# 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 $f$.