# Problem relating convex sets and optimization

I am working on a microeconomics problem, but I have just kind of just boiled down to the following problem involving convex sets. I have a convex set of vectors in $\mathbb{R^n_+}$ of the form $\{x:u(x)\geq t \}$ where $u(x)$ is some function whose level curves are convex to the origin, and this set intersects with the set of vectors $\{x\in \mathbb{R^n_+}:p\cdot x\leq w \}$ for a given vector $p\in \mathbb{R^n_+}$ and some positive real number $w$. Let's call the intersection $B$. I need to show there exists exactly one vector in $B$ that maximizes the value of $x_1$. In $\mathbb{R^2_+}$ you can see the problem easily, and it is easily solved, but I am having big problems trying to generalize the solution to anything higher than $\mathbb{R^2_+}$. I've drawn a lovely picture for you of the problem in $\mathbb{R^2_+}$:

Like I said, I figured out a proof pretty easily in $\mathbb{R^2_+}$, but it falls apart in higher dimensions. I've figured out a few things that I think might be relevant. I know that $B$ is itself convex. I feel like I should be able to use this fact to prove it but I am not 100% sure how. I know that the problem is equivalent to finding the vector in $B$ which maximizes the length of the projection onto the $x_1$ axis. I'm not sure if this is a profitable way to think about it.

Does anyone have any advice, resources, anything that I can look at to try to prove this?

Thanks immensely, and let me know if I've left out any important information.

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What you want to prove is not true. It is true under the additional assumption that $u$ is continuous. Then $B$ is compact as a closed and bounded subset of $\mathbb{R}^n$ and the continuous function that projects $\mathbb{R}^n$ onto the first coordinate has a maximum on $B$. This means that there is vector in $B$ with a largest first coordinate. Convexity is not needed for this argument.

Now a counterexample in $\mathbb{R}^2_+$ with a somewhat pathological utility function. Define a utility function $u:\mathbb{R}^2_+\to\mathbb{R}$ by $$u(x_1,x_2)= \begin{cases} 0 & \text{if } x_1x_2 \leq 1,\\ 1 & \text{if } x_1 x_2>1. \end{cases}$$ This utility function represents convex preferences, for all $t$, the set $\{(x_1,x_2:u(x_1,x_2)\geq t)\}$ is convex. Now let $p=(p_1,p_2)=(1,1)$, $t=1$ and $w=3$. If there would be a bundle with a largest first coordinate, this bundle would certainly lie on the budget line $$\{(x_1,x_2)\in\mathbb{R}^2_+:p_1x_1+p_2x_2=w\}=\{(x_1,x_2)\in\mathbb{R}^2_+: x_1+x_2=3\}.$$ So you would have to find the largest value for $x_1$ such that $x_1(3-x_1)>1$ and there is no such value, since an open constraint can not bind locally.

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Here is a potential counter-example.

Let's suppose your convex set is $\{(x_1,x_2,x_3)\}$ where $0 \le x_i \le x_j+x_k$ for all possibilities of $i,j,k$, and that $p=(2,2,2)$.

Then if $w=24$, the maximum possible value of $x_1$ is $6$, but $(6,1,5)$, $(6,3,3)$, $(6,4,2)$ are each elements of $B$.

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I'm trying to figure out if this is actually a counterexample or if I've left something out of explaining the problem. I believe that I haven't described the convex set well enough. Vectors are bundles of goods and there is a utility function $u(x)$ with level curves that are convex to the origin. The "convex set" is the set $\{x:u(x)\geq t\}$ where t is some positive number. I'm not sure the set you came up with can be expressed as a level curve like that... I should make this more clear in my question I suppose. – crf Nov 29 '11 at 8:15

In ${\mathbb R}^n$ with $n>2$ your "level curves" will be level hypersurfaces. You will have to elaborate somewhat on the idea of "convex to the origin" and whether you can guarantee some strictness there. When you add an $x_3$-axis to your figure and have everything independent of $x_3$ then there will be infinitely many optimal points.

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