1,132 reputation
711
bio website leandromat.wordpress.com
location Brasilia, Brazil
age 36
visits member for 2 years, 9 months
seen May 7 at 20:39
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I am a mathematical physicist at Universidade de Brasília. I am currently interested in Classical Equilibrium Statistical Mechanics, Percolation and Ergodic Theory. Contact: leandro.mat@gmail.com


Aug
4
awarded  Yearling
Jun
15
comment which are positive definite matrix
Yes and worth to mention that in the definition of positivity we only have to verify the inequality for $x\neq 0$.
Jun
15
comment which are positive definite matrix
@Mex Leandro :)
Jun
15
comment Terminology between essentially bounded function and bounded function.
No see the William's answer. It is not possible, in general, to show that $f$ is bounded. Another very simple example can be constructed by using a Dirac delta measure.
Jun
15
answered which are positive definite matrix
Jun
15
comment Terminology between essentially bounded function and bounded function.
Just take the limit in the inequality $|\sigma_{n}(x)|\leq K$
Jun
15
comment Terminology between essentially bounded function and bounded function.
Kns last inequality you mean almost surely ?
Jun
8
awarded  Constituent
Jun
8
awarded  Caucus
Jun
8
awarded  Caucus
May
27
comment Formal proof that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected.
JDH this comment is just to point out a small typo :$A\subset\mathbb{R}^2$. +1 nice answer!
Mar
27
answered Two Dimensional Ising Model and Hamiltonian.
Mar
21
answered Openness and differentiation
Mar
21
answered Where do the higher order terms in Taylor series come from?
Mar
9
comment What distinguishes the Measure Theory and Probability Theory?
"Probability theory is measure theory with soul".
Feb
25
accepted How to prove this subadditivity?
Feb
25
comment How to prove this subadditivity?
It is awesome to get beyond a beautiful answer a new "technique". Thanks a lot Didier Piau for this post and the detailed explanation.
Feb
24
revised How to prove this subadditivity?
edited tags
Feb
23
asked How to prove this subadditivity?
Feb
10
comment The Space $C(\Omega,\mathbb{R})$ has a Predual?
@YemonChoi I think that this cartesian space is always separable in this case, because $E^{\mathbb{Z}^d}$ is compact in the product topology, which is in this case generated by the metric $d_1(\omega,\eta)=\sum_{i\in\mathbb{Z}^d}\frac{1}{2^{\|i\|}} d(\omega_i,\eta_i)$, where $\omega=(\omega_i)_{i\in\mathbb{Z}^d}$ and $d$ is the metric on $E$.