\section{Identifying One-dimensional Operators}\label{Sec 1D}

In this section, we first prove Proposition \ref{prop identify 1D}, and then provide more examples to illustrate how to choose $f$ for $F$ to make the best use of Theorem \ref{thm MC main}. 

The following lemma will be useful. 
\begin{lemma}\label{lemma}
Suppose that $X,Y \in S(n)$ satisfy 
\begin{equation*}
  \begin{pmatrix}
    X & 0 \\ 0 & -Y 
    \end{pmatrix}
    \leq D^2_{x,y} \left(2\vp\left(\frac{|x-y|}{2}, t\right) \right).
\end{equation*}
Then we have $X\leq Y$ and 
\begin{equation*}
    \tr(X-Y) \leq 2\vp''\left(\frac{|x-y|}{2}, t\right) .
\end{equation*}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{lemma}]
Note that the Hessian of $2\vp\left(\frac{|x-y|}{2},t \right)$ has the form $$\begin{pmatrix} P & -P \\
-P & P\end{pmatrix},$$
where $P=2D^2_x\vp\left(\frac{|x-y|}{2},t \right)$. 
Then $X \leq Y$ follows from the fact that the matrix $\begin{pmatrix} P & -P \\
-P & P\end{pmatrix}$ annihilates vectors of the form $\begin{pmatrix} x  \\
x\end{pmatrix}$. 
For any matrix $C$ such that the matrix $\begin{pmatrix} I & C \\
-C & I\end{pmatrix}$ is positive semidefinite, we have 
\begin{eqnarray*}
    \tr(X-Y) &=& \tr \left(\begin{pmatrix} I & C \\
C & I\end{pmatrix} \begin{pmatrix}
    X & 0 \\ 0 & -Y 
    \end{pmatrix} \right) \\
&\leq& \tr \left(\begin{pmatrix} I & C \\
C & I\end{pmatrix} \begin{pmatrix}
    P & -P \\ -P & P 
    \end{pmatrix} \right)  \\
&=& 2 \tr\left( (I-C)P \right).
\end{eqnarray*}
Choosing $C=I-2e \otimes e$ with $e=\frac{x-y}{|x-y|}$ produces 
$$\tr\left( (I-C)P \right) =2 P(e,e)=\vp''\left(\frac{|x-y|}{2},t\right).$$ Thus we have the desired estimate.  
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop identify 1D} ]
(1). For simplicity, we write $e=\frac{x-y}{|x-y|}$. 
For any $x,y \in \R^n $ with $|x-y|=2s$ and $r,v \in \R$ with $v-r=2\vp$,  we have 
\begin{eqnarray*}
   && F(y,r,\vp' e ,Y)-F(x,v,\vp' e, X) \\
   &=& -\tr(Y) -\langle W(y), \vp' e \rangle -Vr-h(y) +\tr(X) +\langle W(x), \vp' e \rangle +V v +h(x) \\
   &= & \tr(X-Y) +\langle W(x)-W(y), \vp' e\rangle +V(v-r) +h(x) -h(y) \\
   &\leq & 2\vp'' +2\|W\|_{L^\infty} |\vp'| +2V\vp+2w_h(s),
\end{eqnarray*}
where we have used Lemma \ref{lemma} and the assumption that $\w_h$ is a modulus of continuity of $h$. Thus $f(\vp,\vp',\vp'')=-\vp'' -\|W\|_{L^\infty} |\vp'| -V\vp -w_h$ satisfies \eqref{eq 1.3 structure}. 

(2). Since the quasi-linear isotropic operator $F$ 

is invariant under rotations, we may choose an orthonormal basis $\{e_i\}_{i=1}^n$ with $e_1=\frac{x-y}{|x-y|}$ to simplify the calculations. 
With respect this basis, we have 

$$D^2_{x,y} \left(2\vp\left(\frac{|x-y|}{2}, t\right) \right) =\begin{pmatrix} P & -P \\
-P & P\end{pmatrix},$$

