Local Non-Satiation Proof

Question

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.

Answer 1

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 $y^\ast\succ x^\ast.$

Answer 2

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.