Reputation
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dg.differential-geometry
Nov
20 |
comment |
Smooth perturbation of a positive self-adjoint operator with compact resolvent
The author gave 3 references to that claim. Do none of them answer you question? @HSM |
Nov
19 |
awarded | Nice Answer |
Nov
19 |
comment |
Completeness of nonharmonic Fourier Series
@Liviu: ah, I see. But to answer the question you asked of the OP: the set of $S = \{s \in \mathbb{Z}: 4\mid s\}$ is a rescaling of harmonic series, but $\Phi^S$ does not span $L^2$. Muntz's theorem tells you something about asymptotic density, but I think a bit more is needed. |
Nov
19 |
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Boomerangs in Polya's orchard
One possible (but really, really poor) bound is this. First make all the discs twice as big. Not you know that any curve that deviates from a straight line in sufficiently small manner will hit a disc in radius $R$. This contradicts a curve reaching distance $2\tilde{R}$ via a curve of radius of curvature $>\tilde{R}$ since within distance $R$ of origin the curve is "well approximated by a line". But this estimate is likely to be way bigger than what actually is there. (What? I am an analyst. :-p.) |
Nov
19 |
comment |
Boomerangs in Polya's orchard
(entirely tangential comment: how the heck do you come up with these titles? most of the time I see one of your questions I open it because I cannot conceive how it is a mathematical question, and then I read it and realize that no other title can be better than the one you chose) |
Nov
19 |
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Completeness of nonharmonic Fourier Series
I read the question and think more along the lines of a finite dimensional scenario: in the case of $\mathbb{R}^k$, the question is asking whether a set of $2k$ vectors $\Phi\subset \mathbb{R}^k$ has the property that any subset of size $k$ is linearly independent. The density that @dime is looking for is more like "as long as we have half of all the vectors, they span the whole thing." |
Nov
17 |
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Smooth perturbation of a positive self-adjoint operator with compact resolvent
See also: mathoverflow.net/questions/43124/… and math.upenn.edu/~kazdan/509S07/eigenv5b.pdf |
Nov
17 |
comment |
Smooth perturbation of a positive self-adjoint operator with compact resolvent
Something nasty does happen when eigenvalues have multiplicities. I believe these are all discussed already in Kato's book. Summary: when eigenvalues are distinct the smoothness of the eigenprojections correspond to that of the parameter. When they are not distinct the eigenvectors need not even be continuous. (You can even see this for symmetric $2\times2$ matrices.) |
Nov
13 |
revised |
Does Gaussian Quadrature actually refer to Gauss-Legendre Quadrature？
edited tags |
Nov
12 |
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Distance between quadratic forms
The area of an ellipse (working in two dimensions) is proportional to the products of the semi-principal axes. The semi-principal axes are the eigenvalues. Note that there's the requirement that $V_h$ and $V_{h'}$ are nested and touch along a circle. You achieve this by rescaling. In terms of what you have written this is requiring $\inf q'(x)/q(x) = 1$. |
Nov
12 |
awarded | Nice Answer |
Nov
12 |
comment |
Distance between quadratic forms
If you insert the missing $\log$ then for quadratic forms on a two dimensional vector space, the definition quoted is the same (up to constant factors) as the definition given in equation 1.2 of the second paper you linked to. |
Nov
12 |
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Distance between quadratic forms
There probably should be a log involved in the definition. Note also that if $q = \lambda q'$ the RHS is 1. This corresponds well with the definition that $[q]$ is the equivalent class of $q$ under the equivalent relation given by conformal scaling. |
Nov
11 |
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Is Hessian operator self-adjoint on infinite dimensional environment?
An important thing that Peter Michor didn't say: it is a lot more natural to think of the Hessian as a bilinear form, instead of as an operator. |
Nov
10 |
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Competitive functions: uniqueness of solution
There probably is a degree theory argument for existence. |
Nov
10 |
reviewed | Leave Open How to divide a square into three similar rectangles |
Nov
6 |
reviewed | Looks OK Where to buy premium white chalk in the U.S., like they have at RIMS? |
Nov
6 |
comment |
Competitive functions: uniqueness of solution
(d) You are almost asking for $f$ to be a $C^1$ diffeomorphism of $[0,1]^n$ to itself. So you need that its derivative $Df$ is everywhere surjective. |
Nov
6 |
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Competitive functions: uniqueness of solution
(a) No, they are not enough. Set $n = 2$ you can easily find an example where $\nabla f_1$ and $\nabla f_2$ are parallel on an open subset of $(0,1)^2$. (Start with $f_1 = 1/2 + x_1 - x_2$ and $f_2 = 1/2 + x_2 - x_1$ and modify it outside $(1/3,2/3)^2$. This can be generalized to higher dimensions.) (b) This question has nothing to do with functional analysis. (c) You should ask this at math.stackexchange.com . |
Nov
6 |
revised |
Competitive functions: uniqueness of solution
edited tags |