distributions? goci 2012-11-01T16:18:37Z 2012-11-01T16:18:37Z <p>I think the usual definition of a Frechet space is a locally convex toplogical vector space with a complete translation invariant metric. How does one go about constructing such a metric for $\mathcal{D}'(\Omega)$ or $\mathcal{D}'(K)$?</p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111186#111186 Answer by goci for Proving continuity on spaces of distributions? goci 2012-11-01T18:04:52Z 2012-11-01T18:04:52Z <p>I am new to the site, so I can't leave comments yet.</p> <p>Responding to Peter Michor, is it possible that you are confusing $\mathcal{D}'(\Omega)$ with $\mathcal{D}(\Omega)$? Here, $\mathcal{D}(\Omega)$ is the space of test functions, i.e., compactly supported $C^\infty$ functions on $\Omega$. The topology on this is exactly like how you described, the colimit of the Frechet topologies on $\mathcal{D}(K)$. However, $\mathcal{D}(\Omega)$ itself is not Frechet. Everything you said is true for these spaces, and also what you said about $\mathcal{E}(\Omega)=C^\infty(\Omega)$.</p> <p>But I doubt that it is true for the dual spaces $\mathcal{D}'(\Omega)$ and $\mathcal{E}'(\Omega)$. I find it very implausible that these spaces are first-countable.</p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111218#111218 Answer by paul garrett for Proving continuity on spaces of distributions? paul garrett 2012-11-01T21:34:47Z 2012-11-03T00:22:09Z <p>Too long for a comment: it is sometimes true that spaces of distributions, with a strong topology, are colimits of Hilbert or Banach spaces. This is certainly true for the $L^2$ BeppoLevi-Sobolev spaces on the circle, for example. Thus, continuity in that strongish topology is just sequential, using the colimit topology (of Hilbert spaces, not weak topologies). Nevertheless, this does not literally prove that the colimit of weak topologies has that property. True, sequential convergence in the colimit of <em>strong</em> topologies implies convergence in the colimit of weak, but this is not at all what the question refers to as "the typical discussion", I think.</p> <p>Edit: in lieu of other answers... the deconstructionist appraisal of many of the sources that claim continuity while proving only sequential continuity are simply "fudging", "cheating", ... though we can observe that on some occasions a proof of a weaker assertion can be so much clearer that it is worth giving... and, since the harder result is true, <em>not</em> giving a proof of it doesn't fail to establish its truth in the larger world, so... no harm done.</p> <p>Nevertheless, all my own experience indicates that defaulting to sequential whatever is not so carefully considered. Indeed, I'd wager that ... too many ... of the "text" sources that collapse back to sequence arguments do so because there is no interest in, and no facility in, arguing anything stronger.</p> <p>As the questioner observes, in the specific situation of the question there is not a countable local basis... </p> <p>Even more strongly, "sequential arguments" either effortlessly generalize to "net" arguments, so there is no problem, or have pointless limitations that should be observed. Either way, no excuse to neglect this, in my opinion. Nevertheless, of course, the actual agenda of a given writer apparently dictates their degree of attention to technical niceties. (!?!) Perhaps the objection/advice would be to acknowledge what one has literally done, versus what one claims/notes is true/provable. </p> <p>Indeed, back to that point, <em>I</em> am not at all interested in the most general argument, but in the argument that most clearly isolates the causal mechanism, and isolates-away-from irrelevant details of (conceivably very interesting) contexts.</p> <p>Oh, wait, what was the question? :) My "answer" is that, no, there is no general assertion that makes sequential arguments apply to situations where there's no local basis... but that, mercifully, this failing is irrelevant to the truth of most (!?!) of the in-the-literature assertions that seem to not recognize this. No, I do not know any general conversion idea/theorem/fact... but that in many cases a proof-idea would be valid for "nets" (so, general, sufficient), but/and is posed in a needlessly small-vocabulary way out of convenience, or impatience, or ignorance of reality.