User jbc - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:29:29Z http://mathoverflow.net/feeds/user/26013 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures/131194#131194 Answer by jbc for Existence of dominating measure for weak*-compact set of measures jbc 2013-05-20T06:29:15Z 2013-05-20T06:42:34Z <p>This more a comment than an answer but will be too long for that. There are two basic approaches to the theory of finite measures---the topogical one and the one based on $\sigma$-algebras. At the core of the first approach lies the duality between $C^b(S)$ (the bounded, continuous functions on a completely regular space $S$) and the space $M^t(S)$ of Radon measures thereon, for the second that between $L^\infty(\Omega)$ (the bounded measurable functions on a measure space $(\Omega,\cal A)$ and the space $\cal M(\cal A)$ of finite $\sigma$-additive real valued-measures. All four of these spaces are Banach spaces in a natural way but these structures are not compatible with the above dualities. This suggests that one should try to provide the function spaces with suitable locally convex topologies which are compatible. In the case of $C^b(S)$ this means a suitable intrinsic locally convex topology which is complete and has the required dual space $M^t(S))$. Due to the joint efforst of several mathematicians (in particular, R.C. Buck and the polish school, e.g., Orlicz and Wiweger) such a topology is known---it is the so-called strict topology (strictly speaking, this is not always complete, but it is so for most of the standard classes of topological spaces).</p> <p>There is such a topology in the second case but I have been unable to find it in the literature and this is the reason that I am submitting this answer. It has various descriptions (in itself, in my opinion, a hint that it is a useful and natural topology). But for me the killer-diller fact is that it has the natural universal property for $\sigma$-additive measures with values in a Banach space (in contrast to the topological case, where we can embed $S$ into the space $M^t(S)$ with corresponding universal property), we can here embed the $\sigma$-algebra $\cal A$ into $L^\infty(\Omega)$ in the natural way.</p> <p>We close with a few remarks. </p> <p>$1$. We emphasise that it is important that the above topologies have intrinsic definitions (i.e., independent of the dual pairs). The fact that they then turn out to be topologies which can be defined via the duality---typically the Mackey topology---is then a theorem to be proved. This is, basically, the reason for the relevance of all this to the OP which essentially asks for a characterisations of the weakly compact subsets of the dual of $L^\infty$.</p> <p>$2$. Another reason for depositing this was to clarify the confusion about which is the relevant topology on the family of measures---it is clear from the formulation that the OP refers to the weak star topology of a subsset of the dual of $L^\infty$ with the topology referred to above.</p> <p>$3$. If one tries to weaken the topology of a Banach space in a non-trivial way (as we are doing here) then there are contraints on the kind of space one can get. For example, one cannot get a barrelled space (closed graph theorem). This means that the resulting topology cannot be a member of one of the traditional classes of well-behaved LCS's (metrisable, inductive limits of Banach spaces, e.g.). Neither can they be nuclear (except, of course, in trivial cases). This, alas, seems to have been a hindrance to their acceptance into the main body of functional analysis. It is my belief that this is a great pity since they are precisely the tool required for extending and enriching the relationship between functional analysis (duality theory) and measure theory (the Riesz representation theorem) which is one of the crown jewels of analysis.</p> http://mathoverflow.net/questions/130396/applications-of-visual-calculus/130409#130409 Answer by jbc for Applications of visual calculus jbc 2013-05-12T12:45:52Z 2013-05-12T12:51:08Z <p>I'm not sure that the following is what you are looking for but I hope that it sheds some useful light on the topic of your query and suggests further applications. Given a curve in the plane with parametrisation $(c_1(u),c_2(u))$ one can consider the transformation $$F(u,v)=(c_1(u)+\sqrt {2 v} \dot c_1(u),c_2(u)+\sqrt {2 v}\dot c_2(u)).$$ (We are actually interested in the network this mapping introduces in the plane---the image of the coordinate network---which has a natural geometrical interpretation related to the OP). A simple computation shows that the determinant of the Jacobi matrix of this mapping is $\ddot c_1(u)\dot c_2(u)-\dot c_1(u)\ddot c_2(u))$. From this we can deduce various useful facts:</p> <p>$1.$ The parametrisation $c$ does not appear explicitly (only its derivatives). This is the reason for the satement at the start of the OP.</p> <p>$2.$ The determinant is independent of $v$ (this was the reason for the strange dependence of $F$ on $v$). In particular we can choose a parametrisation for $c$ for which this is identically $1$ which means that $F$ is area-preserving. This can be used to garner a plethora of results for particular curves.</p> <p>$3.$ The case of the cycloid has some special features which explains some results and methods in the works quoted If we use the standard parametrisation $(t-\sin t,1-\cos t)$, then the above determinant is $1-\cos t$ which is just the height of the given point above the $x$-axis.</p> <p>Much more could be said about this topic, but we would like to close with the remark that these facts were not just pulled out of thin air---behind them there lies an important concept, that of a Samuelson configuration, which was introduced by the economics laureate Paul Samuelson (not under that name, of course) in his Nobel acceptance speech, i.e., over 40 years ago.</p> http://mathoverflow.net/questions/119362/unit-sphere-in-r-infty-is-contractible/130405#130405 Answer by jbc for Unit sphere in R^\infty is contractible? jbc 2013-05-12T12:20:03Z 2013-05-12T12:20:03Z <p>Don't want to be too pedantical but, to set the record straight, both topologies coincide in this situation---one of the consequences of the Banach-Dieudonne theorem.