in the context of elliptic equations of the form $$\text{div}(Df(Du))= 0,$$ where $f,u\colon\mathbb{R^n}\to\mathbb{R}$, $D$ denotes the gradient of $f,u$, and $f$ is a convex function (which implies that the Hessian $D^2f$ is positive semidefinite), I have recently come across the following notation $$C_1 Id \leq D^2f\leq C_2 Id,$$ where $C_1>0,C_2>0$ denote positive constants, and $Id$ is the identity matrix. Does this notation basically mean that the smallest Eigenvalue of $D^2f$ is strictly positive? Or is it understood in the sense that $(D^2f-C_1 Id)$ is still a positive semidefinite matrix? Even so, what does the second inequality $D^2f\leq C_2 Id$ denote?
Furthermore, what has the preceding inequality to do with ellipticity?
Thanks for the help!
The general notation here is the partial order on symmetric $n \times n$ matrices. This is defined as follows. Given two symmetric matrices $S, T \in \mathbb{R}^{n \times n}$ we write $S \le T$ if and only if their quadratic forms are ordered, i.e. $Sx \cdot x \le T x \cdot x$ for all $x \in \mathbb{R}^n$. It's easy to check that $S \le T$ if and only if $0 \le T-S$ if and only if $S-T \le 0$. Now, all symmetric matrices are diagonalizable, and there is a connection to this partial order. Indeed, upon writing $$ Sx \cdot x = \sum_{k} \lambda_x |(x,v_k)|^2 $$ for $S v_k = \lambda_k v_k$ and $\{v_1,\dotsc,v_n\}$ an orthonormal basis, it's simple to see that $S \ge 0$ if and only if $\lambda_k \ge 0$ for all $k$.
When we have $S \le T$ we can get some crude information from the Rayliegh quotients: $$ \lambda(S)_{min} = \min\{ S x \cdot x : |x|=1\} \text{ and } \lambda(S)_{max} = \max\{ S x \cdot x : |x|=1\}. $$ This and $S \le T$ yield $$ \lambda(S)_{min} \le \lambda(T)_{min} \text{ and } \lambda(S)_{max} \le \lambda(T)_{max}. $$ In fact, there is a stronger version of the Rayleigh quotient called the Courant minimax theorem that yields a precise ordering of all of the eigenvalues: $\lambda(S)_k \le \lambda(T)_k$ for all $k$, when the eigenvalues are increasing in $k$. This extra information isn't so useful in this context, though. In the particular case of $c I \le S \le CI$ our crude work tells us that $$ c \le \lambda(T)_{min} \le \lambda(T)_{max} \le C, $$ which then gives useful information on where the spectrum of $T$ lies. The converse is also true here: $c \le \lambda(T)_{min} \le \lambda(T)_{max} \le C$ implies that $c I \le T \le C I$.
What does any of this have to do with elliptic PDE? For second order elliptic operators with constant coefficients, $$ L u = \text{div}(S \nabla u) $$ for $S$ a symmetric $n \times n$ matrix, ellipticity is equivalent to $S \ge c I$ for some $c >0$. When we have non-constant coefficients, then ellipticity is defined as $$ L u = \text{div}(S \nabla u) $$ where now $S(x) \ge c I$ for every $x$ in the domain under consideration. In this situation we often play games with the quadratic form $$ Q(u,v) = \int S \nabla u \cdot \nabla v, $$ and when doing so it is extremely convenient to assume that $S \le C I$ as well since this means $Q$ will be well-defined on standard $L^2-$based Sobolev spaces. For your nonlinear operator $$ N(u) = \text{div}(DF(Du)) $$ what's going on is that ellipticity can be enforced / defined by the condition that $D^2 F \ge c I$ since then the linearization around any background $u_0$ yields the linearized operator $$ L_{u_0} v = \text{div}(D^2F(u_0) Dv) $$ and we're guaranteed ellipticity since $D^2F(u_0) = S \ge c I$. The corresponding upper bound $D^2F \le C I$ is added for the same reason as above.