Consider the partial Loewner order $\le\_L$ for symmetric matrices: let $A,B$ be symmetric matrices of the same dimension, we say $A\le\_L B$ if $B-A$ is positive definite. Now let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a Lipschitz function, i.e., there exists $L\ge 0$ such that $ |f(x)-f(y)|\le L|x-y|$. I was wondering whether the following statement holds: there exists $C\ge 0$ such that for all $\epsilon\in (0,1]$ and $x\in \mathbb{R}^n$,
$$
f(x)x^\top+xf(x)^\top \le\_L \epsilon I+\frac{C}{\epsilon} xx^\top.
$$




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The claim holds for $n=1$. In this case, $\le\_L$ is equivalent to the Euclidean order. By the Lipschitz continuity of $f$, $ f(x)\le |f(0)|+L|x|$ for all $x\in \mathbb{R}$. Then
by Young's inequality, for all $\epsilon\in (0,1]$,
\begin{align\*}
xf(x) \le |f(0)||x|+L|x|^2\le \epsilon+ \left(L+\frac{|f(0)|^2}{4\epsilon}\right)x^2
\le \epsilon+ \frac{4L+|f(0)|^2}{4\epsilon}x^2.
\end{align\*}
It is not clear to me how to extend the argument to a multidimensional setting.



Lipschitz continuity and quadratic growth in Loewner order
$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.


The inequality in question is
$$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$
for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.


To obtain a contradiction, suppose that this is true. Take any $t\in(0,1)$. For all $x=(x\_1,x\_2)\in\R^2$, let $f(x):=f\_t(x):=y+tx$, where $y:=(-x\_2,x\_1)$, so that $|y|=|x|$ and $y$ is orthogonal to $x$. Then $f$ is $2$-Lipshitz.


For each $x\in\R^2$, let now $u=f(x)$.
Then \eqref{1} becomes
$$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$
For any nonzero $x\in\R^2$, dividing both sides of this inequality by $t|x|^4$ and then
letting $|x|\to\infty$, we get
$$2(1+t^2)\le\frac C\ep\,t$$
for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$


