User delio mugnolo - MathOverflow

Interior regularity for elliptic equations

2013-02-20T17:18:10Z

The monograph "Non-Homogeneous Boundary Value Problems and Applications" by Lions and Magenes is infamous for developing a truly extensive regularity theory for elliptic problems on domains, and for doing so under amazingly restrictive assumptions on the regularity of the boundary - basically, the domain is always assumed to be $C^\infty$.

Here comes one of their classical results, obtained by combining Theorems 2.7.3 and 2.7.4 in their first volume, when specialized to the case I am interested in:

Theorem. The trace operator $\gamma$ is bounded from $$ D^\frac{1}{2}(\Omega):=\left\lbrace u\in H^\frac{1}{2}(\Omega): \Delta u\in \Xi^{-\frac{3}{2}}(\Omega)\right\rbrace $$ to $L^2(\partial \Omega)$, and furthermore $$\begin{pmatrix} \Delta \atop \gamma\end{pmatrix} $$ is an isomorphism from $D^\frac{1}{2}(\Omega)$ to $\Xi^{-\frac{3}{2}}(\Omega)\times L^2(\partial \Omega)$.

(Here $\Xi^{-\frac{3}{2}}(\Omega)$ is a rather ugly interpolation space, which however is nice enough to contain $L^2(\Omega)$).

In particular, it follows that

Corollary. $u\in H^\frac{1}{2}(\Omega)$ whenever $u$ satisfies $\Delta u=f$ for some $f\in L^2(\Omega)$.

Many results of Lions-Magenes' have been extended to the case of $C^{1,1}$-domains, or even to general convex bounded domains, most notably in Grisvard's "Elliptic problems in nonsmooth domains", but I was not able to find an extension of the above Corollary. What I am interested in is simply the case of the hypercube $\Omega:=(0,1)^N$, that is I am asking the following

Question. Let $u$ solve $\Delta u=f$ for some $f\in L^2\left((0,1)^N\right)$. Is it true that $u\in H^\frac{1}{2}\left((0,1)^N\right)$?

radon-nikodým property of $\ell^\infty$

2013-05-07T17:41:28Z

I am wondering whether $\ell^\infty(\mathbb N)$ has the Radon-Nikodým property. Of course $\ell^1(\mathbb N)$ does, but I was unable to find out whether (e.g.) duals of spaces with the R-N property have the R-N property themselves.

Thank you in advance.

UPDATE: A Banach space is Asplund if and only if its dual has the Radon-Nikodým property. On the other hand, a separable space is Asplund if and only if its dual is separable, too. This rules out the possibility that $\ell^\infty(\mathbb N)$ has the R-N property. But is there any other (more) elementary argument for this assertion?

Elements of the history of mathematics

2013-01-22T10:19:53Z

Is it known who actually wrote Bourbaki's Elements of the History of Mathematics?

projection of sobolev spaces onto cones

2013-03-19T11:29:51Z

Consider the Sobolev space $W^{k,p}(\Omega)$ for $k\in \mathbb N$, $p\in [1,\infty]$ and some open domain $\Omega\subset \mathbb R^n$ $^*$. Then it is known that $W^{k,p}(\Omega)$ is an ordered Banach space, and indeed a lattice-ordered Banach space if $k=1$, but not a Banach lattice because the norm is not monotone on the positive cone $W^{k,p}(\Omega)_+$. Furthermore, it is known that in the reflexive range $W^{k,p}(\Omega)$ is isomorphic, but in general not lattice isomorphic to an $L^p$ space. It is also clear that the positive cone of $W^{k,p}(\Omega)$ is a closed subset, hence at least in the reflexive range each element of $W^{k,p}(\Omega)$ can be projected onto $W^{k,p}(\Omega)_+$.

That said:

Does anybody have an idea of how said orthogonal projection looks like?

$^*$ I would be happy already with an answer in the case of $n=1$, $k=1$, $p=2$, $\Omega$ bounded.

Answer by Delio Mugnolo for H_0^1 and C_infinity on the interior, does that imply classical limit is 0 on the boundary?

