User jjcale - MathOverflow

Answer by jjcale for Open problems in PDEs, dynamical systems, mathematical physics

2013-03-20T22:18:13Z

For a list of 15 open problems in mathematical physics (in 2000) see Simon's Problems, http://mathworld.wolfram.com/SimonsProblems.html.

Some of these problems are solved, as mentioned in the link.

Answer by jjcale for on an inequality of Brezis-Lieb

2013-01-23T19:46:51Z

No, choose $\Omega={z \in \mathbb{C} : |z| \leq 1 } ,\ f(z)=Re\ z^{n}$ and let $n\rightarrow \infty$ .

Answer by jjcale for Does Physics need non-analytic smooth functions?

2012-11-26T22:53:50Z

In the BCS-theory of superconductivity the energy gap that seperates the ground state from the excited states is a non-analytic function of the exchange energy.

Here one doesn't use a taylor expansion of the interacting Hamiltonian around the non-interacting Hamiltonian. Instead one uses a Bogoliubov transformation.

Answer by jjcale for Projections in Banach spaces

2012-11-18T11:41:52Z

Here a simple example :

Let X be the cartesian product of $L^{\infty}$ and $L^{1}$ on the interval $[0,1]$, let $P_{t}$ the canonical projection on the subspace of functions with support $[0,t]$ and choose $Q(f_{1},f_{2}) = (0,f_{1})$ .

Answer by jjcale for Examples of theorems with proofs that have dramatically improved over time

2012-05-03T16:40:53Z

Example of a bounded linear operator on a Banach space without non-trivial closed invariant subspace.

The first example was given bei Enfo in 1975. Enflo submitted the full article in 1981 and the article's complexity and length delayed its publication to 1987 (see http://en.wikipedia.org/wiki/Per_Enflo). Simpler examples were constucted for example by Beauzamy and Charles Read.

Answer by jjcale for Does there exist a continuous function of compact support with Fourier transform outside L^1?

2012-05-01T16:54:14Z

Choose $f(x)=\frac{(\frac{1}{2} -x)}{log(x)}$ for $0 < x \leq \frac{1}{2}$ and $0$ otherwise.

Then $Im \int_{0}^{\infty} dk \ e^{-\varepsilon k} \int_{-\infty}^{\infty} dx \ f(x) e^{-ikx} = \int_{0}^{\frac{1}{2}} dx \ \frac{(\frac{1}{2} -x)}{log(x)} \cdot \frac{-x}{\varepsilon^{2}+x^{2}} \rightarrow \infty$ for $ \varepsilon \downarrow 0$ .

Therefore the Fourier transform of $f$ is not in $L^{1}$ .

Answer by jjcale for Quasinilpotent example

2012-01-07T20:07:34Z

In http://www.jstor.org/pss/2047905 you can find a weighted shift operator that has this property.

Answer by jjcale for On the Paley-Wiener theorem

2011-12-11T10:15:34Z

The answer is yes :
Let $h$ be an even real valued Schwartz function whose Fourier transform has compact support. Then choose $f(y) = \int_{-\infty}^y x h(x)^{2} dx$ .

Answer by jjcale for Perturbations of an operator that disconnect the spectrum

2011-11-23T23:13:31Z

For Hilbert spaces, the conjecture follows from fact 4 and the answer to question http://mathoverflow.net/questions/81652/complement-of-a-subspace-which-is-a-cartesian-product applied to the kernel of the map $H\times H\ni (v,w) \mapsto Av + (I-A)w \in X$ .

Complement of a subspace which is a cartesian product

2011-11-22T20:24:54Z

Let $H$ be a Hilbert space and $U$ a closed subspace of $H\times H$ . Does then exist closed subspaces $V$ and $W$ of $H$ such that $H\times H = U \oplus (V\times W)$ ?

Answer by jjcale for Most helpful math resources on the web

2011-11-16T19:39:24Z

For people who are interested in prime factorization : www.mersenneforum.org

quasinilpotence and finite spectrum II

2011-10-14T19:14:38Z

Let A be a quasinilpotent operator on a Hilbert space and let every operator of the algebra generated by $A$ and $A^{*}$ have finite spectrum. Does then follow, that A is nilpotent ?

quasinilpotence and finite spectrum

2011-10-04T21:19:28Z

Let A be a quasinilpotent operator on a Hilbert space and let $A^{*}A$ have finite spectrum.
Does then follow, that A is nilpotent ?

Answer by jjcale for quasinilpotence and finite spectrum

2011-10-06T18:44:48Z

I found a counterexample :

Let $e_{1},e_{2},...$ be ON basis of the Hilbert space and define A by $Ae_{2n-1} = \sqrt{1-\frac{1}{n^{2}}} \ e_{2n} \ + \ \frac{1}{n} \ e_{2n+1}$ , $\ \ $n=1,2,3,... ,
$Ae_{2n} = 0$ , $\ \ $n=1,2,3,...

Then A is a partial isometry and therefore $A^{*}A$ a projection.
Furthermore A is quasinilpotent but not nilpotent.

Answer by jjcale for Theorems that are 'obvious' but hard to prove

2011-08-26T20:46:00Z

A differentiable manifold M that is homeomorphic to the n-sphere is also diffeomorphic to the n-sphere . Obvious, but wrong ! (But right for 1-, 2-, 3-, 5-, 6- and 12-spheres).