</p> http://mathoverflow.net/questions/17189/is-there-a-good-reason-why-a2b-b2a-1-when-ab1/130399#130399 Answer by jbc for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1? jbc 2013-05-12T09:45:39Z 2013-05-12T09:45:39Z <p>Since this question has been bumped up I would like to state what I think is its natural framework: We have the inequality $f(a,b) \leq 1$ for $a+b=k$ when $k$ lies between $\frac 12$ and $1$. Otherwise, the inequality is $f(a,b) \leq \frac {k^k}{2^{k-1}}$ (with $f(a,b) = a^{2b}+b^{2a}$). This version is not just more comprehensive but it illustrates the dichotomy in where the maximum occurs (at the symmetric point $(\frac k2,\frac k2)$ or at the boundary $(k,0)$). The two cases considered above ($k=\frac 12$ and $k=1$) are precisely the transitional ones. One can also get estimates from below (usually by the constant $1$ but in a small neighbourhood around the critical interval $[\frac 12,1]$ the sharp version involves values which are given implicitly as the solution of transcendental equations).</p> <p>(P.S. I had already given some of this information in a comment but, since it elicited no reaction, I have taken the liberty to repeat it here despite the fact that it isn't really an answer but, hopefully, does shed some light on the problem and its solution).</p> http://mathoverflow.net/questions/129992/radon-nikodym-property-of-ell-infty/130003#130003 Answer by jbc for radon-nikodým property of $\ell^\infty$ jbc 2013-05-07T18:20:07Z 2013-05-07T18:20:07Z <p>Since a closed subspace of a space with RNP clearly also has RNP, in order to get the requested elementary example, it suffices to display a measure on the unit interval with values in $c_0$ which is absolutely continuous with respect to Lebesgue measure but whose derivative does not take its values there. This can be done via the usual trick with the sequence $\frac 1 n \cos n x$. I guess this example is in the classical text by Diestel and Uhl which is probably still the best reference for the RNP.</p> http://mathoverflow.net/questions/128901/tensor-product-of-c-algebras-of-bounded-uniformly-continuous-functions-on-metri/128924#128924 Answer by jbc for Tensor product of C*-algebras of bounded, uniformly continuous functions on metric spaces jbc 2013-04-27T14:43:32Z 2013-04-27T14:43:32Z <p>The spectrum of the $C^\ast$-algebra of bounded, uniformly continuous functions on a uniform space is the Samuel compactification. So your query can be restated in the form: is the Samuel compactification of a product naturally identifiable with the product of the Samuel compactifications of the individual spaces. This is almost certainly wrong. The corresponding result for the Stone-Cech compactification of completely regular spaces is about as wrong as it could be. Presumably, you can give an explicit counterexample in the uniform case by using the fine uniformity. This is yet another example of the fact that if you want to extend duality (Riesz representation theory, Gelfand duality) from the compact case to the non-compact one (for completely regular spaces, uniform spaces, etc.) then you are well advised to leave categories of Banach spaces or algebras and use more general ones (mixed topologies, Saks spaces and algebras). A fairly systematic account can be found in the monograph "Saks Spaces and Applications to Functional Analysis".</p> http://mathoverflow.net/questions/128451/extending-uniformly-continuous-functions-on-subspaces-to-non-metrizable-compactif/128453#128453 Answer by jbc for Extending uniformly continuous functions on subspaces to non-metrizable compactifications jbc 2013-04-23T08:44:03Z 2013-04-23T08:44:03Z <p>The closure of $X$ in $Z$ is compact , so there is no hope if $f$ is not bounded. If it is bounded then so is its extension to the closure of $X$ in $Y$ and this gives a bounded uniformly continuous function on the closure which can be extended to a bounded continuous function on $Y$ by the Tietze extension theorem. This, in turn, extends to a continuous function on the Stone-čech compactification. You can then restrict the latter to the closure of $X$ in $Z$.</p> http://mathoverflow.net/questions/127906/gelfand-representation-and-functional-calculus-applications-beyond-functional-ana/128315#128315 Answer by jbc for Gelfand representation and functional calculus applications beyond Functional Analysis jbc 2013-04-22T06:38:45Z 2013-04-22T06:38:45Z <p>If I understand the question correctly, then the celebrated Corona problem must be such an example. It can be formulated as a classical interpolation problem for bounded, holomorphic functions on the disc but is equivalent to the fact the the open unit disc is dense in the spectrum of the Banach algebra $H^\infty$. The answer is, of course, yes (a famous deep result of Carleson) but the corresponding problem for more complicated domains in the plane or in higher dimensions is still an active area of research with some partial results available.</p> http://mathoverflow.net/questions/128015/meromorphic-functions-as-distributions/128019#128019 Answer by jbc for Meromorphic Functions as Distributions jbc 2013-04-18T22:06:56Z 2013-04-18T22:06:56Z <p>It is folklore that any meromorphic function on the real line can be regarded as a distribution in a natural way. One defines $x^{-n}$ to be the $n$-th derivative of the locally integrable function $\log |x|$, with appropriate coefficient. In order to treat the general case, one uses the principle of recollement des morceaux---if the real line is covered by a family of open subsets and one has a family of distributions, one on each set, which satisfies the obvious compatibility condition, then one can combine them to a single distribution. This is a theorem in the Schwartz book---the proof uses partitions of unity. I heard this argument in a course given regularly by the portuguese mathematician J. Sebastião e Silva at the University of Lisbon (I attended it in 1969). The same globalisation argument, combined with the comment of Sönke Hansen above, gives the result for the complex plane. I have no information on the second part of the query.</p> http://mathoverflow.net/questions/127601/does-the-derivative-of-log-have-a-dirac-delta-term/127650#127650 Answer by jbc for Does the derivative of log have a Dirac delta term? jbc 2013-04-15T18:36:22Z 2013-04-15T18:42:48Z <p>Questions about distributions, usually involving the $\delta$ distribution ("function"), are of frequent occurrence on this and related sites. Since they are often treated in a rather cavalier fashion, I would like to attempt to answer this query in some detail. The basic problem lies in the interpretation of the functions $\frac 1 x$ and $\log x$ (notabene not $\log |x|$) as distributions. As a preliminary remark, it is not surprising that theoretical physicists are guided by their physical intuition and not by mathematical rigour in dealing with such questions and it is perfectly acceptable for mathematicians to proceed in the same way in formulating such conjectures. However, since this is a mathematical forum, one does have the right to expect that the final formulation of the solution conforms to the usual standards of mathematical rigour (as is implied in the OP). In fact it is rather disquieting that this is often not the case, since this task can be achieved by elementary methods which have been on record in the primary and secondary literature for at least 50 years.