2013-03-21T16:21:25Z

Yes. By Theorem 9.17 in Brezis' book, let $\Omega$ be just a bit regular$^*$ and let $u\in W^{1,p}(\Omega)\cap C(\bar{\Omega})$. Then $u\in W^{1,p}_0(\Omega)$ if and only if $u(z)=0$ for all $z\in \partial \Omega$.

$^*$ $C^1$ will do, but even less regularity might be really necessary, I guess.

Answer by Delio Mugnolo for Manifold-Valued Sobolev Spaces

2013-03-19T11:39:32Z

A (slightly) alternative approach to that suggested by Piero D'Ancona, and indeed quite similar to that proposed by Alexander Shamov, is to consider a partition of unity associated with a given atlas and define $H^1(M)$-functions locally wrt to said partition of unity. You can find a more detailed explanation of this approach in §1.7.3 in the first volume of the monograph by Lions-Magenes.

origin of the notion of "network" in graph theory

2013-03-11T21:50:23Z

In current graph theory, a "network" is a precisely defined object: It is a directed graph associated with a function, the "capacity", which is defined on the edge set and has certain specific properties, and for which "flows" - other functions with yet other properties - may be defined.

However, "network" is also a common word. The similarity between everyday's life "networks" (especially electric ones) is overwhelming, and it is not surprising: there is a long history of interplays between graph theory and electric engineering, probably beginning with a famous paper by Kirchhoff (but please correct me if there are even earlier connections). I have the feeling - but again, I might be wrong - that (electric) networks first made it into pure graph theory through the articles of Brooks, Smith, Stone and Tutte, and in particular through The dissection of rectangles into squares (1940); but they still used the word quite informally.

My question is:

When was this similarity formalized? When did networks become more than merely a metaphor or a source of heuristics and assume today's precise definition?

I have an upper bound: in both famous 1956 papers on the Max-Flow-Min-Cut theorem (Ford-Fulkerson and Elias-Feinstein-Shannon), the definition is given already quite clearly, if casually.

Not quite sure whether this question can have an answer at all, as the boundaries are clearly fluid.

Answer by Delio Mugnolo for Limit of a function in a weighted Sobolev space

2013-03-12T21:00:19Z

Not quite sure. If you are in $H^{2-s}(\mathbb R^3)$ you can embed in $W^{1,\frac{6}{1+2s}}(\mathbb R^3)$, and from there you can get into $C^{0,\alpha}({\mathbb R}^3)$ by Morrey's inequality, for all $\alpha,s\in (0,1)$ such that $s+\alpha=\frac12$. So, I would say there's plenty of space for you beneath $H^2(\mathbb R^3)$ to still get a continuous function in the end.

Applications of line graphs

2013-03-07T07:52:10Z

I am trying to collect a few examples of applications of line graphs in sciences other than mathematics. To be more precise: I am thinking of models where there is a clear conceptual added value in switching the paradigm from a description focused on agents (nodes) to a description focused on relations (edges).

I know: put it like this it sounds very much like I am thinking of anthropological/sociological models, and indeed they are the most natural I could come up with$^{*}$, but I do believe that similar ideas appear elsewhere, too, modulo considering graph operations/decorations that allow for more flexible descriptions.

Inter/transdisciplinary examples would be particularly appreciated.

$^{*}$ It is not by chance that line graphs were so much pushed by Frank Harary after all, of all possible graph theorists :)

existence of a minimum for a convex functional on a non-reflexive space

2013-03-06T06:53:20Z

Let $X$ be a Banach space; $K\subset X$ nonempty, closed and convex; and $f:K\to \mathbb R$ lower semicontinuous, convex functional. Let also $f$ be coercive, i.e., $f(x)\to +\infty$ as $\|x\|\to +\infty$.

Now, it is well-known that:

If $X$ is reflexive, then $f$ has a minimum.

