Quasi-Concave Function $u(x,y)$ has Decreasing Marginal Rate of Substitution

22 Views Asked by At

Let $u:\mathbb{R}^2 \to \mathbb{R}$ be $C^2$, strictly increasing in both arguments, and strictly quasi-concave: $\{(x,y)\vert u(x,y) \leq \bar{u}\}$ is strictly convex for all $\bar{u}$.

By the Implicit Function Theorem, for any $x_0,y_0$, and $u_0=u(x_0,y_0)$, we have (for $x$ near $x_0$) $y_{u_0}^\prime(x) = -\frac{u_x(x_0,y_0)}{u_y(x_0,y_0)}$. This justifies the definition of the marginal rate of substitution between $x$ and $y$ at $(x_0,y_0)$ as $$\text{MRS}(x_0,y_0) \equiv \frac{u_x(x_0,y_0)}{u_y(x_0,y_0)}.$$

I believe the above assumptions should imply $\text{MRS}(x_0,y_0)$ is strictly decreasing in $x_0$.

In other words, the slope of the $u$-indifference curve at $(x_0,y_0)$ levels-off as we increase $x_0$.

But my question is how to show this?

Do I use the characterization of quasi-concavity in terms of positive-definite Hessian $\nabla^2 u(x,y)$ restricted to the space orthogonal to $\nabla u(x,y)$?

Thank you.