</p> <p>For the example in question (and, indeed, for most of the examples in such forums), one need only be cognisant of the following simple facts about distributions.</p> <p>$1$. Every continuous function on the reals, better, every locally integrable function, determines in a natural way a distribution.</p> <p>$2$. There is a notion of convergence for distributions. For our purposes, it suffices to know that if a sequence of continuous functions converges uniformly on compacts, then it converges in the sense of distributions. In fact, local $L^1$ convergence suffices.</p> <p>$3$. The dream theorem of every freshman calculus student holds---if a sequence of distributions converges, then the sequence obtained by differentiating term by term also converges.</p> <p>We can now turn to the above query. Note that the problem lies in the fact that while the function $\log |x|$, being locally integrable, determines a distribution, the same is not true a priori for $\frac 1 x$ and $\log x$. In these two cases, we have to proceed in a more delicate manner.</p> <p>Firstly, note that the function $\log |x|$ is locally integrable and so is a distribution. More importantly the same is true for its derivative in the distributional sense. It is then rather natural to define this derivative to be the distribution $\frac 1 x$.</p> <p>The case of the distribution $\log x$ is rather more subtle. In this case we resort to the complex logarithm. y thus we define, for non-zero $\epsilon$, the distribution $\log(x+i\epsilon)$ to be $\log|x|+i \arctan \frac \epsilon x$ (i.e., we are using the principal branch of the complex logarithm). We now define the distribution $\log x$ to be the limit of this distribution as $\epsilon$ tends to zero. At this point, we see that we get different values, depending on whether we consider the limit $\epsilon \to 0_+$ or $\epsilon \to 0_-$. If we now differentiate this equation we obtain the required formula.</p> <p>To conclude, a few remarks.</p> <p>The above approach is due to the portuguese mathematician J. Sebastião e Silva who developed it in the context of his axiomatic approach to the theory of distributions.<br> Sadly, the definitive monograph on his approach that he was preparing 40 years ago was never completed, due to his premature passing.<br> However, the elementary part has been presented in the book "An introduction to the theory of distributions" by Campos Ferreira which is based on lectures at the University of Lisbon (ca. 1970) and this contains the material described here.</p> <p>The above is a special case of the family of distributions $x^\lambda$ for general $\lambda$ (even complex) which is developed in detail in the above reference. They are also described in the standard four volume text of Gelfand and Silov.</p> <p>The distribution $\frac 1 x$ is also treated by Laurent Schwartz who used the Hadamard principal value (the latter was his great uncle by marriage). The connection with the above approach can easily be established by considering the truncated function $\log_\epsilon$ (which is set equal to zero on the interval $]-\epsilon,\epsilon[$), letting $\epsilon$ tend to zero and differentiating.</p> <p>The ambiguity in the definition of the distribution $\log x$ is no more disconcerting than that in the definition of the logarithm function for complex arguments. In a more sophisticated approach this distribution would not be defined on the real line but on the natural domain of definition of the latter, i.e., on the unversal covering of the punctured plane. </p> http://mathoverflow.net/questions/127045/fixed-point-theorems/127331#127331 Answer by jbc for Fixed point theorems jbc 2013-04-12T08:35:18Z 2013-04-12T08:35:18Z <p>It would be a pity not to mention the work of F. Browder, in particular his study of non linear pde's, the main tool being FPT's on Banach spaces. This is documented in many of his publications, perhaps most memorably in his "Nonlinear operators and nonlinear equations of evolution".</p> http://mathoverflow.net/questions/127250/a-graduate-course-on-sturm-liouville-theory/127265#127265 Answer by jbc for A graduate course on Sturm Liouville theory? jbc 2013-04-11T16:42:04Z 2013-04-11T16:42:04Z <p>The classic "Methods of Mathematical Physics" by Courant and Hilbert has a wealth of material on the Sturm-Liouville problem and its connections to various themes of mathematical physics. In fact, one could almost say that the treatise is constructed around this topic. Little or no knowledge of Functional Analysis is required. It is, of course, not the most up to date version but can be recommended as a first introduction--- it contains a multitude of explicit examples (Green functions, eigenfunction expansions, connections with the classical pde's of mathematical physics via separation of variables, special functions etc.) and is a joy to read.</p> http://mathoverflow.net/questions/127045/fixed-point-theorems/127085#127085 Answer by jbc for Fixed point theorems jbc 2013-04-10T12:38:08Z 2013-04-10T18:55:42Z <p>There is a celebrated fixed point theorem of A. Borel with applications to algebraic geometry (Ann. of Math. (2)64(1956)).</p> http://mathoverflow.net/questions/126464/is-there-an-analog-of-determinant-for-linear-operators-in-infinite-dimensions-as/126483#126483 Answer by jbc for Is there an analog of determinant for linear operators in infinite dimensions as that of finite dimensions? jbc 2013-04-04T06:31:54Z 2013-04-07T19:11:27Z <p>The answers above point out that one cannot define a determinant in a meaningful way on the algebra of bounded operators on a Banach space, unless finite-dimensional. However, this does not preclude the possibility of doing this for suitable subclasses and this is precisely what Alexander Grothendieck did in his celebrated (amongst functional analysts) article "Théorie de Fredholm" (Bull. Soc. Math. vol. 84---freely available online). This is one of those articles which changed the face of mathematics forever. The closely related question of which operators have a trace has been investigated in great detail, by, for example, Albrecht Pietsch and Hermann König.