The proof goes essentially like this: One takes a sequence $(x_n)$ such that $(f(x_n))\to \inf_{x\in K} f(x)$, which is then necessarily bounded and hence, due to reflexivity of $X$, also weakly convergent, say to $x_0$. Then Hahn-Banach yields that $x_0$ is actually in $K$, and then lower semicontinuity implies that $f(x_0)\le \inf_{x\in K} f(x)$.

My question:

Does $f$ have a minimum if $X$ is merely a separable dual space?

I do not see how the above proof could be modified: On one hand one may still use Banach-Alaoglu to find a weak*-limit of the sequence. But then I do not see how to conclude.

I am interested at minimization of a certain functional in $\ell^1$. Thanks, any help will be appreciated.

Answer by Delio Mugnolo for Compactness of Sobolev embedding for domains of finite measure

2013-03-03T21:58:36Z

You can find a rather general answer in the book of Adams and Fournier, Theorem 6.16. You do have the embedding you wish, under a mild measure theoretical assumption.

power bounded adjacency matrices

2013-02-27T22:11:39Z

A bounded linear operator $T$ on a Banach space $X$ is called power bounded if $\|T^k\|\le M$ for some $M>0$ and all $k\in \mathbb N$.

A classical result of Lorch says that if $X$ is reflexive, then each power bounded operator $T$ on it is mean ergodic, meaning that the sequence $$ C_n x:= \frac{1}{n} \sum_{k=1}^n T^k x,\qquad n=1,2,\ldots, $$ of its Cesaro sums converges for all $x\in X$. (Actually, also the converse implication holds).

Now, take a simple digraph (possibly infinite, but uniformly locally finite) and consider its adjacency matrix $A$. It would be nice to know whether $(A^n)_{n\in \mathbb N}$ converges, but this seems to be usually hopeless. Hence an more pragmatic approach would be to investigate whether $A$ is at least mean ergodic. By the above result by Lorch, it would be sufficient to show power boundedness.

Hence my question:

Is it possibly known whether $A$ is power bounded for certain classes of digraphs?

EDIT: As Robert Israel points out in his answer, the question is trivial if the graph is not oriented. Therefore, I have restricted the question to digraphs.

Answer by Delio Mugnolo for real symmetric matrix has real eigenvalues - elementary proof

2013-02-26T21:34:00Z

Just found in Godsil-Royle's Algebraic graph theory: One first proves that two eigenvectors associated with two different eigenvalues are necessarily orthogonal to each other (pretty standard), then observes that if $u$ is eigenvector associated with eigenvalue $\lambda$, then $\bar u$ is eigenvector associated with eigenvalue $\bar\lambda$. Now the eigenvalues $\lambda,\bar\lambda$ cannot be different, for otherwise by the above observation $0=u^T u=\|u\|^2$ although $u\not=0$.

(It does contain complex numbers, but is still amazingly straightforward).

Answer by Delio Mugnolo for What are some interesting almost equitable partitions which are not equitable?

2013-02-21T06:53:45Z

Well, my favourite example: There has been some research on the Laplacians on infinite graphs lately, say here and here. The authors deduce a few decompositions for the Laplacians on (usually infinite) trees and "perturbed trees" under certain symmetry assumptions, and they even draw a lot of picture to show them more clearly. You will easily see that most of their graphs have natural a.e. partitions (usually induced by the distance to the root) which are not equitable.

algorithmic almost equitable partitioning

2013-02-19T18:55:50Z

Let $G$ be a graph -- possibly infinite, but I will be glad to learn a positive result even in the finite case. Then the trivial partition (i.e., one cell coinciding with the whole $G$) is clearly almost equitable (a definition can be found here, for example), but this is of course very coarse. Another trivial almost equitable partition is instead much too fine: it is the partition consisting of singletons only - this is indeed even equitable.

My question: Is there any way of algorithmically constructing further (non trivial) almost equitable partitions -- that is, almost equitable partitions with a larger number of smaller cells, but not too small? Or almost equitable partitions with a smaller number of larger cells, but not too large?

