I am stuck at a very simple question. Let $V(y)$ be the set of all $x \in \mathbb{R}^n$ that can produce at least $y$. We are given that $V(y)$ is convex set.
Given $w$, factor prices, let
$$x^* = \arg \min wx$$ $$\text{such that } x \in V(y)$$
Show that $x^* \notin V(y’)$ for $y’>y$
I tried using contradiction but only proved that if the above doesn’t hold then $x^*$ can produce infinite $y$. Can I claim that as a contradiction (to what?) and say that proof is complete?
I don't think you need convexity. However, I think you do need to assume some monotonicity condition. The following should work (but might not be the minimal set of assumptions that provides the result).
Consider the production possibility set $V(.)$. $$ V(y) = \{x \in \mathbb{R}^n_+| x \text{ can produce } y\}. $$ We assume that $V(y)$ is a closed non-empty subset of $\mathbb{R}^n_{+}$. Let $w \in \mathbb{R}^n_{++}$ and define: $$ c(y) = \arg\min_x wx \text{ s.t. } x \in V(y). $$ As $V(y)$ is closed, this problem is well defined. The value of $c(y)$ gives the minimal cost to produce $y$. Define $X(y) = \{x \in V(y)| w x = c(y)\}$ as the set of all optimal solutions.
Assumption 1: If $y' >y$, then $V(y') \subseteq V(y)^\circ$ where $A^\circ$ is the interior of the set $A$ (relative to $\mathbb{R}^n_+$).
Assumption 2: if $y > 0$ then $0 \notin V(y)$.
Assumption 1 requires the production possibility sets to be strictly nested. Assumption 2 requires that we can not produce something from nothing.
Lemma 1: If assumptions 1 and 2 are satisfied, then $y' > y$ implies $c(y') > c(y)$.
Proof: Let $y' > y$. And let $x^\ast \in X(y')$. Then as $x^\ast \in V(y')$ we have by Assumption 1 that $x^\ast \in V(y)^\circ$. Then we know there is a $\varepsilon > 0$ such that $B_\varepsilon(x^\ast) \cap \mathbb{R}^n_{+} \subseteq V(y)$. As $x^\ast \ne 0$ (by Assumption 2), we can find an $x' \in B_\varepsilon(x^\ast) \cap \mathbb{R}^n_+$ such that $x' < x$ and $x' \in V(y)$. Then: $$ c(y) \le w x' < w x^\ast = c(y') $$ which demonstrates the proof.
Theorem 2: Let $y' > y$ and $x^\ast \in X(y)$ then $x^\ast \notin V(y)$.
Proof: Towards a contradiction assume that $y' > y$, $x^\ast \in X(y)$ and $x^\ast \in V(y)$. Then $c(y) = w x^\ast$ and $c(y') \le w x^\ast$. However, this implies that: $$ c(y') \le w x^\ast = c(y), $$ which contradicts Lemma 1.
An attempt with adding an extra assumption:
Let $0 \in V(0)$ and $0 \notin V(y), \forall y>0$
Since $V(y)$ is set of all $x$ that can at least produce $y$, we have that $V(y’) \subseteq V(y), \forall y’>y$
Now let $x^*$ (as defined in question) $\in V(y’)$.
From definition: $wx^* \leq wx, \forall x \in V(y)$. Therefore, we also have that
$$wx^* \leq wx, \forall x \in V(y’)$$
So, $$x^* = \arg \min wx$$ $$\text{such that } x \in V(y’)$$
Now this can be extended backward as well. Say $y’’<y$ and
$$x’’ = \arg \min wx$$ $$\text{such that } x \in V(y’’)$$
Following the logic from above, $x’’$ also minimizes cost for $y, y’$.
Taking it all the way back, it must be that:
$$x^* = \arg \min wx \,\,\, \forall x \in V(0)$$
However, due to the additional assumption (inaction), minimum cost for producing 0 is 0. So $x^*=0$ (for $w>0$), which is also a contradiction to the assumption stated below.
