I am looking for an example of a continuous, non-affine function $u\colon X\to \mathbb{R}$ and a continuous, non-negative function $\epsilon\colon X\to\mathbb{R}_{\geq 0}$ such that the following hold (for $x,y,z\in X$ and $\alpha\in(0,1)$):
\begin{align}&\text{(i) If } u(x)\geq u(y)\text{ then } \epsilon(x)\geq\epsilon(y) \text{ and } \bigg( u(y)-u(z)> \epsilon(y) \implies u(x)-u(z)> \epsilon(x)\bigg)\newline&\text{(ii) } u(\alpha x + (1-\alpha)y)+\epsilon(x)\geq u(x)\iff u(y)+\epsilon(\alpha y + (1-\alpha) x)\geq u(\alpha y + (1-\alpha) x)\newline&\text{(iii) } \text{If $ u(x)<u(y)$ then there is $z\in X$ such that either $\bigg(u(z)-u(x)>\epsilon(z)\text{ and }u(z)-u(y)\leq\epsilon(z)\bigg)$ or $\bigg(u(y)-u(z)>\epsilon(y)\text{ and }u(x)-u(z)\leq\epsilon(x)\bigg)$holds.}\end{align}
where $X\subset\mathbb{R}^n$ is any convex set.Edit: I modified the conditions a little. They now look more complicated, but are a bit less restrictive. Also note that picking $\epsilon$ constant and $u$ affine would satisfy $(i)-(iii)$.