咨询：∑pᵢ≤∑log(1+axᵢ)等约束条件的凸性判定

Let’s break down exactly why your constraints form a convex set—this relies on basic properties of convex functions and sets, which we’ll unpack clearly:

1. Non-negativity Constraints ($p_i \geq 0$, $x_i \geq 0$) #

Each of these is a simple half-space constraint. For example, $p_i \geq 0$ includes all points where the $i$-th $p$ variable is non-negative. Half-spaces are inherently convex: take any two non-negative values, and any weighted average (convex combination) of them will also be non-negative.

Since we’re intersecting all these half-spaces (one for every $p_i$ and $x_i$), and intersections of convex sets are always convex, this part of your constraints is convex.

2. The Sum Inequality: $\sum_{i=2}^{n}p_i\leq \sum_{i=1}^{n-1}\log(1+ax_i)$ #

Let’s rewrite this constraint to make convexity easier to verify:
$$\sum_{i=2}^{n}p_i - \sum_{i=1}^{n-1}\log(1+ax_i) \leq 0$$
Let’s call the left-hand side $f(p, x)$. For the constraint $f(p, x) \leq 0$ to define a convex set, $f$ needs to be a convex function (because the lower level set of a convex function is always convex). Let’s split $f$ into two components:

Component 1: $\sum_{i=2}^{n}p_i$ #

This is a linear function of the $p_i$ variables. Linear functions are both convex and concave—they satisfy the convexity condition perfectly, since the line connecting any two points lies exactly on the function itself.

Component 2: $-\sum_{i=1}^{n-1}\log(1+ax_i)$ #

First, look at the individual term $\log(1+ax_i)$ where $a>0$ and $x_i\geq0$. Let’s check its convexity using the second derivative:

• First derivative with respect to $x_i$: $\frac{a}{1+ax_i}$
• Second derivative: $-\frac{a^2}{(1+ax_i)2}$

Since the second derivative is negative for all valid $x_i$, $\log(1+ax_i)$ is a concave function of $x_i$. When we multiply a concave function by -1, it becomes convex. Summing convex functions keeps the result convex, so $-\sum_{i=1}^{n-1}\log(1+ax_i)$ is convex.

Combining the Components #

$f(p,x)$ is the sum of two convex functions (Component 1 and Component 2). The sum of convex functions is always convex, so $f(p,x)$ is convex. This means the set of all $(p,x)$ where $f(p,x)\leq0$ is a convex set.

Final Conclusion #

Your full constraint set is the intersection of two convex sets:

• The convex set from all non-negativity constraints
• The convex set from the sum inequality

Since intersections of convex sets are always convex, your entire constraint system is convex.

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