How to calculate a Fredholm index numerically

2011-08-24T19:37:33Z

How can one calculate the index of a Fredholm operator numerically ?

In numerically calculations one uses always finte dimensional spaces. But linear operators on finite dimensional spaces have always index zero.

Answer by jjcale for Examples of non-rigorous but efficient mathematical methods in physics

2011-08-24T18:56:36Z

The Hypernetted-chain approximation used in statistical mechanics.

Was for instance used in the theory of the fractional quantum hall effect by Laughlin in order to estimate the energies of elementary excitations of Laughlins wave function.

Answer by jjcale for Does a quantitative version of Fredholm theory exist?

2011-08-24T18:17:51Z

The second resolvent equation you are studying is an important tool in quantum scattering theory.

See M. Reed and B. Simon, "Methods of Modern Mathematical Physics", Vol IV, Academic Press

Answer by jjcale for Bounding a spectral gap: what proof techniques exist?

2011-08-22T21:19:10Z

For VBS quantum antiferromagnets in one dimension see also :

Ian Affleck, Tom Kennedy, Elliott H. Lieb and Hal Tasaki, Valence bond ground states in isotropic quantum antiferromagnets. Comm. Math. Phys., Volume 115, Number 3 (1988)

and

Stefan Knabe, Energy gaps and elementary excitations for certain VBS-quantum antiferromagnets, Journal of Statistical Physics Volume 52, Numbers 3-4, 1988

Answer by jjcale for How is the physical meaning of an irreducible representation justified?

2011-08-22T20:12:45Z

You can use group representation theory to reduce the dimensions of the problem. And you can explain energy degeneracies . That's all !

Existence of a special density

2011-08-19T20:53:32Z

Does the following function $f:\mathcal{P}(\mathbb{N})\rightarrow\{0,1\}$ exist :
$f(\mathbb{N})=1$,
$f(A\cup B)=f(A)+f(B)$ for $A\cap B=\emptyset$,
$f(A)=0$ for finite $A$

Comment by jjcale

2013-06-01T17:37:42Z

Your argument only shows that the action is not norm continous, and this means that the generators of the group action are unbounded operators.

Comment by jjcale

2013-05-24T18:58:36Z

See en.wikipedia.org/wiki/Measurement_problem. Or read Roger Penrose's "The road to reality".

Comment by jjcale

2013-05-23T17:32:21Z

An example are maxwells equations, see en.wikipedia.org/wiki/Maxwell%27s_equations.

Comment by jjcale

2013-05-15T18:19:11Z

@Rodrigo: I used PARI/GP : f(n)=n+n/vecmin(factor(n)[,1])-1 g(n)=while(1,my(m);m=f(n);if(n==m,break);print(m);n=m;) g2(n)={my(m);my(i=0);while(1,m=f(n);i++;if(n==m,break);n=m;);[m,i]} g2max(n)={ my(m=[0,0]); for(i=2,n, my(p=g2(i)); if(p[2] > m[2],m=[i,p[2]]); ); m; }

Comment by jjcale

2013-05-05T07:09:03Z

Computed for n up to 10000000 : record holder 6730914 with orbit length 226 .

Comment by jjcale

2013-02-13T19:44:34Z

Comment by jjcale

2012-11-21T21:04:56Z

Maybe you should ask this question in the forum mersenneforum.org .

Comment by jjcale

2012-11-18T12:32:49Z

@fedja : Because $\|(P_{t+\varepsilon}-P_{t})Q(f_{1},f_{2})\| \leq \varepsilon \|f_{1}\|_{\infty} \leq \varepsilon \|(f_{1},f_{2})\|$

Comment by jjcale

2012-11-17T20:27:05Z

The restriction of $Q$ to $(I-P_{0})P_{1}X$ must be quasinilpotent.

Comment by jjcale

2012-10-24T19:35:23Z

@Delio Mugnolo : The Resolvent is only analytic outside the spectrum. If for instance the spectrum consists of the non negative real numbers, then it might be possible that on a subspace of the Hilbert space the resolvent can be analytically continued from the upper half plane to the lower half plane across the spectrum. If A is self adjoint then the resolvent itself can never have a pole outside the real axis, but the analytic continuation may have a pole there, and that's a resonance. And the imaginary part of the pole is interpreted as the inverse of the lifetime of the state

Comment by jjcale

2012-10-23T19:28:46Z

@Delio Mugnolo : To my knowledge a resonance is not a pole of the resolvent but a pole of an analytic continuation of the resolvent.

Comment by jjcale

2012-10-14T16:53:18Z

A normal matrix N is in this class iff all absolute values of its eigenvalues are at least 1.

Comment by jjcale

2012-09-13T19:14:09Z

Here is a reference : arxiv.org/pdf/1007.1796.pdf

Comment by jjcale

2012-08-30T19:28:17Z

Every nonzero element of the spectrum of a compact operator on a Banach space is an eigenvalue of this operator. So this is a classical result in the infinite dimensional case.

Comment by jjcale

2012-08-29T19:45:32Z

Arne Beurling did important work during World War II : He deciphered a german secret teletypewriter, see en.wikipedia.org/wiki/Arne_Beurling .