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In their paper A Direct Approach to the Determination of Gaussian and Scalar Curvature Functions, Kazdan and Warner claim something along the lines of: if $g$ is a metric in $W^{2,p}$ ($p>n$) whose sc …

The Kazdan-Warner trichotomy states that for $n\ge 3$, a compact $n$-manifold falls into one of three categories: (A) Every (smooth) function is a scalar curvature. (B) The manifold is strongly scal …

Let $(M^m,g)$ be a compact Riemannian manifold with smooth nonempty boundary, and $N^n\subseteq \Bbb R^d$ a boundaryless isometrically embedded Riemannian manifold. For $1\le p<\infty$ we define as us …

Firstly, you have the wrong inequality (there is a small typo in the paper). Young's inequality is typically written for nonnegative numbers, but for any $a,b\in\Bbb R$ we have \begin{align*} -ab&\le …

This question is not to be confused with the similarly titled question here. In the above lined question, I gave a complete answer, but noticed that things are apparently not so simple in the ellipt …

Christos Mantoulidis showed me how to construct examples in (C) in all dimensions. Namely, if $\Sigma_g^2$ denotes a genus $g$ surface with $g\ge 2$, then $\Sigma_g^2\times T^{n-2}$ does is in class ( …

I will essentially explain the comment under Theorem 8.14 in Gilbarg-Trudinger. I will assume the result stated there: given smooth boundary data and RHS, the Poisson equation has a unique smooth so …

On a Riemannian manifold $(M,g)$, let $\mathcal L^p(M,k)$ denote the space of measurable $k$-tensors $T$ (i.e., the coordinate components in any chart are Lebesgue measurable) for which the norm $$|| …

In many places, Ricci flow with surgery is done with orientable manifolds. Morgan and Tian do not require orientability, but instead they impose the condition that $M^3$ have no embedded $\Bbb RP^2$ w …

You said you know why the $L^2$ norms are equivalent, so let's look at the gradient term $$\int |\nabla^{g_1}u|^2,$$ where the norm associated to $|\cdot|$ doesn't matter. But since $u$ is a function, …

It turns out that this statement can be rephrased as: The Minkowski-Bouligand dimension of $M$ is equal to $k$ (this is immediate from the definition). Another word for this is the box dimension, and …