EDIT: As Aaron Meyerowitz suggests in the comments, the mathoverflow entry linked above does in fact call almost equitable what is usually called equitable. So, let me for reference write down here what is the correct definition, even in the general case of a weighted graph: Let $G$ be a (possibly infinite) graph with node set $V$, where each edge $e=(v,w)$ has a weight $\mu_{vw}\in (0,\infty)$ and each node $v$ has a weight $\nu_v\in (0,\infty)$. Given a subset $W\subset V$ and a $v\in V$, one denotes by $d_W(v)$ the weighted degree of $v$ in $W$, i.e., $$ d_W(v):=\frac{1}{\nu_v} \sum_{w\in V \hbox{ s.t. }w\sim v} \mu_{vw}. $$ Then, a (possibly infinite) partition $(V_i)_{i\in I}$ of $V$ is called almost equitable if for all $i,j\in I$, $i\neq j$, there is a number $c_{ij}$ such that $d_{V_j}(v)=c_{ij}$ for all $v\in V_i$.

Answer by Delio Mugnolo for On exponential formula

2013-02-12T10:19:03Z

In general, this is not true (think of the first derivative acting on $L^1(\mathbb R)$).

However, this might be true in some special cases: most notably, by the spectral theorem, if $A$ is a self-adjoint operator. Namely, in that case you can simultaneously diagonalize all resolvent operators, whose eigenvalues you explicitly know - so that you are essentially back to the scalar case. More generally, this is true for some (large) class of sectorial operators on Hilbert spaces, cf. Thm. 5.1 in Operator-Norm Approximation of Semigroups by Quasi-sectorial Contractions by Cachia-Zagrebnov, JFA 2001. Their answer has been extended to general Banach spaces in Some remarks on operator-norm convergence of the Trotter product formula on Banach spaces by Blunck, JFA 2002, see also Corollary 3.6 in Euler’s Exponential Formula for Semigroups by Cachia, Semigroup Forum 2004, if you are interested in an error estimate as well.

Answer by Delio Mugnolo for Generator of a generated $C_0$ semigroup.

2013-01-29T16:36:47Z

Both the semigroup law and strong continuity follow directly from the semigroup law satisfied by $S_t$ and ${S_t}^*$ and by their strong continuity (here it is fundamental that $\mathcal H$ is a Hilbert space, strong continuity of the adjoint semigroup being false in the case of general Banach spaces). Also determining the generator is just an easy exercise in differentiation of a vector-valued function of one real variable (applying the chain rule).

Answer by Delio Mugnolo for How to define Laplacian on $L_2$

2013-01-28T22:35:49Z

Let me comment a bit on Daniel Spector's answer. Yet two further possible alternative approaches are based on the notion of weak derivative and derivative almost everywhere. For the sake of simplicity, I only discuss the case of functions defined over $(0,1)$.

1) An $L^2(0,1)$ function $f:(0,1)\to \mathbb R$ is said to be weakly differentiable if there exists a $g\in L^2(0,1)$ such that $$ \int_0^1 f(x)h'(x)dx=-\int_0^1 g(x)h(x)dx\qquad \hbox{for all }h\in C^1[0,1], $$ and in this case $g$ is uniquely defined and called the weak derivative of $f$. So far so good. Now, you can define in the very same way the second (weak) derivative $f''$ of $f$ and define the (weak) Laplacian as the mapping $f\mapsto f''$.

2) If a function is Lipschitz continuous, then by Rademacher's theorem it is differentiable almost everywhere. Hence, a Lipschitz continuous functions that is differentiable with a Lipschitz continuous derivative is twice differentiable a.e. It is then natural to associate with each function its second derivative. This is possible at least a.e., giving rise to another notion of Laplacian that is well-behaved if one wants to solve integrated versions of elliptic problems.

