minimum of a function $f(x_1,...x_n)$

Question

Let $A\in \mathbb R^{m\times n}, b\in \mathbb R^m, f(x)=\sum_{j=1}^n c_jx_j$ for $c\in\mathbb R^n$ and $M=\{x\in\mathbb R^n: Ax<b\}$ not empty and bounded. Is it true that $\min\{f(x)-\gamma\sum_{j=1}^m \ln((b-Ax)_j):x\in M\}\leq \inf\{f(x):x\in M\}$ for $\gamma>0$? I dont know how to prove or disprove it or can I say nothing about it?

Answer 1

It is not true that $$\inf\{f(x)-\gamma\sum_{j=1}^m \ln((b-Ax)_j):x\in M\}\leq \inf\{f(x):x\in M\}. \tag{1}$$

Let $n=1$, $m=2$, $A = (1, -1)^T$ and $b = (0.8, -0.1)$. Then, according to the comment, $Ax < b$ means $0.1 < x < 0.8$. Let $f = 0$ (by choosing $c_1 = 0$). Then, clearly $$\inf\{f(x):x\in M\} = 0,$$ but for all $x \in (0.1, 0.8)$, there is some $\delta > 0$ small enough that $$\log(0.8 - x) < -\delta < 0, \quad \log(-0.1 + x) < -\delta < 0,$$ which are the two terms appearing in the sum. It follows that for all $x \in M$ and $\gamma > 0$, $$f(x) - \gamma(\log(0.8 - x) + \log(-0.1 + x)) > 2\gamma\delta > 0.$$ Thus the infimum on the left-hand side of (1) is at least $2\gamma\delta$.