关于二元函数$f(x,y)=\min(y(1-x),xya)$的凹性判定咨询

Great question! Let's break this down clearly: regardless of the value of $a \geq 0$, the function $f(x,y) = \min(y(1-x), axy)$ (with $x \geq 0, y \geq 0$) is not concave with respect to $(x,y)$. Here's a concrete proof using a counterexample:

Counterexample for any $a \geq 0$ #

When $x > 1$, $y(1-x)$ becomes negative (since $y \geq 0$), while $axy$ is non-negative (since $a,x,y \geq 0$). In this region, $f(x,y) = y(1-x)$, so we can use properties of this sub-function to find a violation of the concave function definition.

1. Choose two points:

   • $A = (2, 3)$: $f(A) = \min(3(1-2), a \cdot 2 \cdot 3) = \min(-3, 6a) = -3$ (since $6a \geq 0 > -3$)
   • $B = (4, 1)$: $f(B) = \min(1(1-4), a \cdot 4 \cdot 1) = \min(-3, 4a) = -3$ (same reasoning, $4a \geq 0 > -3$)

2. Compute the midpoint $C = 0.5A + 0.5B = (3, 2)$:

   • $f(C) = \min(2(1-3), a \cdot 3 \cdot 2) = \min(-4, 6a) = -4$ (again, $6a \geq 0 > -4$)

3. Check the concave function condition:
   A function is concave if for all $\lambda \in [0,1]$,
   $$f(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2) \geq \lambda f(x_1,y_1) + (1-\lambda)f(x_2,y_2)$$
   Substituting our values:

   • Left-hand side: $f(C) = -4$
   • Right-hand side: $0.5f(A) + 0.5f(B) = 0.5(-3) + 0.5(-3) = -3$

   Since $-4 < -3$, the condition fails. This means $f(x,y)$ does not satisfy the definition of a concave function.

Why this happens #

The function $f$ is the minimum of two functions:

• $g(x,y) = y(1-x)$: Its Hessian matrix is $\begin{pmatrix}0 & -1 \ -1 & 0\end{pmatrix}$, which has eigenvalues $1$ and $-1$—this makes it indefinite, so $g$ is neither concave nor convex.
• $h(x,y) = axy$: When $a > 0$, its Hessian matrix $\begin{pmatrix}0 & a \ a & 0\end{pmatrix}$ is also indefinite (eigenvalues $a$ and $-a$). When $a=0$, $h$ is linear (both concave and convex), but $g$ still introduces non-concavity in the $x>1$ region.

Since $f$ inherits the behavior of $g$ in regions where $g(x,y) < h(x,y)$, the non-concavity of $g$ propagates to $f$, making the entire function non-concave.

内容的提问来源于stack exchange，提问作者Frank Moses