Answer by Delio Mugnolo for Convergence of Dirichlet Forms

2013-01-27T19:33:54Z

Kato shows in §VI.3 of his book "Perturbation theory for linear operators" that in particular if a sequence of Dirichlet forms (or, more generally, of bounded closed sesquilinear forms) converges to a limiting form $a$ (in a certain quite natural sense), then the associated operators converge to the operator associated with $a$ in the norm resolvent sense. This in turn implies norm convergence of the associated semigroups, e.g. because of the representation of the semigroups via the backward Euler scheme applied to the resolvents. I guess this answers your question about the diffusion processes associated with the forms.

on an inequality of Brezis-Lieb

2013-01-23T18:55:04Z

In their 1983 JFA-paper Brezis and Lieb have shown, among many other things, a Poincaré-type inequality: in the case of a harmonic function $f$ on a bounded domain $\Omega$, their inequality ((3.14) in the paper) states that the $L^2(\Omega)$-norm of $f$ can be estimated by the $L^2(\partial \Omega)$-norm of its trace on $\partial \Omega$ (times a constant only depending on $\Omega$). My question: Is it possible to reverse this inequality, viz estimating the $L^2(\partial \Omega)$-norm of the trace of $f$ by the $L^2(\Omega)$-norm.

This is indeed possible if $\Omega\subset R$ ($\Omega$ an interval), but this clearly follows simply from the fact that on an interval both the space of harmonic functions and the space of their boundary values are $2$-dimensional (then using equivalence of any two norms on a finite dimensional space). I have no clue whether this may extend to higher dimensions - in fact, I am pessimistic.

Thanks in advance.

Answer by Delio Mugnolo for Multigraphs and Social Network Analysis

2012-12-19T21:26:30Z

To comment a bit on D.S. Stones' comment: I think that negative edges are very natural if you think of your graphs as oriented. In fact, this kind of objects are the very fundament of so-called cognitive maps, which have been around for quite a while now - although I cannot really judge how popular they still are, nowadays. http://sipi.usc.edu/~kosko/FCM.pdf

Answer by Delio Mugnolo for Symmetric Operators Robin Boundary Conditions

2012-11-21T19:52:08Z

just three words: integration by parts.

Answer by Delio Mugnolo for Bound on Hilbert transform

2012-11-09T09:32:49Z

I am not sure this is what is commonly known under the name "Hilbert transform" - which by the way is known to be bounded on $L^p$ iff $p\in (1,\infty)$.

Answer by Delio Mugnolo for Exponential stability in nonlinear differential equations

2012-11-03T01:53:05Z

I don't understand if you are thinking of the finite dimensional case only; but Chapter 9 in Lunardi's monograph is a rich source of information about exponential convergence of solutions to nonlinear equations.

a variation on the theory of equitable partitions for graphs

2012-09-19T23:31:50Z

Assume you have a graph with an equitable partition with respect to cells $V_1,\ldots,V_n$. Accordingly, you can take the cellwise average value of a function on the node set - in other words, the projection of the node space onto the susbpace of cellwise constant functions. Let us call $P$ this projector.

Then, it is well-known (see e.g. Bollobas' book, Prop. VIII.3.15; or Brouwer-Haemer's book, §2.3) that there is a matrix $C$ such that $P$ intertwines with $A$ and $C$, i.e., $AP=PC$, where $A$ is the adjacency matrix of the graph; in fact, $C$ can be investigated as the adjacency matrix of a certain auxiliary "quotient" graph, with certain nice connections between the spectra of $A$ and $C$.

Now, what I'd like to know is what happens if we consider $(I-P)$ instead of $P$, or - if you prefer - the projector onto the null space of $P$, instead of its range. Is there a matrix D such that $(I-P)$ intertwines with $A$ and $D$? Can $D$ be interpreted as the adjacency matrix of a certain auxiliary graph, again?

(If necessary, in the above question you can gladly replace the adjacency matrix by the discrete laplacian, the normalized laplacian, or the signless laplacian).

Answer by Delio Mugnolo for Dual space pairing question (Sobolev space, Bochner space)

2012-10-28T19:47:59Z

It's more than just v(w), right?

No, it's not, as far as I know. It is just a (historically motivated) alternative notation for $v(w)$.