</p> http://mathoverflow.net/questions/5299/measurable-functions-and-unbounded-operators-in-von-neumann-algebras/125379#125379 Answer by jbc for Measurable functions and unbounded operators in von Neumann algebras jbc 2013-03-23T14:16:48Z 2013-03-23T20:33:06Z <p>This is a very old question but I would like to mention what I think is the most natural response, namely the fact that we can borrow the concept of rings of quotients from algebra. In the commutative case, this is transparent: if $f$ is measurable, then we can express it as the quotient of the two bounded, measurable functions $\frac f{1+|f|^2}$ and $\frac 1 {1+|f|^2}$. Conversely, if $f$ and $g$ are bounded measurable functions and the set where $g$ vanishes is negligible, then $\frac f g$ is a measurable function (we use the usual sloppy notation to talk about equivalence classes of measurable functions).</p> <p>The non-commutative case is, of course, more delicate but a standard result which is often used to extend the spectral theorem from the bounded to the unbounded case and which can be found in Riesz-Nagy, states that if $T$ is a closed, densely defined, unbounded linear operator in Hilbert space, then both $T(I+T^\ast T)^{-1}$ and $(I+T^\ast T)^{-1}$ are bounded and $T$ is their quotient (in the more subtle sense used for unbounded operators). This suggests that a suitable definition for an unbounded operator $T$ to be associated with a von Neumann algebra $\cal A$ would be for these two operators to lie in $\cal A$.</p> http://mathoverflow.net/questions/124852/riesz-representation-theorem-for-vector-valued-fields/125377#125377 Answer by jbc for Riesz representation theorem for vector-valued fields jbc 2013-03-23T13:52:29Z 2013-03-23T13:52:29Z <p>Since my comment seems to have been misunderstood, I would like to take the opportunity to expand on it. The basic categories are $\bf {Ban}_1$ and $\bf W$ of Waelbroeck spaces, i.e., Banach spaces with linear contractions as morphisms, resp., Banach spaces provided with an additional compact, linear topology on the unit ball (details can be found in the book by Cigler, Losert and Michor on categories of Banach spaces). If $E$ is a Banach space, then its dual $E'$ is a Waelbroeck space and indeed the two categories are dual to each other. The important example for us will be the pair $C(K)$ of continuous functions on a compactum and its dual, the space $M(K)$ of Radon measures thereon. The latter has, as does every Waelbroeck space, a natural complete, locally convex topology---the finest to agree with the given compact one on the unit ball---and in our example, this (and not the norm) is usually the natural one. We now denote the family of finite dimensional subspaces of a Banach space $E$ by $\cal F$ and use the fact that $E$ is the inductive limit of this family, regarded as an inductive spectrum in $\bf {Ban_1}$ in the natural way (this is very simple and can be found explicitly in the above reference). It follows fairly easily that $C(K;E)$, the Banach space of continuous functions with values in $E$, can be identified with the inductive limit of the spectrum ${C(K;F), F \in \cal F}$. General abstract nonsense shows that the dual of the latter is the projective limit (in the sense of the category $\bf W$) of ${M(K;F') : F \in \cal F}$ (we are using the trivial extension of the Riesz representation theorem to the case of functions with values in a finite dimensional space). One can then identify the elements of this projective limit with measures which take their values in $E'$ and which are bounded and Radon for the topology mentioned above to obtain the desired representation of the dual of $C(K;E)$.</p> <p>If one is prepared to use the extension of the Riesz representation theorem which covers the case of a completely regular space $S$ and identifies the space of bounded, Radon measures on $S$ as the dual of the space $C^b(S)$ of bounded continuous functions with the strict topology (see the monograph "Saks spaces and Applications to Functional Analysis"), then one can obtain a suitable version of this duality which works for completely regular spaces.</p> http://mathoverflow.net/questions/124771/convergence-of-probability-measure-and-the-weak-convergence/125319#125319 Answer by jbc for Convergence of probability measure and the *-weak convergence ? jbc 2013-03-22T19:51:17Z 2013-03-22T19:51:17Z <p>It is a standard result of descriptive topology that the space of Borel probability measures on a polish space $S$ is itself a polish space with the weak star topology as the dual of $C^b(S)$ (Nota bene, not as a Banach space but with the so-called strict topology, i.e., the finest locally convex topology which agrees with compact convergence on the unit ball). This result is to be found, e.g. in the text on descriptive topology by Kechris.</p> http://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function/123876#123876 Answer by jbc for Riemannian surfaces with an explicit distance function? jbc 2013-03-07T14:31:03Z 2013-03-07T20:19:07Z <p>This is an old question but since it has been bumped up I would like to mention two classes of Riemann metrics (on the upper half-plane, resp. on the punctured plane) where your conditions can be met, at least partially. In the first case these are the metrics of the form $ds^2 =y^\beta(dx^2+dy^2)$ and in the second case $ds^2=r^\beta(dx^2+dy^2)$. The basis for this lies in the remarkable properties of the class of functions of the form $f(t)=p (\cos (d(t-t_0)))^{\frac 1 d}$ (we have included the parameters for a reason). Then we have the following facts:</p> <p>$1$. If we consider the family of curves with parametrisations of the form $(F(t),f(t))$ where $F$ is a primitive of $f$ (called the MacLaurin catenaries in the arXiv article 1102.1579), then these are, for a fixed $d$, the geodesics for the first class of Riemann metric above (where the exponent $\beta$ depends in a simple way on $d$).</p> <p>$2$. Similarly, the family of curves with polar equation $rf(\theta)=1$ (no, this is not a misprint) are, for a fixed $d$, the geodesics for the second class of surface (again there is a simple, but different, relation between $d$ and $\beta$).</p> <p>$3$. The lengths along these curves can be calculated explicitly (this involves computing the integrals of functions of the form $f^\alpha$ with $f$ as above and Mathematica can handle this---the primitives involve hypergeometric functions).</p> <p>We refer to the above mentioned article for the details and the rationale of the above representations and remark only that the reason behind all of this is that, for suitable choices of parameters, these functions are the solutions of the euler equations for calculus of variation problems of the form: minimise the functionals $\int f^\gamma(f^2+f'^2)^{\frac 1 2} dt$, resp. the same functional with restraint $\int f(t) dt = constant$ under suitable boundary conditions. The essential fact is that the functions of the above type are precisely those for which the expression $f^2+f'^2$ is proportional to a power of $f$. In fact, $f^2+f'^2=p^2 f^{2-2d}$. (We included the parameters to ensure that we have obtained all solutions). (Remark: the case $d=0$ is an exception---here we use the functions $f(t) = Ae^{bt}$).