# Seemingly simple consumer theory problem

A function $$v: \mathbb{R}^K_+ \xrightarrow{} \mathbb{R}_+$$ is is said to be a valuation function if

1. The value of function $$v$$ at $$x = \textbf{0}$$ is $$0$$: $$v(\textbf{0}) = 0$$
2. $$v$$ is continuous on the domain $$\mathbb{R}^K_+$$, strictly increasing and strictly concave on $$\mathbb{R}^K_{++}$$
3. For any $$p \in \mathbb{R}^K_{++}$$, the set $$A(p) = \{x \in \mathbb{R}^K_+| v(x) \geq p \cdot x\}$$ is compact, the set $$A^*(p) = A (p)-\{\bf{0}\}$$ is nonempty and lies inside $$\mathbb{R}^K_{++}$$.

Show that a if $$v$$ is a valuation function, for every $$p \in \mathbb{R}^K_{++}$$, for every $$x \in \mathbb{R}^K_{++}$$, there exist $$t>0$$ small enough that $$v(tx) > p\cdot tx$$

In other word, for every $$p>\bf{0}$$, if you zoom the set $$A(p)$$ sufficiently close at the origin, it contains all the points near the origin in first quadrant.

Remark 1: This is the problem I reduce from a microeconomics problem I am working on. I want to extend the Inada condition to multivariate case in a topological way. My prototype function is $$f(x,y)=x^{\alpha}y^{\beta}$$ where $$\alpha+\beta <1, \alpha>0,\beta>0$$. This family of functions satisfy (1),(2),(3) and the property I want to prove, say (4). I am trying to show either (1),(2),(3) => (4) or should I include (4) as an additional axiom?

Remark 2: A few examples of $$A(p)$$

1. An example of A(p) when p is small 1. An example of A(p) when p is large (4) is demonstrated by the geometric fact that the contours approach vertical and horizontal axis near $$0$$.

It follows indeed from the first three conditions, though I did not find a simple proof. Here is a messy one:

Observe first that, $$v(tx)>p\cdot tx$$ with $$t>0$$ is equivalent to $$v(tx)/t>px$$. By replacing $$p$$ by some multiple, we can see that the right-side can take any value without changing the left value. The function $$t\mapsto v(tx)$$ is strictly concave, strictly increasing, and has value $$0$$ at $$0$$. The problem is then equivalent to showing that its right derivative is infinite for all $$x\gg 0$$. That the right derivative exists follows from the concavity of $$v$$.

As it turns out, it is enough to find a single $$x^*\gg0$$ such that $$t\mapsto v(tx^*)$$ has an infinite right derivative at $$0$$.

To see this, let $$x\gg 0$$. There is some $$\delta>0$$ such that $$x\gg \delta x^*$$. Take some $$K>0$$. For $$t>0$$ small enough, we have $$v\big(\delta t~x^*\big)/(\delta t)>K/\delta$$ or $$v(\delta t x^*)/t>K.$$ Since $$v$$ is strictly increasing on the interior, $$v(tx)/t>v(t\delta x^*)/t>K$$ for $$t>0$$ small enough. Therefore, the function $$t\mapsto v(tx)$$ must have infinite right derivative at $$0$$.

It remains to find the damn $$x^*$$. Let $$\Delta$$ be the $$K-1$$-dimensional unit simplex. For each $$N$$, there must exist some $$x\in\Delta$$ such that the function $$t\mapsto v(tx)$$ has a right derivative of at least $$N$$ at $$0$$. Otherwise, for $$p=(N,N,\ldots,N)$$ we would have $$A(p)-\{0\}=\emptyset.$$ Also, $$A(p)-\{0\}\subseteq\mathbb{R}^K_{++}$$ implies that such an $$x$$ must be positive in each coordinate. So let $$x_n\in\Delta$$ be such that $$t\mapsto v(tx_n)$$ has a right derivative of at least $$n$$ at $$0$$. Now let $$x^*=\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}.$$ That the relevant series converges follows from any norm being bounded on the compact set $$\Delta$$ and absolute convergence implying convergence. Clearly, $$x^*\gg 0$$ since $$x_{2^i}\gg 0$$ for all $$i$$. It remains to show that $$t\mapsto v(tx^*)$$ has an infinite right derivative at $$0$$. Since $$t^{-1}~v\bigg(t\sum_{i=1}^\infty\frac{1}{2^i}x_{2^i}\bigg)>t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)$$ for all $$n$$ by $$v$$ being strictly increasing on the interior, it suffices to show that for each $$K>0$$, there exists some $$n$$ such that $$t^{-1}~v\bigg(t\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)>K$$ for $$t>0$$ small enough. Let $$S_n=\big(\sum_{i=1}^n 1/2^i\big)^{-1}$$ and note that $$t^{-1}v\bigg(tS_n\sum_{i=1}^n\frac{1}{2^i}x_{2^i}\bigg)=t^{-1}v\bigg(\sum_{i=1}^n S_n\frac{1}{2^i}t x_{2^i}\bigg)\geq\sum_{i=1}^n S_n\frac{1}{2^i}t^{-1}v(tx_{2^i})$$ by the concavity of $$v$$. Since $$\lim_{t\downarrow 0}t^{-1}v(t x_{2^i})\geq 2^i$$, this sum is almost $$S_n n$$ for $$t>0$$ small enough. Now $$S_n$$ converges to $$1$$, so $$S_n n$$ is larger than $$K$$ for $$n$$ large enough and $$t>0$$ small enough.

• +1 for finding "the damn $x^*$." Mar 29, 2021 at 16:41
• The construction of $x^*$ is very clever. The statement of existing $x$ that make the map $t \to v(t x)$ has arbitrarily large right derivative at 0 is simple. But the statement of existing $x$ that make the map $t \to v(t x)$ has infinite right derivative at 0 is not simple. Thanks for a very nice proof. Mar 30, 2021 at 6:18
• @KhánhToàn if the answer answered your question do not forget to accept it (you can do that by clicking on the tick under votes)
– 1muflon1
Mar 30, 2021 at 7:15
• @1muflon1 Oops I never know about that. Thanks for telling me. Mar 30, 2021 at 7:53