Answer by Delio Mugnolo for Question on "publication List" for applying to post-doctoral jobs

2012-10-26T07:51:51Z

Answer to 2): In my experience, I think an abstract is not appropriate in a list of publications.

Answer to 1): A submitted article that nobody can see is basically a non-existing article - the committee members do not even now whether it is a deep 300-page-essay or a 2-page-note. More generally, I believe that mentioning preprints or submitted articles in an applications is only somewhat intereting if the committee members can use these pieces of information to extract a pattern about your research, otherwise is not really useful, perhaps even dangerous ("this guy has posted his paper on XYZ in the arXiv in 2009 and in 2012 it has not yet been published, uhm").

Let me add that in the last few months the policies of Elsevier and Springer about arXiv have become explicitly supportive - see their home pages. This has been enough for me to convince all my co-authors to post our manuscripts in the arXiv - even those co-authors who would have been skeptical a few years ago. Besides, virtually all publishers allow self-archiving - I would be surprised if your coauthors would object even to your posting a joint paper on your own web page.

Answer by Delio Mugnolo for Easy question on Sobolev spaces

2012-10-25T08:00:54Z

Axel,

please read my above comment. If by $W^p$ you mean $W^{p,2}$, then the imbedding follows from the very definition of Sobolev space - or rather, if you wish to keep your definition, from the Meyer-Serrin theorem which essentially establishes the equivalence between your definition and mine.

If by $W^p$ you mean $W^{1,p}$, then $\phi$ is an embedding only if you have a finite measure space (and in this case the assertion follows from the embedding of $L^q$ ind $L^p$ by Hölder's inequality), otherwise it is in general wrong (think of the Sobolev spaces over $\mathbb R^n$).

Answer by Delio Mugnolo for 1 or -1 as an eigenvalue of graph

2012-10-18T08:15:54Z

first (trivial) answer: the spectrum of a bipartite graph is symmetric wrt to 0; hence, +1 is an eigenvalue iff -1 is an eigenvalue.

second (trivial) answer: an individual edge has eigenvalue +1 (and hence also -1).

further (much less trivial) answers: take a look at this article by chan and godsil, where several conditions answering your questions are presented, e.g. at page 76. in a slightly different version you can find on the internet (google is your friend), the same authors show that -1 is an eigenvalue if the graph has a perfect 1-code. also, in this monograph you can find some relevant information, e.g. the observation that +1 is an eigenvalue of so-called collinearity graphs.

Answer by Delio Mugnolo for The conormal derivative of a function

2012-10-15T21:54:51Z

You may define a conormal derivative of $u$ in a very weak sense, just as a distribution. Or you can indeed take a normal derivative in a stronger recalling that each $u\in H^\frac{3}{2}(\Omega)$ s.t. $\Delta u\in L^2(\Omega)$ has a weak normal derivative in $L^2(\Omega)$, see e.g. the classical book of Lions-Magenes if you allow for a smooth boundary of $\Omega$; things are more delicate if $\Omega$ is rougher, say, merely Lipschitz, but can still be dealt with, cf. e.g. this article.

Here a weak normal derivative is defined as follows: If $u\in H^1(\Omega)$, then $g\in L^2(\partial \Omega)$ is called its weak normal derivative if the Gauss-Green formula $$ \int_\Omega \nabla u \nabla \phi +\int_\Omega \Delta u \phi = \int_{\partial \Omega} g \phi $$ holds for all $\phi \in H^1(\Omega)$. ($g$ need not exist for general $u\in H^1(\Omega)$; but if it exists, it is clearly unique).

Comment by Delio Mugnolo

2013-05-22T19:53:40Z

@Vidit Nanda I do not believe that quoting an older paper for its notation is any better than copying and pasting its content to the new one. A reader might easily conclude that one is simply trying to increase her/his own h-index...

Comment by Delio Mugnolo

2013-05-08T07:40:06Z

ok, now I understand.

Comment by Delio Mugnolo

2013-05-08T04:34:54Z

András, I do not understand this comment, either. This shows that $\ell^1$ has the R-N property, of course. But why does anything about $\ell^\infty$, or weighted version thereof, follow?