</p> <p>The first class of curves was introduced in the article mentioned above, the second are the so-called spirals of MacLaurin and were introduced by this scottish mathematician in the 18th century. Of course, several members of the first class (i.e., for special choices of $d$) are familiar ---e.g. Dido circles, straight line, catenaries, cycloids, special types of parabolas. some of which have been mentioned in the above responses---and the MacLaurin spirals (sometimes called sinusoidal spirals) include, as special cases, some of the most famous curves of classical geometry (Teixeira Gomes' standard work on special curves includes many sections on this subject). Both have a startling array of special properties, all depending on the above property of the functions $f$ (for a unified exposition see, again, the aforementioned arXiv article).</p> <p>We end with a caveat. For some of these spaces we can measure the distance between two points simply as the length of the geodesic joining them (we can, of course, always do this locally). However, for some values of $\beta$ there are points which cannot be joined by geodesics and then one would presumably need a more delicate argument. This has already been pointed out in the case of parabolas in the above responses and for catenaries the question is intricate enough for Hancock to have devoted a complete article in Annals of Mathematics to it.</p> http://mathoverflow.net/questions/123760/topological-characterization-of-the-closed-interval-0-1/123791#123791 Answer by jbc for Topological characterization of the closed interval $[0,1]$. jbc 2013-03-06T17:19:47Z 2013-03-06T20:45:49Z <p>A space which is homeomorphic to the closed unit interval is called a simple arc in the monograph "Dynamic topology" by Whyburn and Duda and there is a characterisation of it on p. 70 of this book. This assumes that the given space is a metric space, a condition which can be avoided by using the Urysohn metrization theorem. </p> <p>On request, the characterisation is as follows: a space is a simple arc if and only if it is a non-degenerate (i.e., with more than one point) compact, connected set which is second countable and such that each point (with the exception of two specified ones---the endpoints) is a cut point. (A point in a connected space is a cut point, if its complement is disconnected).</p> http://mathoverflow.net/questions/101635/functional-differentation/121988#121988 Answer by jbc for Functional differentation jbc 2013-02-16T14:42:06Z 2013-02-16T14:42:06Z <p>As pointed out by Liviu Nicolaescu, there is a Taylor theorem for functionals on infinite dimensional spaces which could be used here but there are problems in applying it to your case. Non linear functionals on function spaces occur regularly in applications and, since they can be very complicated, it is natural to seek a tractable concrete representation and the one you display is very commonly used. As a general principle, if such a (strictly speaking incorrect) formulaiton is useful, then there is probably a rigorous formulation which lies behind it and it is usually useful to seek it out. In your case, I think that the required result is a pair of theorems by the French mathematicians Fr\'echet and Gateaux (incidentally, the two names associated with the introduction of the concept of differentiability in infinite dimensional spaces), dating from before the first world war.</p> <p>Before describing them, a couple of generalities. The prototype of such results is the question of approximating continuous, one-dimensional functions (say, on the unit interval) by simpler ones, e.g., polynomials. There are two natural ways to do this. The first one is probably more familiar to non-mathematicians---the Taylor series---whereas the second one (the Weierstrass approximation theorem) tends to be known only to professional mathematicians. On the surface, the first one seems preferable---the approximation is by an infinite power series whose coefficients can be explicitly described. However, it has two very important disadvantages---it only applies to very smooth functionals and the explicit expression for the coefficients is not usually very useful for computations since it involves (potentially arbitrarily) high derivatives of the function. The second one is more cumbersome to state since it does not present a series approximation but simply a sequence of polynomials which converge to the given function but is usually the better version for applications. Indeed, whole libraries have been written about methods of approximating functions by polynomials (Google under approximation theory or constructive function theory).</p> <p>Of course, the same applies to your problem which is at the next metalevel---approximating functionals which depend on functions. Again there are two methods. Using Taylor-type expansions has the same disadvantages, together with the one mentioned in the above answer that the kernel functions involved in the representation need not, in general, be continuous functions, but could be something much more pathological (measures or distributions) and so not likely to be useful for computations.</p> <p>This brings us finally to the result mentioned above. This is a complete analogue of the Weierstra\ss theorem and states that any continuous functional on $C([0,1])$, say, can be approximated by a sequence of functionals of the type in your formulation (with finite sums).</p> <p>Since these results seem to be forgotten lore and have not , to my knowledge, found their way into the secondary (English language) textbook literature, I can only refer to the original articles, "Sur les fonctionnelles continues" (F) and "Sur les fonctionneles continues et les fonctionnelles analytiques" (G), both of which are, fortunately, available online. I think that a modern, accessible write-up in English would be a service to engineers. I should remark that a modern analyst would be able to prove these results (in more general form) in a couple of lines, using the abstract version of Weierstra\ss' theorem, the Stone-Weierstra\ss theorem. However, the methods used by the above-mentioned authors could be used, together with a version of the Weierstra\ss theorem in higher dimensions, to cobble together a version with an explicit expression for the approximating functionals (using, e.g., spline functions as did G, trigonometric approximations as did F and Bernstein polynomials.)