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I have been having trouble with how to go forward with a proof for about three days now. I know the basic structure of the proof, but can't seem to construct it.

Basically, I am trying to do a proof by contradiction for the following:

Say $u: x \rightarrow \mathbb{R}$ has no local maxima. Let $p \in \mathbb{R^l}_{++}$ and $w>0$. Show that if $x^*$ is a solution to the maximization problem:

$$\max_x \ (u(x)) \ \text{s.t.} \ x \in B(p) = [x \in \mathbb{R^l}_{++} : p \cdot x \leq w]$$ then for all $y \in \mathbb{R^l_{++}}$ such that $u(y) \geq u(x^*)$, then it must be that $p \cdot y \geq w$

So I'm supposed to do this proof by contradiction, (suppose we have $y \in \mathbb{R^l_{++}}$ such that $u(y) \geq u(x^*)$, then it must be that $p \cdot y < w$) and use the fact that $u$ doesn't have a local max implies that the function $u$ is locally non-satiated:

$$\forall y \in \mathbb{R^l_{+}, \forall \epsilon > 0, \exists {y'} \in \mathbb{R^l_{+}}} \ \text{s.t.} \ \|y - y'\| < \epsilon \ \text{and} \ y' \succ y$$

But I've been stuck for a while now. Any help would be appreciated.

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There, I changed the condition we want to prove. It now says show then it must be $p \cdot y \geq w$ – Kitsune Cavalry Sep 30 at 0:05
You are correct. Thanks a lot! – Kitsune Cavalry Sep 30 at 1:12

2 Answers 2

up vote 2 down vote accepted

Assume for a contradiction that there exists $y$ such that $p\cdot y<w$ and $u(y)\geq u(x^\ast)$. Let $$\epsilon^\ast= \frac{w-p\cdot y}{ \sum_i p_i}.$$ Then, for all $y'$, if $\ \|y - y'\| < \epsilon^\ast$, we would have $\left|y_i-y_i'\right|<\epsilon^\ast$. Notice that the cost of any bundle $y'$ would satisfy $$ \begin{eqnarray} p\cdot y' &=& p_1 y'_1 + p_2 y'_2 + \dots +p_n y'_n\\ &<&p_1 (y_1+\epsilon^\ast) + p_2 (y_2+\epsilon^\ast) + \dots +p_n (y_n+\epsilon^\ast)\\ &=& p\cdot y+\epsilon^\ast\sum_i p_i \\ &=&p\cdot y + \frac{w-p\cdot y}{ \sum_i p_i} \sum_i p_i\\ &=& w \end{eqnarray} $$ which shows that for $\epsilon^\ast$, every $y'$ satisfying $\ \|y - y'\| < \epsilon^\ast$ is affordable.

However, since $u$ is locally non satiated, there exists $y^\ast$ with $\ \|y - y^\ast\| < \epsilon^\ast$ such that $y^\ast\succ y$. Since $y^\ast$ is affordable, i.e. $y^\ast\in B(p)$ this leads to a contradiction with the assumption that $x^\ast$ solves the utility maximization problem because transitivity implies that $y^\ast\succ x^\ast.$

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This looks really good. Thank you very much. How did you think to set $\epsilon$ equal to that expression though? I always like doing these proofs differently where you find a value of $\epsilon$ > 0 at the end that causes the contradiction. So doing it this way always fascinates me. – Kitsune Cavalry Sep 29 at 5:09
At the expense of discounting my effort, I should admit that this might be one of the questions that most likely an econ phd would have seen a version of. Well, I probably earn humility points for addressing that, which is not totally bad. And for your question, I would say the method is the same, reverse engineering. But when you reverse engineer your reverse engineering, then it starts looking like Memento, which is cool, but nothing more than an illusion. – ramazan Sep 29 at 5:20
Fair enough. I'm currently a Masters student and the rest of the proofs I've had to do so far were not bad. This one was a whole other doozy. I've also figured out how to engineer epsilon for myself, so that's all settled. Thanks a lot once again! – Kitsune Cavalry Sep 29 at 5:26
Somewhere in there you write "since $u$ is locally non-satiated..." This means that local non-satiation is used as a premise. But it is the very property that you want to prove. What am I missing? – Alecos Papadopoulos Sep 29 at 19:40
You have a valid point. However, the question is phrased in a way that u has the local non-satiation property, which is why that part of the proof is skipped. On the other hand, showing that is straightforward since it is almost by definition that not having local non-satiationness implies no local maxima. – ramazan Sep 29 at 20:41

I am not sure how the contradiction method of proof should work here. Here is my take:

Ad absurdum, assume that $x^*$ solves the utility maximization problem, and that we have $y \in \mathbb{R^l_{++}}$ such that $u(y) \geq u(x^*)$, and that that $p \cdot y < w$. So $y$ is feasible, and since $u()$ represents a (rational and continuous) preference relation, $u(y) \geq u(x^*) \implies y \succeq x^*$.

Moreover, there is some budget left. Since $x^*$ is not equal to the zero vector, at least one of the elements in the bundles must be a good (not all of them can be bads), say the first one, $y_1$. Define the bundle $y'$ to be equal to $y$ bar $y_1$ and set the quantity $y_1'>y_1$ such that $p \cdot y' = w$.

So $y'$ is feasible, and $y'> y \implies y' \succeq y$ since $y_1$ is a good (but we do not assume that local non-satiation holds, and this is why the preference relation is weak. $y_1$ being a good only guarantees that increasing its quantity does not diminish utility). So also $y' \succeq x^*$.
Since by assumption $x^*$ solves the utility maximization problem, it has to be the case that $u(y') \leq u(x^*) \leq u(y)$.

So, under the ad absurdum assumption, we have that

$$ y' > y, \;\;\;\; u(y') \leq u(y) \tag{1}$$

Note that this relation that we ended up with, does not involve anymore anything related to the utility maximization problem and/or the budget constraint. It is a conclusion describing a property of $u()$ as a function, under the ad absurdum assumption. Note also that $y'$ is not unique -we can form many such bundles, and if we accept that other elements of the bundles are also goods and not bads, we can spread the remaining available budget to increase more than one element...

Concluding the proof by showing that $(1)$ (and how it can be further extended) contradicts the premise that $u()$ has no local maxima is now easy.

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I'll get back to this one Wednesday hopefully. – Kitsune Cavalry Sep 30 at 0:18
I'm not really smart enough to completely figure whether this makes sense, but I think the intuition is right. I like using epsilon proofs since I had a little math background, but this is cool too. Thank you! – Kitsune Cavalry Oct 1 at 5:01

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