# positivity of Lagrange multiplier on infinite dimensional vector spaces

I'm reading an economic article "Skewing the odds: Taking risks for rank-based rewards"

The author consider the following maximization problem: let $P$ be a continuous increasing function on $[0,\infty$) with $P(0)=\underline{\nu}$ and $P(\infty)=\overline{\nu}$ with $0\leq \underline{\nu}<\overline{\nu}<\infty$. Let $\mu \in (0,\infty)$ be a constant. Suppose in addition $P(\mu)<\overline{\nu}$. Then we consider the following optimization problem:

$$\begin{array}{r l} \text{maximize} & \int_0^\infty P(x)dF(x) \\ \text{subject to} & \int_0^\infty dF(x) \leq 1\\ & \int_0^\infty xdF(x)dx \le \mu. \end{array}$$

The meaning of $P$ is the contest payoff function, the second constraint is capacity constraint.

Then the associated Lagrangian is $$L(dF,\alpha,\beta) = \int_0^\infty [P(x) -(\alpha+\beta x)] dF(x) + \alpha+\beta.$$ Then by the Lagrange multiplier theorem, there exist $\alpha^0,\beta^0$ such that \begin{aligned} & \sup \left\{\int_0^\infty P(x)dF(x) : \int_0^\infty dF(x) \leq 1, \int_0^\infty xdF(x) \leq \mu \right\}\\ &=\sup_{dF\geq 0} L(dF,\alpha^0,\beta^0). \end{aligned}

Here is my question. According to the author, $\beta^0>0$. The reasoning is the following:

Since $P(\mu)<\overline{\nu}$, the contestant does not have sufficient capacity to gurantee the largest possible payoff $\overline{\nu}$. Thus, increased capacity has value. Hence the capacity constraint cannot be slack, i.e., the capacity constraint must be binding at the optimum and hence $\beta^0>0$.

I don't know what theorem does the author used. First, I thought that this is related to complementary slackness, but I'm not sure.

One of my colleague tried to prove this $\beta^0>0$ by using enveloping lemma(He is major in economics), but it seems that there is some mathematical issues on that argument.

How can I prove this?

• I can't say that I follow the author's discussion, but Aat a quick glance, there are many details the author would need to fill in to ensure that Luenberger's result can be applied (what is the normed space in question, what is the positive cone with interior point, etc.). – copper.hat Dec 24 '17 at 19:46

It is the Lagrangian Multiplier Theorem that guarantees $$\beta^0>0$$. Otherwise, $$\beta^0$$ should be $$0$$, and if that happens,
\begin{aligned} &M:= \sup \left\{\int_0^\infty P(x)dF(x) : \int_0^\infty dF(x) \leq 1, \int_0^\infty xdF(x) \leq \mu \right\}\\ &=\sup_{dF\geq 0} L(dF,\alpha^0,\beta^0) = \sup_{dF\geq 0} L(dF,\alpha^0,0) \\ &=\sup \left\{\int_0^\infty P(x)dF(x) : \int_0^\infty dF(x) \leq 1\right\}=:M'. \end{aligned} Note that $$M <\overline{\nu}$$. It is because if $$M=\overline{\nu}$$, then $$dF$$ must be supported on $$[x^*,\infty)$$ with full measure 1, where $$x^* := \inf\{x\geq 0|P(x)=\overline{\nu}\}$$, but then we must have $$\mu \geq x^*$$, contradicting $$P(\mu)<\overline{\nu}$$.
At the same time we notice that without the second constraint, $$M'$$ has the value $$\overline{\nu}$$ (just take $$dF$$ as dirac delta at arbitrary point $$x$$ and take $$x\to \infty$$.) This shows that $$M and $$\beta^0 =0$$ is contradicted.