</p> http://mathoverflow.net/questions/121261/to-what-extent-does-trajectory-determine-gravity-sources/121697#121697 Answer by jbc for To what extent does trajectory determine gravity sources? jbc 2013-02-13T10:25:01Z 2013-02-13T13:18:00Z <p>Although this question has been answered many times, I would like to add the following simple way to generate examples. It is clear that in the plane any central force, no matter how complicated, generates circular orbits as special cases. There are many ways to distribute matter in three space to generate such a force field. For example, any distribution of mass (which can, of course, consist of point masses) along the $z$-axis will produce such a field in the $x,y$-plane. As a side remark, such configurations occur in practice---e.g., as the gravitational field generated by a wire.</p> <p>I might add for those interested in the theme of trajectories that there is a considerable body of work on the question of which families of curves can occur as the trajectories of force fields, including a monograph "Differential-geometric aspects of dynamics", by Edward Kasner and his associates, much of which is easly accessible online. This is, however, not directly relevant to the query.</p> http://mathoverflow.net/questions/110461/direct-proof-of-injectivity-of-l-infty/121613#121613 Answer by jbc for Direct proof of injectivity of $L_\infty$ jbc 2013-02-12T15:14:52Z 2013-02-12T15:14:52Z <p>I am not sure whether the following would meet your requirements but I vaguely remember having heard it in a course many years ago and I think that it is sufficiently distinct from the above proofs to justify a brief mention. The crucial common property of the three spaces in question---the real line, the sequence space and the function space---is that they are Dedekind-complete Banach lattices for which the norm is intimately connected to the order structure---the unit ball coincides with the interval $[-1,1]$. This fact allows one to mimic directly the standard proof of the classical Hahn-Banach theorem in the two more advanced cases.</p> http://mathoverflow.net/questions/39670/is-a-compactly-generated-hausdorff-space-functionally-hausdorff/121569#121569 Answer by jbc for Is a compactly generated Hausdorff space functionally Hausdorff? jbc 2013-02-12T08:28:46Z 2013-02-12T08:28:46Z <p>Since you implicitly ask for sufficient conditions for a compactly generated space to be completely Hausdorff, it might interest you that one such is that the family of compacta be countably generated. As an appendix to the negative example given by Qiaochu Yuan, any Hausdorff space which is not completely Hausdorff provides a counterexample, since a space is completely Hausdorff if and only if the associated compactly generated space is completely Hausdorff. Both of these facts are in the 1969 paper "Topologies et compactologies" by Henri Buchwalter.</p> http://mathoverflow.net/questions/118515/derivative-indicator-function/121156#121156 Answer by jbc for Derivative indicator function jbc 2013-02-08T07:01:50Z 2013-02-10T06:28:42Z <p>You are asking for the derivative of a non-linear function on an infinite dimensional space. (You do not specify the latter but the space locally integrable functions seems a natural candidate). The derivative can only exist in a very weak sense so it is natural to go for the directional derivative which has two inputs---the point $x$ where the derivative is computed and the direction $y$ in which the rate of change takes place. A back of the envelope calculation suggests that the derivative should then be $$-\sum \frac {Y(a)}{x(a)}\delta_a$$ where we use capitals to indicate the primitives which occur in the formulation and the sum is taken over the $a$ in the pre-image of $c$ under $X$.</p> <p>Some general remarks: the question was posed in such a vague manner that it is not really possible to give a precise, rigorous answer. Presumably you have some concrete application in view and I suggest that you check the above formula there to see if it leads to the expected result (ones I looked at were the functions $t$---with primitive $t^2$---and $\sin t$ for the function $x$).</p> <p>As is evident from the above formula, the derivative is in a much weaker sense than the usual concepts of functional analysis---one requires special conditions on $x$ (beyond just smoothness) for the above expression to make sense and the limit of the difference quotients used to define the derivative does not take place in the underlying function space but in a larger space (of distributions---hence the Dirac functions in the formula. I presume that this is the reason for the reference to distributions in your question).</p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/121377#121377 Answer by jbc for An example of a non-paracompact tvs (over the reals, say) jbc 2013-02-10T06:18:08Z 2013-02-10T06:18:08Z <p>This is an old question which has been answered satisfactorily but the following remark might shed some light on it and put it in a broader context: we can embed any completely regular space as a closed subspace of a separated locally convex space in a canonical manner. We simply provide the free vector space over the set with the finest locally convex structure so that the natural inclusion is continuous. Hence if any topological property is stable under closed subspaces (e.g., normality, paracompactness), then whenever we can find a completely regular space which fails it, we can find a locally convex space which fails it.</p> http://mathoverflow.net/questions/120485/fourier-analytic-proofs/120498#120498 Answer by jbc for fourier analytic proofs jbc 2013-02-01T06:57:32Z 2013-02-01T06:57:32Z <p>There are two very extensive monographs on this subject, more precisely, its relationship to convex geometry: by Groemer (Geometric applications of Fourier series and spherical harmonics) and Koldobsky (Fourier analysis in convex geometry).</p> http://mathoverflow.net/questions/120421/schwartz-kernel-theorem-for-topological-spaces/120460#120460 Answer by jbc for Schwartz kernel theorem for topological spaces jbc 2013-01-31T20:38:33Z 2013-01-31T20:38:33Z <p>As has been pointed out by Peter Michor above, this is false, even for compact spaces. However, it is perhaps of interest that one can characterise in a natural way those operators from $C(K_1)'$ to $C(K_2)$ which are represented by kernels of the above type. They are the linear operators whose restrictions to the unit ball of $C(K_1)'$ are continuous for the weak star topology. This can be proved directly with standard methods but the following background to the result might shed more light. There is a natural locally convex topology on the dual of $C(K_1)$ (indeed on the dual of any Banach space) which is complete and compatible with the duality with $C(K)$. It has many descriptions, perhaps the simplest being as the topology of uniform convergence on the compact subsets of $C(K_1)$. Then $C(K_1)'$ with this structure contains $K_1$ as a topological subspace (in the natural manner---identifying a point in $K_1$ with the corresponding $\delta$ measure). This embedding of $K_1$ has the universal property that each continuous mapping from $K_1$ into a Banach space $F$ can be lifted in a unique way to a linear mapping on $C(K_1)'$ which is continuous in the above sense. If we take $F$ to be $C(K_2)$ then the required result pops out---use the fact that there is a natural identification between $C(K_1\times K_2)$ and the continuous functions from $K_1$ into the Banach space $C(K_2)$.