<p>Let <span class="math-container">$u(x,y)=f(x)+y$</span> be a quasilinear utility. Now we rotate it by 45 degrees, (such that the <span class="math-container">$x-$</span>axis becomes the direction of <span class="math-container">$(1,1)$</span>)</p>

<p><span class="math-container">$v(x,y)=f(x-y)+x+y$</span>. Is <span class="math-container">$v$</span> also a quasilinear utility? What is its name? It share many similarity with quasilinear utility such as parallel indifference curves. Being a rotation of quasilinear utility seems like the iff condition of parallel indifference curve.</p>

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2019-06-02T17:09:43.573Z
It's not clear what you mean by rotation by 45 degrees. $u$ has two arguments, so in which direction are you rotating it. In any case, a (differentiable) utility function $g(x_1,\ldots,x_n)$ is quasi-linear in $x_i$ if and only if $\frac{\partial g}{\partial x_i}$ is a constant. We have $\frac{\partial v}{\partial x}=f'(x-y)+1$, which is constant only if $f$ is linear. A similar point holds for $y$.
Ubiquitous
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2019-06-02T19:39:02.353Z
@Ubiquitous In my case, $\nabla_v u$ is constant for some directional vector $v$. If $v$ is a basis vector, then it reduces to your definition of quasilinear utility: $\nabla_i g$ for some $i$.
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Rotation of Quasilinear Utility
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# Rotation of Quasilinear Utility

(Asked by: username_0 on 2019-06-02)

Let $u(x,y)=f(x)+y$ be a quasilinear utility. Now we rotate it by 45 degrees, (such that the $x-$axis becomes the direction of $(1,1)$)

$v(x,y)=f(x-y)+x+y$. Is $v$ also a quasilinear utility? What is its name? It share many similarity with quasilinear utility such as parallel indifference curves. Being a rotation of quasilinear utility seems like the iff condition of parallel indifference curve.

**Tags:** microeconomics, utility, decision-theory, theory
> **Comments:**
> **username_1**: It's not clear what you mean by rotation by 45 degrees. $u$ has two arguments, so in which direction are you rotating it. In any case, a (differentiable) utility function $g(x_1,\ldots,x_n)$ is quasi-linear in $x_i$ if and only if $\frac{\partial g}{\partial x_i}$ is a constant. We have $\frac{\partial v}{\partial x}=f'(x-y)+1$, which is constant only if $f$ is linear. A similar point holds for $y$.  
> **username_0**: @Ubiquitous In my case, $\nabla_v u$ is constant for some directional vector $v$. If $v$ is a basis vector, then it reduces to your definition of quasilinear utility: $\nabla_i g$ for some $i$.  

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