Comment by Delio Mugnolo

2013-05-08T04:33:20Z

András, I am not sure I understand. That Proposition is about $c_0$, right?

Comment by Delio Mugnolo

2013-05-07T17:54:18Z

Uh, thanks a lot. I found exactly the same answer simultaneously. I am not familiar with the theory of Asplund spaces, so why is $\ell^1$ "obviously" no Asplund space? For the reason I mention or is there some simpler explanation?

Comment by Delio Mugnolo

2013-04-21T03:10:04Z

Is your $H^2$ the same as $H^2(\mathbb R^3)$? If so, and due to uniqueness of your decomposition, is $Q$ not always 0 and $\phi=u$?

Comment by Delio Mugnolo

2013-04-12T21:08:23Z

well, and in large parts of the rest of europe, too.

Comment by Delio Mugnolo

2013-03-26T09:16:31Z

thanks. i do not yet see why you write "hence, there is no easy formula" - several special solutions of complicated nonlinear pdes are known, and of course that minimum is the inf of an lsc functional (perhaps even smoother), but i do agree that "morally" this will be hard.

Comment by Delio Mugnolo

2013-03-25T20:23:10Z

btw, i know too little of calculus of variations to appreciate your reference to the obstacle problem. i guess that what you call the "obstacle problem" (or a solution thereof) is simply the abstract formula which in functional analysis is used to characterize the orthogonal projection. but of course this characterization is in general purely abstract, that is, you come nowhere close to an explicit expression.

Comment by Delio Mugnolo

2013-03-25T20:20:39Z

gerw, thanks for your answer/comment. sure, i can take the vector with smallest distance. the problem is that i'd like to have an explicit, analytic expression of how this element of minimal distance look like. let me be more explicit: if we were in $L^2$, and not in $H^1$, then for all $f\in L^2$ its orthogonal projection onto the cone of positive-valued $L^2$-functions would be simply $f^+$, the positive part of $f$: that is, the pointwise maximum between $f(x)$ and $0$ (a.e.). but i have no idea how an analogous formula could look like, if one considers the $W^{1,2}$-distance instead.

Comment by Delio Mugnolo

2013-03-24T21:39:53Z

Sorry, I just realized that there is a bug (?) in MO-code so that it will accept \bar but not \overline. This induced a wrong version of Brezis' result. You can now see the actual version. Concerning your new question: Your setting should be formulated in a more precise way (what kind of elliptic problem are you thinking of?) Anyway, relatively mild assumptions imply that the solution is in $C^2(\bar{\Omega}$, cf.Thm. 9.25 in Brezis. (Btw: for better later readability you should in the future highlight your edits, otherwise the first answers might look bizarre/off-topic/wrong afterwards)

Comment by Delio Mugnolo

2013-03-21T13:46:51Z

I guess what you are looking for are Krein spaces and their (better behaved) special cases, Pontryagin spaces. Heinz Langer and his school have worked a lot on them.

Comment by Delio Mugnolo

2013-03-18T04:13:11Z

in germany there are indeed some places, called the "fachhochschule" (or sometimes simply "hochschule"), that are some kind of professional, low-level universities with less or no research attitude; and something ever more practical called "berufsakademie". however, in order to be hired there, you generally need a long working experience in industry. by the way, the less advanced the courses, the less are german students expected to be acquainted with english language: it will be almost impossible for you to get a teaching position in germany without a good command of english.

Comment by Delio Mugnolo

2013-03-16T04:54:37Z

Peter Michor's suggestion is correct, but you might want to check Gårding's inequality for a possibly even easier argument.

Comment by Delio Mugnolo

2013-03-16T04:49:49Z

Ben Crowell, I am not quite sure the two ways of using that world are actually unrelated. The solvability features of a wave equation (as opposed to, say, a heat equation) actually depend on the fact that the "hyperbolic" wave equation is a PDE whose symbol is a function with certain properties - the same properties that one would need to define a riemannian metric of negative curvature.