most recent 30 from http://mathoverflow.net 2013-05-23T16:57:29Z http://mathoverflow.net/feeds/user/20341 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88413/a-criterion-for-the-sum-of-two-closed-sets-to-be-closed Fabien Besnard 2012-02-14T09:47:22Z 2012-02-14T19:04:20Z

<p>Let $V$ and $I$ be two closed subsets of a Banach space $A$.</p> <p>The set $V$ is a convex cone, and $I$ is a linear subspace of $A$. I also know that $V\cap I={0}$.</p> <p>I would like to know whether $I+V$ is closed. I've seen that there is a criterion of Dieudonné which I can't use here because I know that neither $V$ nor $I$ is locally compact.</p> <p>So my question is : are there any other criteria that I could try to use ? </p>

http://mathoverflow.net/questions/31354/theorems-in-euclidean-geometry-with-attractive-proofs-using-more-advanced-methods/87270#87270 Fabien Besnard 2012-02-01T20:16:13Z 2012-02-01T20:16:13Z

<p>The classification of conics would be an example, but I don't know if you count matrix reduction as a "more advanced concept".</p>

http://mathoverflow.net/questions/86289/help-me-find-good-math-questions-for-my-students/86306#86306 Fabien Besnard 2012-01-21T14:14:23Z 2012-01-21T14:14:23Z

<p>About 1), maybe you could try <a href="http://en.wikipedia.org/wiki/Logistic_function" rel="nofollow">http://en.wikipedia.org/wiki/Logistic_function</a></p> <p>and</p> <p><a href="http://en.wikipedia.org/wiki/Logistic_map" rel="nofollow">http://en.wikipedia.org/wiki/Logistic_map</a></p> <p>It might also solve 2) because there are a lot of possible developments so you can structure a problem with easy questions first and a few harder ones, opening on fancy math stuff like chaos, fractals and the like in the end.</p>

http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/85810#85810 Fabien Besnard 2012-01-16T13:46:37Z 2012-01-16T13:46:37Z

<p>So your question seems to be : what is the connection between the eigenvectors of $A_1=Dbb^TD^T$ and $A_2=Db_\perp b_\perp^TD^T$ ?</p> <p>Well it's easy to find these eigenvectors. First case : $Db,Db_\perp$ linearly independant.</p> <p>Then the eigenspaces of $A_1$ are ${\mathbb R}Db$, and $(Db)^\perp$, and similarly for $A_2$. Since $D$ is skew-symmetric, in particular it does not preserve orthogonality and there is no connection between the eigenvectors of $A_1$ and $A_2$. The second case is obvious.</p>

http://mathoverflow.net/questions/85065/unexpected-applications-of-the-fact-that-nth-degree-polynomimals-are-determined-b/85123#85123 Fabien Besnard 2012-01-07T09:42:56Z 2012-01-07T09:42:56Z

<p>I'm not sure it is helpful, but the proof of the <a href="http://en.wikipedia.org/wiki/Gaussian_quadrature" rel="nofollow"> fundamental theorem of the Gaussian quadrature method </a> uses the fact you mention.</p>

http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/84893#84893 Fabien Besnard 2012-01-04T17:45:22Z 2012-01-04T17:45:22Z

<p>Hi Bramiozo.</p> <p>What you write is a bit confusing. Here are some questions in order to understand better :</p> <p>1) When you say "the two parts can also be written", this is an hypothesis, is it ? (anyway without further hypothesis there is no connection between $v_1$ and $v_2$)</p> <p>2) "Suppose $K_1$ and $K_2" etc. : this is another question then ?</p> <p>3) Is $b$ a vector ? Is $b_\perp$ orthogonal to $b$ ?</p> <p>I guess your matrices are real. If yes, you should right symmetric instead of self-adjoint.</p>

http://mathoverflow.net/questions/14376/why-is-addition-of-observables-in-quantum-mechanics-commutative/84809#84809 Fabien Besnard 2012-01-03T16:25:10Z 2012-01-03T17:50:59Z

<p>If I correctly understand, you are misinterpreting the meaning of the product and sum of observables.</p> <p>When you say "We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values."</p> <p>This cannot possibly describe the usual sum A+B and product AB of operators. For the product, it is not even hermitian unless A and B commute. Agreed, A+B is hermitian, but the spectrum of A+B does not contain the result of the sum of a measurement of A followed by a measurement of B (in either way), again unless A and B commute. For a counter-example take $A=\pmatrix{1&amp; 0\cr 0&amp;-1}$ and $B=\pmatrix{0&amp;1\cr 1&amp;0}$.</p> <p>I hope I correctly understood your question.</p>

http://mathoverflow.net/questions/102178/the-space-of-pure-states/102310#102310 Fabien Besnard 2013-04-16T12:25:46Z 2013-04-16T12:25:46Z

To complement Julia's answer, maybe it's worth to add that what is usually called the pure state space of A, that is the weak* closure of the space of pure states $P(A)$, is of course compact, since it is closed in the (compact) state space. This is an elementary comment, but maybe there can be some confusion at times between the pure state space and the space of pure states.

http://mathoverflow.net/questions/88413/a-criterion-for-the-sum-of-two-closed-sets-to-be-closed/88438#88438 Fabien Besnard 2012-02-20T16:32:52Z 2012-02-20T16:32:52Z

Ok, I managed to show that if the unit spheres of the subspaces $V-V$ and $I$ are a positive distance apart then $V+I$ is closed under the hypotheses I have given. The subspace $V-V$ need not be closed, only $V$ has to be. The sufficient condition was all I cared for, and this one works in the case I have at hand. So I thank you and accept your answer.

http://mathoverflow.net/questions/88413/a-criterion-for-the-sum-of-two-closed-sets-to-be-closed Fabien Besnard 2012-02-14T15:51:20Z 2012-02-14T15:51:20Z

Maybe I should add that by &quot;criterion&quot; I mean a sufficient condition.