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I have a utility function $u(x,z)$ from $\mathbb{R}_+$ to $\mathbb{R_+}$, where $x,z \in \mathbb{R}_+$.

I would like to turn the following statement into math: "the utility function $u$ is increasing as the Euclidean distance between $x$ and $z$ is increasing".

Can I write $\frac{ d u(g) }{d g}>0$, where $g=d(x,z)\in \mathbb{R_+} $ is the Euclidean distance between $x$ and $z$?

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  • $\begingroup$ I think you mean $x,z\in\mathbb R_+$ in the first paragraph. $\endgroup$
    – Herr K.
    yesterday
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    $\begingroup$ $\frac{\mathrm du(g)}{\mathrm dg}>0$ makes sense, but if $g$ is the only argument in $u$, then you should probably change $u$'s domain from $\mathbb R_+^2$ to $\mathbb R_+$. Perhaps define another function $v(x,z)=u(g(x,z))$. $\endgroup$
    – Herr K.
    yesterday

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As Herr K. pointed out, if you write $\frac{du(g)}{dg}$, then $u$'s domain has to be $\mathbb{R}^+$ and $x,z$ have to be real numbers.

You'd rather ask:

When $u : \mathbb{R}^+ \to \mathbb{R}^+$ increases along with $d(x,z) := |x-z|$, can we say $\frac{du(g)}{dg} > 0$?

Yes, if $u(x)$ is differentiable at all points in its domain.

Since $\text{range}(\{d(x,z) : x,z \in \mathbb{R}^+\}) = \mathbb{R}^+$, $\frac{du(g)}{dg}$ is equivalent to writing $u'(x)$ where $x \in \mathbb{R}^+$. For you to be able to define this way, $u$ has to be differentiable throughout its domain which may not always be the case.

A better way to verify if $u$ increases with $d$ would be to check if $u$ is a (strictly) increasing function which doesn't require differentiability.

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  • $\begingroup$ Can you explain why $u$'s domain has to be $\mathbb{R}^+$? Does this not imply we are in a one-dimensional space? I would like to make the statement in respect to two vectors. i.e the Euclidean distance between 2 points in a two-dimensional space, $\mathbb{R}_+^2$. The distance would then be found using the square-root formula. $\endgroup$
    – Eli J
    yesterday
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    $\begingroup$ @EliJ For $x,z \in \mathbb{R}^+$, $g(x,z) = |x-z| \in \mathbb{R^+}$. When you have $g$ inside $u$, then the range of $g$ forms the domain of $u$. And the range is $\mathbb{R}^+$ as I wrote in the post. $\endgroup$
    – Kur_Kush
    yesterday
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    $\begingroup$ @EliJ As for using "two vectors" where each vector lies in the 2D plane, you can use the function $v(x,z) := (u \circ g)(x,z)$ where $v : \mathbb{R_+}^2 \times \mathbb{R_+}^2 \to \mathbb{R_+}$. $\endgroup$
    – Kur_Kush
    yesterday

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