Ngô Quốc Anh

February 22, 2014

An application of the Brezis-Li-Shafrir estimate to the limiting case of the prescribing Gaussian curvature problem in the negative case

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 14:56

On a Riemannian surface M, let consider the following PDE

\displaystyle -\Delta u +\alpha = R(x)e^u

naturally arising from the prescribed Gaussian curvature problem. A simple variable change, one can assume that \alpha is a negative constant, see this. It follows from a very well-known result due to Kazdan and Warner that it is necessary to have

\displaystyle \int_M R(x) dx<0.

In addition, Kazdan and Warner also showed that if \int_M R(x) dx<0 and R changes sign, then there exists a number \alpha_0 \in (-\infty, 0) such that the above PDE is solvable for all \alpha > \alpha_0 but not if \alpha<\alpha_0. In fact, the number \alpha_0 can be characterized as follows

\displaystyle \alpha_0 = \inf\{\alpha : \text{the PDE is solvable}\}.

This can be easily seen from the the following comprising property: If the PDE is solvable for some \alpha_1, it is also solvable for any \alpha_2 \geqslant \alpha_1.

However, Kazdan and Warner did not tell us what happens when \alpha=\alpha_0. In an attempt to see what really happens when \alpha=\alpha_0, Chen and Li made use of the Brezis-Li-Shafrir estimate to answer in the following way: The PDE is also solvable even when \alpha=\alpha_0. The purpose of this note is to talk about the beautiful Chen-Li argument, see this.

The idea is to approximate the equation for \alpha_0 by a sequence \{\alpha_k\}_k of negative real numbers in the following sense \alpha _k\searrow a_0 as k \to \infty. Their proof consists of three steps as follows:


February 9, 2014

Monotonicity of trace inequalities involving Hermitian matrices and weights

Filed under: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:21

This note aims to prove a very interesting property concerning the monotonicity of trace inequalities of square matrices.

Before going further, let us denote by M_n the set (or the algebra) of n \times n complex square matrices. Then, by M_n^{\rm h} we mean the set of Hermitian matrices in M_n. We also denote by M_n^+ the set of positive semi-definite matrices in M_n^{\rm h}. In other words, there holds

\displaystyle M_n^+ \subset M_n^{\rm h} \subset M_n.

As usual, the notation A \leq B means B-A \in M_n^+ for any A, B \in M_n^{\rm h}.

The following inequality is basically due to Hoa-Tikhonov [here].

Theorem 1. Let n\geq 2 and let a function f :\mathbb R^+ \to \mathbb R be Borel measurable. The inequality

\displaystyle \text{trace}(Af(A)) \leq \text{trace}(Af(B))

holds for all A,B \in M_n^+ with A \leq B if and only if the function g(x)=xf(x) is convex on \mathbb R^+.

The proof of the above theorem is rather simple but elegant. The idea is to transform the condition 0 \leq A \leq B into the relation A^\frac{1}{2} = U B^\frac{1}{2} for some U \in M_n with \|U\| \leq 1. Then the theorem follows immediately from the Jensen trace inequality for contractions.

It is also interesting to note that the super-additivity property, i.e.

\text{trace}(f(A)) + \text{trace}(f(B)) \leq \text{trace}(f(A+B)) \quad \forall A,B \in M_n^+

is equivalent to the convexity of the function f.


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