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Prove that $e$ is the unique equilibrium point
This is an exercise from a past exam:
Let $U: X \rightarrow \mathbb{R} $ be a $\mathcal{C}^1$ function and let $\Omega\subset \mathbb{R}^n$ be an open, bounded set with $\partial\Omega = \{0\}$. Let $f\in C^1(\overline{\Omega},\mathbb{R})$ satisfies the following conditions:
$\langle f(x)-f(y),x-y\rangle \leq 0$ for all $x,y \in \overline{\Omega}$
$f$ is a homogeneous function of degree one
Prove that $e \in \Omega$ is the unique equilibrium point of $U$.
I am suppose to use this fact to find a solution to the following first order differential equation:
$\frac{dx}{dt}=f(x)-x$ for $x\in\Omega$.
The function $U(x) = e^x$ is convex in $x$ and $\langle f(x)-f(y),x-y \rangle \leq 0$ is equivalent to
$\frac{x-y}{\|x-y\|^2}\cdot (f(x)-f(y))\leq 0$
by the homogeneity of $f$. But I am unable to get anywhere with it, I can’t see how I can use that fact that $e \in \Omega$ is a unique equilibrium point to get that result. I have also tried it for the function $U(x) = e^{x^2}$, but even worse.
Could someone help me, please?
A:
Let $x$ be an equilibrium. Then $x$ is a solution of the initial value problem $x’=f(x)$ with $x(0)=x_0$. Since $x_0$ is in $\Omega$, by
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