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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be convex decreasing function. I read from somewhere that we have the following inequality:

For any $x\leq y$ and $t\geq 0$, $f(x+t) - f(x) \leq f(y +t) -f(y)$.

So I am wondering whether we can extend the inequality into $n$-dimensional case. That is, let $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex and decreasing function. Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ and $\mathbf{x}\leq \mathbf{y}$ (component-wise comparison). Let $\mathbf{t} \geq \mathbf{0}$. Does the following inequality hold?

$$g(\mathbf{x} + \mathbf{t}) - g(\mathbf{x}) \leq g(\mathbf{y} + \mathbf{t}) - g(\mathbf{y}).$$

Note that $g$ is a decreasing function means that $g(\mathbf{x}) \geq g(\mathbf{y})$ for $\mathbf{x}\leq \mathbf{y}$ (component-wise comparison).


Some updates:

I have been thinking about this problem for a few days. Let's consider a simpler instance.

Let $f(x_1,x_2)$ be a function of two variables. It was proved that $f$ is twice-differentiable, convex, and decreasing (element-wise comparison). Condier the following inequality:

$$f(2,3) + f(3,2) \leq [f(2,2) + f(3,3)]$$ Can we prove that the above inequality always holds given only that $f$ is convex and decreasing? Or it depends on the specific form of the function $f$?

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  • $\begingroup$ it's $\mathbf{t} \geq \mathbf{0}$. Thanks. $\endgroup$
    – Paradox
    Jun 30, 2018 at 17:16

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