where 
\begin{equation*}
    P=\begin{pmatrix} 
    \frac 1 2 \vp''& 0 & \cdots & 0\\
0 & \frac{\vp'}{|x-y|} & \cdots & 0 \\
0 & 0 & \cdots & \frac{\vp'}{|x-y|}
    \end{pmatrix}.
\end{equation*}
Let
$$A=\begin{pmatrix}
\a(|\vp'|) & 0 & \cdots & 0\\
0 & \b(|\vp'|) & \cdots & 0 \\
0 & 0 & \cdots & \b(|\vp'|)
\end{pmatrix}$$
and 
$$C=\begin{pmatrix}
-\a(|\vp'|) & 0 & \cdots & 0\\
0 & \b(|\vp'|) & \cdots & 0 \\
0 & 0 & \cdots & \b(|\vp'|)
\end{pmatrix}.$$
It's easy to see that 
$\begin{pmatrix}
A & C \\
C & A
\end{pmatrix}$ is a positive semidefinite matrix.
Therefore, we have 
\begin{eqnarray*}
&& F(\vp e_1, Y) -F(\vp e_1, X) \\
&=& -\tr (AY)+\tr(AX) \\
&=& \tr\left[ \begin{pmatrix} A & C \\
C & A\end{pmatrix} \begin{pmatrix} X & 0 \\
0 & -Y\end{pmatrix}\right]  \\
&\leq&  \tr\left[ \begin{pmatrix} A & C \\
C & A\end{pmatrix} \begin{pmatrix} P & -P \\
-P & P\end{pmatrix} \right] \\
&=& 2\tr\left[ (A-C)P \right] \\
&=&  2 \a(|\vp'|)\vp''
\end{eqnarray*}
Thus, $f=-\a(|\vp'|)\vp''$ satisfies \eqref{eq 1.3 structure}. 

(3). This is a special case of Part (5). By Lemma \ref{lemma}, we have 
\begin{equation*}
    F(Y)-F(X) \leq -\l \tr(Y-X) \leq 2 \l \vp''.
\end{equation*}
So one can take $f(\vp'')=-\l \vp''$. 

For simplicity, we write $e=\frac{x-y}{|x-y|}$ in (4) and (5). 

(4) By Lemma \ref{lemma}, we have $X\leq Y$. In view of $v-r=2\vp>0$, properness of $F$ implies that 
\begin{equation*}
    F(t,r, \vp' e, Y) -F(t,v,\vp' e, X) \leq 0.
\end{equation*}
It follows that $f \equiv 0$ satisfies \eqref{eq 1.3 structure}. 

(5). Using the assumptions on $F$, we estimate that 
\begin{eqnarray*}
&& F(t,y,r, \vp' e, Y) -F(t,x, v,\vp' e, X) \\
&\leq& L|x-y| + F(t,x,r, \vp' e, Y) -F(t,x,v,\vp' e, X) \\
&\leq& L|x-y| +K(v-r) + F(t,x,v, \vp' e, Y) -F(t,x,v,\vp' e, X)\\
& \leq& L|x-y| +K(v-r) + \lambda(|\vp'|,t) \tr (X-Y) \\
&=& 2Ls + 2K\vp+ 2\lambda(|\vp'|,t) \vp'',
\end{eqnarray*}
where we used $|x-y|=2s$, $v-r=2\vp$ and Lemma \ref{lemma} in the last inequality. 
Thus $f(s,\vp,\vp',\vp'')=-\l(|\vp'|,t)\vp''-K \vp -Ls$ satisfies \eqref{eq 1.3 structure}. 


\end{proof}


Next, we provide a few more examples. 
\begin{prop}\label{prop 1D identify}
The following operators $F$ and $f$ satisfy the structure condition \eqref{eq 1.3 structure}. 
\begin{enumerate}
\item 
    \begin{equation*}\label{}
        F(p,X)=-\tr \left[ \left(I- \frac{p \otimes p}{1+|p|^2}\right) X \right],
\end{equation*}
    \begin{equation*}
        f=-\frac{\vp''}{1+(\vp')^2}.
    \end{equation*}
    
    
    

    
    
    
    
    
    
    
    
    
    
    \item
    \begin{equation*}\label{}
        F(p,X)=-\tr( A(p,t) X ),
\end{equation*}
where $A(p,t)=a_{ij}(p,t)$ and there exists a continuous function $\a(R,t)$ such that 
\begin{equation*}
0< \a(R,t) \leq R^2 \inf_{|p|=R, (v \cdot p) \neq 0} \frac{v^T A(p,t)v}{(v \cdot p)^2}
\end{equation*}
    \begin{equation*}
        f=-\a(\vp',t)\vp''.
    \end{equation*}
     
    
    
    
    
    
    
    
    
    
    
    
    
    
\end{enumerate}
\end{prop}

Theorem \ref{thm MC main} covers \cite[Theorem 2.1]{AC09} with the operators in part (1) and  \cite[Theorem 3.1]{AC09} with the operators in part (2). 

\begin{proof}[Proof of Proposition \ref{prop 1D identify}]
(1). This is a special case of part (1) in Proposition \ref{prop identify 1D} with $\a=\frac{1}{1+|p|^2}$ and $\b=1$. 

(2). The assumption implies $A(p,t) \geq \a(|p|,t) I$. So for $Z\geq 0$, we have 
\begin{equation*}
    F(p,X+Z)-F(p,X)=-\tr(A(p,t)Z) \leq -\a(|p|,t)\tr(Z). 
\end{equation*}
This becomes a special case of Part (5) of Proposition \ref{prop identify 1D}. 
\end{proof}
{"config": "arxiv", "file": "2006.16631/4OneD.tex"}