</p> <p>(As in the case that one's life will not be completely blissful after buying good tires for an automobile, but, nevertheless, it is a wise choice to buy good tires, if one has an auto, and it <em>is</em> true that good tires provide various benefits.)</p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111355#111355 Answer by jbc for Proving continuity on spaces of distributions? jbc 2012-11-03T09:35:55Z 2012-11-05T09:41:31Z <p>The space of distributions of compact support on a manifold and that of distributions on a compact manifold (perhaps with boundary) are not metrisable as has been pointed out here. They are, however, what is called Silva spaces (inductive limits of sequences of Banach spaces with compact interlocking mappings) and these have the desired properties---continuity of linear mappings (even non-linear ones, I think) are determined on sequences----and many more. This was know to the portuguese mathematician J. Sebastiao e Silve (circumflex missing---can't manage it) who created the theory in the 50's. An accessible account can be found in the first volume of Koethe's (problem with an umlaut now) treatise on toplogical vector spaces. The case of the distributions on a non-compact manifold (e.g. the real line) is more intricate. This space is what I called a $DLF$-space (for obvious reasons) in my thesis (unpublished). In this case, however, help is at hand in the form of the concept of partitions of unity for inductive and projective limits of locally convex spaces (de Wilde). This implies that that we can get it as a complemented subspace of a countable product of Slva spaces. I suspect that this suffices to justify the sequential arguments but, to my knowledge, nobody has investiagted this in detail.</p> http://mathoverflow.net/questions/111162/proving-continuity-on-spaces-of-distributions/111643#111643 Answer by Jochen Wengenroth for Proving continuity on spaces of distributions? Jochen Wengenroth 2012-11-06T14:05:38Z 2012-11-07T08:42:53Z <p><em>Every linear mapping $T:\mathcal D'\to \mathcal D'$ which is $\sigma(\mathcal D',\mathcal D)$-sequentially continuous is $\sigma(\mathcal D',\mathcal D)$-continuous.</em></p> <p>Perhaps, this is the same as Peter Michor's answer, but let me try to explain that the major point is the <em>bornologicity</em> of $\mathcal D'$ endowed with the strong topology $\beta (\mathcal D',\mathcal D)$ (uniform convergence on the bounded subsets of $\mathcal D$) which follows e.g. from a beautiful theorem of Laurent Schwartz (<em>the strong dual a every complete Schwartz space is bornological</em>).</p> <p>Moreover, since $\mathcal D$ is barrelled, the strong and the weak-$*$ dual have the same bounded sets.</p> <p>If now $T: (\mathcal D',\sigma(\mathcal D',\mathcal D)) \to (\mathcal D',\sigma(\mathcal D',\mathcal D))$ is sequentially continuous then every bounded sequence of $\mathcal D'$ is mapped to a bounded sequence and hence $T$ is bounded on bounded sets of $\mathcal D'$. As $(\mathcal D',\beta(\mathcal D',\mathcal D))$ is bornological $T:(\mathcal D',\beta(\mathcal D',\mathcal D)) \to (\mathcal D',\beta(\mathcal D',\mathcal D))$ is continuous and therefore $T:(\mathcal D',\sigma(\mathcal D',\mathcal D''))\to (\mathcal D',\sigma(\mathcal D',\mathcal D''))$ is continuous. By reflexivity, you get the desired weak-$*$ continuity.</p> <p>In general, you are perfectly right with your skepticism against careless arguments involving sequential continuity. For example, it may very well be that for a linear partial differential operator with constant coefficients $T=P(D): \mathcal D'(\Omega) \to \mathcal D'(\Omega)$ the inverse of $P(-D): \mathcal D(\Omega) \to \mathrm{Im} (P(-D))$ is sequentially continuous without being continuous (the former condition is equivalent to $P$-convexity, the latter to strong $P$-convexity)</p>