$a\geq 0$, $x\sim y$ implies $x+a\sim y+a$ so the preference is linear? $\succsim$ is a countinuous and convex weak order.

$x,y,a$ are vectors in $\mathbb R^n$

We say $a\geq0$ if all directions of the vector $a$ is greater or equal to zero.

We want to prove (or disprove by counterexample) that: 

Suppose $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, 

Then the preference is linear. 



One definition of linear preference is that $x\sim y$ implies $x+a\sim y+a$ for any $x,y,a$.
This is not true.

Let $n=1$ and define $u(x)=\min{\{x,0\}}$. Let $\succsim$ be the preference relation represented by $u$. This preference relation is continuous and convex. We also have $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R$. But let $x=0$, $y=1$, and $a=-1$. Then $x\sim y$, but $y+a=0\succ -1=x+a$, thus $x+a\nsim y+a$ and $\succsim$ is not linear.