</p> <p>There is a natural extension to a corresponding result for bounded continuous functions on completely regular spaces but one needs some more elaborate functional analytic structures to give a precise version.</p> http://mathoverflow.net/questions/120291/metrization-of-weak-convergence-of-signed-measures/120334#120334 Answer by jbc for Metrization of weak convergence of signed measures jbc 2013-01-30T16:26:04Z 2013-01-30T16:41:02Z <p>As mentioned in several replies, you require that the compact space be metrisable for a positive answer to your question. Then the claim is, in fact, true for bounded sequences. The natural setting for this question is the Banach-Dieudonn\'e theorem. If $E$ is a separable Banach space, then, as has been mentioned in previous answers, the unit ball of the dual space is metrisable under the weak star topology. The natural topology on the dual is then the finest locally convex topology on the whole space which agrees with the latter on this ball---it is even the finest such topology (i.e., not necessarily locally convex or even linear) and can also be characterised as the topology of uniform convergence on compact subsets of $E$. This topology has several nice properties---it is complete and has the same bounded subsets as the norm topology---but it is not metrisable. However, its restrictions to the bounded sets of $E'$ are metrisable. This is the essential content of the above-mentioned result. The consequence which is relevant to the question posed is the fact that it has the same bounded convergent sequences as the weak star topology on $E'$ defined by a dense countable subset of $E$ and this {\it is} metrisable. In order to answer the question posed, it suffices to specialise to the case where $E$ is $C(K)$ with $K$ compact and metrisable.</p> <p>I should perhaps mention that the theorem of Banach-Dieudonn\'e holds for any (i.e., not necessarily separable) Banach (or even, in suitable form, Fr\'echet) space and this provides information for the case where $K$ is not metrisable.</p> http://mathoverflow.net/questions/97497/one-sided-cauchy-principal-value/116064#116064 Answer by jbc for One-sided Cauchy principal value jbc 2012-12-11T10:02:18Z 2012-12-11T10:02:18Z <p>This is an old question but since it concerns the non-symmetric case and this has not been addressed by the above answers, the following solution might be of interest. The portuguese mathematician Sebastiao e Silve gave an elementary definition of definite integals of distributions which makes, for example, functions such as $\frac 1 {x^2} \sin\frac 1x$ integrable on $[0,\infty[$. The first ingredient is the simple fact that each distribution on the line (or a subinterval) has a primitive. The definite integral is then defined as in elementary calculus, using the concepts of the value of a distribution at a point, respectively limits of a distribution at a point (including one-sided limits as required in this question). Of course, these need not exist in the general case and so there are restrictions required for the existence of definite integrals, as one would expect. The precise definitions and examples can be found in Campos Ferreira's book on Distributions (Pitman) which is essentially based on courses Sebastiao e Silva gave in the 60's. As a sample, for a distribution $f$ defined near infinity, write $\lim_{s \to \infty} f(s)= \lambda$ if there is an integer $p$ and a continuous function $F$ defined on a neighbourhood of infinity so that $f=D^p F$ (derivative in distributional sense) and $\frac {F(s)}{s^p} \to\frac{\lambda}{p!}$ as $s \to \infty$.</p> <p>We remark that in addition to this application of these concepts, there are many situations where they are, despite their simplicity, of some consequence. For example, the comment that one often reads in the literature, that distributions don't have values at points is unnecessarily pessimistic. Most distributions which are of practical value do have values at most points---simple example, the Dirac $\delta$ function, which , of course, has values at all points, apart from its singularity. This is a very simple example, but there are much more subtle ones where the assignment of a value is not, a priori, obvious. </p> http://mathoverflow.net/questions/100469/opinions-on-the-multiplication-of-measures/114742#114742 Answer by jbc for Opinions on the Multiplication of Measures jbc 2012-11-28T08:28:45Z 2012-11-28T13:21:32Z <p>Not a definitive answer, just some thoughts that I hope you might find useful. Firstly, there is a celebrated example of Schwartz that shows that you can't do this globally in the sense of getting a ring structure on the space of measures which extends that on the space of continuous functions. It can, however, often be fruitful to try a local approach, i.e., find pairs of measures which {\it can} be multiplied in a sensible way. For example, the product of two dirac measures is only a problem if their singularities coincide. One can, trivially, always multiply a continuous function and a measure (and with a little bit of integration theory this can be extended to more general functions). One way to deal with less trivial situations is to say that if we can cover $I$ (for the sake of simplicity, I will assume that the measures are defined on the unit interval $I$) with a finite family of relatively open subintervals so that on each interval at least one of the the measures is a continuous function (which one depending on the interval, of course), then we can define the product. This covers the above case of the product of dirac measures with distinct singularities. One would have to know more about the kind of measures that you want to multiply in order to decide if this is of much help. </p> http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures/131194#131194 Comment by jbc jbc 2013-05-20T10:42:11Z 2013-05-20T10:42:11Z By &quot;compatibility&quot; above I of course meant &quot;non-compatibility&quot;. It is perhaps worth mentioning that in