Someone gave me a proof of this, but I am not sure if it is correct.

Let $B(p,w) = \{x: p\cdot x \leq w\}$ (the budget set). Then:

\begin{align} x(p,w) &= \arg \max_{x\in B(p,w)} u(x)\\ &=\arg \max_{\alpha x\in \alpha B(p,w)} u(\alpha x) \\ &=\arg \max_{y\in B(p,\alpha w)} u(y) \\ &=\frac{1}{\alpha} \arg \max_{y\in B(p,\alpha w)} u(y) \\ &=\frac{1}{\alpha} x(p, \alpha w) \end{align} Where the result follows from taking $\alpha=\frac{1}{w}$.

Is this proof correct (I am not sure of the middle three equalities)? Where is homotheticity used?

EDIT: A monotone preference relation $\succsim$ on $X= \mathbb{R}^{L}_{+}$ is homothetic if all indifference sets are related by proportional expansion along rays; that is, if $x \sim y$, then $\alpha x \sim \alpha y$ for any $\alpha \geq 0$.

Also, recall a continuous $\succsim$ on $X = \mathbb{R}_{+}^{L}$ is homothetic iff it admits a utility function that is homogenous of degree one; $u(\alpha x) = \alpha u(x)$.

  • $\begingroup$ It is used in the step between equations 2 and 3. But if you don't believe the guy that his proof is correct why would you believe me...? $\endgroup$
    – Giskard
    Commented Jan 19, 2016 at 12:56
  • $\begingroup$ @denesp Like I said I am unsure of the middle three steps. I am looking for someone to rationalize these steps so I can confirm for myself that the proof is correct. (In addition, you specifically have provided some useful answers for me before! I already trust you more than the other guy...) $\endgroup$
    – möbius
    Commented Jan 19, 2016 at 13:02
  • $\begingroup$ Upon rereading the proof I now think homotheticity is used in several places. Can you edit your question to show the exact definition of homothetic functions you use? $\endgroup$
    – Giskard
    Commented Jan 19, 2016 at 13:42
  • $\begingroup$ @denesp I have edited as requested using the only definitions I have studied for homotheticity. Hopefully it is sufficient. $\endgroup$
    – möbius
    Commented Jan 19, 2016 at 16:25

1 Answer 1


An indirect proof. Suppose $$ x(p,w) = w\cdot x(p,1) $$ does not hold. This is equivalent with stating $$ U(x(p,w)) \neq U(w\cdot x(p,1)). $$ (To be precise: $x(p,w)$ and $x(p,1)$ may be set valued. In this case we are talking about two elements at least one of which is not included in both sets.)

Case 1.
$$ U(x(p,w)) > U(w\cdot x(p,1)) $$ As $U$ is homothetic $$ U(x(p,w)) = U(w \cdot \frac{1}{w}\cdot x(p,w)) = w \cdot U(\frac{1}{w}\cdot x(p,w)). $$ Using this we have $$ w \cdot U(\frac{1}{w}\cdot x(p,w)) = U(x(p,w)) > U(w\cdot x(p,1)) = w\cdot U(x(p,1)) $$ and thus $$ U(\frac{1}{w}\cdot x(p,w)) > U(x(p,1)) $$ However as $\frac{1}{w} \cdot x(p,w)$ is clearly an element of $B(p,1)$ this is impossible as $x(p,1)$ gives maximal utility in that budget set.

Case 2.
$$ U(x(p,w)) < U(w\cdot x(p,1)) $$ As $w \cdot x(p,1)$ is clearly an element of $B(p,w)$ this is impossible as $x(p,w)$ gives maximal utility in that budget set.

Thus we have proven that $$ U(x(p,w)) = U(w\cdot x(p,1)) $$ which is equivalent with $$ x(p,w) = w\cdot x(p,1). $$


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