Preference relation value representation without uncertainity Let $\succeq $ be a weak order in $\mathbb{R}^n$. If


$ x\geq y \Rightarrow x\succeq y$
for any $x\succ y\succ z$ there exists a unique $\lambda \in (0,1)$ such that $y\sim \lambda x+(1-\lambda)z$, 


then there exists a value representation of $\succeq$.

Could anyone give me an example where the Value representation theorem(with out uncertainity) in the $n$ dimensional Euclidean space can be applied and a counter example that cannot be applied?
First, consider the preference relation $\succeq$ defined by
\begin{equation*}
x \succeq y \text{ if } x_1 + \cdots + x_n \geq y_1+\cdots+y_n
\end{equation*}

This preference relation satisfies your assumptions. Indeed, suppose that $x \succ y \succ z$. You can check that $y \sim \lambda x + (1-\lambda) z$ for a unique $\lambda$ defined by
\begin{equation*}
\lambda = \dfrac{y_1+\cdots+y_n-(z_1+\cdots+z_n)}{x_1+\cdots+x_n-(z_1+\cdots+z_n)} 
\end{equation*}

And these preferences obviously admit the utility representation $u(x)=x_1+\cdots+x_n$.

Now, consider the lexicographic preference relation $\succeq$ defined (I take $n=2$ for simplicity) by
\begin{equation*}
x \succ y \text{ if } (x_1&gt;y_1) \text{ or } (x_1=y_1,x_2&gt;y_2)
\end{equation*}
Lexicographic preferences have two important properties:


$x \sim y$ if and only if $x=y$
they have no utility representation


If $x=(1,1)$, $y=(0,1)$ and $z=(0,0)$, we have $x \succ y \succ z$. In addition, any $\lambda&gt;0$ is such that $\lambda x + (1-\lambda) z \succ y$. Thus assumption 2. is not valid, and the conclusion of the theorem (existence of a utility representation) is not valid either.