| bio | website | leandromat.wordpress.com |
|---|---|---|
| location | Brasilia, Brazil | |
| age | 36 | |
| visits | member for | 2 years, 9 months |
| seen | May 7 at 20:39 | |
| stats | profile views | 226 |
I am a mathematical physicist at Universidade de BrasÃlia. I am currently interested in Classical Equilibrium Statistical Mechanics, Percolation and Ergodic Theory. Contact: leandro.mat@gmail.com
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Aug 4 |
awarded | Yearling |
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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 $x\neq 0$. |
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Jun 15 |
comment |
which are positive definite matrix @Mex Leandro :) |
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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. |
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Jun 15 |
answered | which are positive definite matrix |
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Jun 15 |
comment |
Terminology between essentially bounded function and bounded function. Just take the limit in the inequality $|\sigma_{n}(x)|\leq K$ |
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Jun 15 |
comment |
Terminology between essentially bounded function and bounded function. Kns last inequality you mean almost surely ? |
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Jun 8 |
awarded | Constituent |
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Jun 8 |
awarded | Caucus |
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Jun 8 |
awarded | Caucus |
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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! |
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Mar 27 |
answered | Two Dimensional Ising Model and Hamiltonian. |
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Mar 21 |
answered | Openness and differentiation |
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Mar 21 |
answered | Where do the higher order terms in Taylor series come from? |
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Mar 9 |
comment |
What distinguishes the Measure Theory and Probability Theory? "Probability theory is measure theory with soul". |
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Feb 25 |
accepted | How to prove this subadditivity? |
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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. |
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Feb 24 |
revised |
How to prove this subadditivity? edited tags |
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Feb 23 |
asked | How to prove this subadditivity? |
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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 $E$. |