the topological situation, the universal property works in the other direction. $S$ embeds into $M^t(S)$ in such a way that every continuous, bounded function on $S$ with values in a Banach space lifts in a unique fashion to a continuous linear mapping with the appropirate (which, again, is not the norm). http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures/131194#131194 Comment by jbc jbc 2013-05-20T10:06:10Z 2013-05-20T10:06:10Z Compatibility in the first part means that the corresonding dual spaces are too large. Thus the Banach space duals of the function spaces consist in both cases of the finitely additive measures, not the countably additive or Radon ones. The unversal property is that every countably additive meaure on the $\sigma$-algebra with values in a Banach space (for which see Diestel and Uhl &quot;Vector measures&quot;) lifts to a unique continuous linear mapping on $L^\infty$ with the topology mentioned in my answer. http://mathoverflow.net/questions/130963/on-uniform-convergence-of-sequences-of-bounded-holomorphic-functions-with-formal Comment by jbc jbc 2013-05-17T17:19:38Z 2013-05-17T17:19:38Z I can't give a reference but there is a large number of related results which follow from the following general considerations: the unit ball of $H^\infty$ is compact for the topology of compact convergence and so the latter coincides there with any weaker Hausdorff topology. In your case, this would be the weak topology induced by evaluation of the derivatives at $z_0$. http://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-require/93201#93201 Comment by jbc jbc 2013-05-17T07:08:07Z 2013-05-17T07:08:07Z The precise reference for Garnir's result is to be found in MR0477688 (and the word &quot;proved&quot; is missing in the previous comment). http://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-require/93201#93201 Comment by jbc jbc 2013-05-17T07:01:50Z 2013-05-17T07:01:50Z For the record, in 1973 the Belgian mathematician Henri Garnir, combining results of Schwartz on a measurable graph theorem and Solovay as mentioned here, that it is consistent with ZF without AC that every linear map from an ultrabornological space (in particular, Banach, Fr&#233;chet or an inductive limit of a sequence of Banach spaces which covers the spaces mentioned here) into any locally convex space is continuous. This in turn implies that the algebraic dual of such a space coincides with the topological dual and that the same holds for biduals for most spaces of interest. http://mathoverflow.net/questions/119362/unit-sphere-in-r-infty-is-contractible/130405#130405 Comment by jbc jbc 2013-05-12T12:22:57Z 2013-05-12T12:22:57Z Sorry. This was meant to be a comment on the above comment. Tried to delete but couldn't. Maybe somebody more powerful could erase it. http://mathoverflow.net/questions/127906/gelfand-representation-and-functional-calculus-applications-beyond-functional-ana/128315#128315 Comment by jbc jbc 2013-04-23T13:24:14Z 2013-04-23T13:24:14Z My understanding of the proof is that Carleson demonstrated the validity of the interpolation property with hard analysis (which, by the way, led to the important concept of Carleson measures). The Banach algebra bit, i.e., the proof of the equivalence of this with the denseness (which is apparently due to Newman), uses the fact that the Gelfand-Neumark transform allows one to consider bounded analytic functions on the open disc and their absolute values as continuous functions on a compactum, the point being that such functions attain their infimum. http://mathoverflow.net/questions/128451/extending-uniformly-continuous-functions-on-subspaces-to-non-metrizable-compactif/128453#128453 Comment by jbc jbc 2013-04-23T12:34:20Z 2013-04-23T12:34:20Z If you have $Y$ and $Z$ as in your question with the extension property for each subspace $X$ and each bounded, uniformly continuous function thereon, then by using $X=Y$ one sees that $Z$ has the uiversal property for each bounded, uniformly continuous function on $Y$. This means that $Z$ is the so-called Samuel compactification of $Y$. So the only leeway is in the case of $Y$ for which the two notions of compactification diverge. http://mathoverflow.net/questions/128015/meromorphic-functions-as-distributions/128019#128019 Comment by jbc jbc 2013-04-20T05:04:26Z 2013-04-20T05:04:26Z Yes, indeed it does. http://mathoverflow.net/questions/128015/meromorphic-functions-as-distributions/128019#128019 Comment by jbc jbc 2013-04-18T22:27:09Z 2013-04-18T22:27:09Z I forgot one more ingredient---you can always multiply a distribution with a smooth function. So you consider an open cover which is such that each element contains only one pole. The restriction of the meromorphic function to such a set is an analytic function times one of the form $(x-x_0)^{-n}$ and so a distribution. Now globalise. http://mathoverflow.net/questions/127045/fixed-point-theorems/127126#127126 Comment by jbc jbc 2013-04-11T16:08:43Z 2013-04-11T16:08:43Z Another contribution to the theme &quot;FTP's and Nobel Prizes in economics&quot;. The Arrow-Debreu theory of equilibrium in economics uses the Brouwer FTP and its extension by Kakutani in an essential way. Both are laureates and this theory is generally regarded as one of their most significant contributions. http://mathoverflow.net/questions/127045/fixed-point-theorems/127048#127048 Comment by jbc jbc 2013-04-11T15:54:31Z 2013-04-11T15:54:31Z It is worth mentioning the sensationally short proof given by Lomonosov of his theorem that every continuous linear mapping on a Banach space which commutes with a non-zero compact operator has a non-trivial invariant subspace. This was then the strongest positive result on the invariant subspace problem (and might still be for all I know) and the key ingredient was the Schauder-Tychonoff FTP. http://mathoverflow.net/questions/127045/fixed-point-theorems/127047#127047 Comment by jbc jbc 2013-04-11T15:45:35Z 2013-04-11T15:45:35Z Another application is a nice proof of the Inverse Function Theorem. http://mathoverflow.net/questions/127045/fixed-point-theorems Comment by jbc jbc 2013-04-10T09:01:27Z 2013-04-10T09:01:27Z Not a FPT but a book: &quot;Fixed point theory&quot; by Granas and Dugundji. http://mathoverflow.net/questions/99777/does-x-embed-in-y-and-y-embed-in-x-always-imply-that-x-isomorphic-on Comment by jbc jbc 2013-04-04T06:45:51Z 2013-04-04T06:45:51Z The answers below show that this is false but it is perhaps worth mentioning that, on the positive side, there are versions under additional conditions which are correct and these are frequently used in the isomorphic theory of Banach spaces. They go under the collective name of &quot;Pelczynski decomposition method&quot;.