# Multivariable concave function $(n - 1) f(x) >= \sum_{i=1}^{n} f(x_{-i})$

Define the multi-dimension concave function $$f(x): \mathbb{R}^n_+ \rightarrow \mathbb{R}_+$$ where $$x \in \mathbb{R}^n_+$$, here I use $$\mathbb{R}_+$$ to represent the range $$[0, \infty)$$ and we let $$f(\mathbf{0}) = 0$$ . Suppose we have a vector $$x^* = [1, ..., 1]$$, let $$x^*_{-i} = [1, ..., 0, ..., 1]$$ whose $$i^{th}$$ element is 0. Note the value of $$x^*$$ can be any positive value, I just use $$1$$ for convenience.

Let $$a = f(x^*)$$ and $$b = \sum\limits_{i=1}^{n} f(x^*_{-i})$$, I have an intuition that $$$$(n - 1) \cdot a \geq b$$$$ because of the concavity of $$f(x)$$. But I cannot think out a proof for this. Does anyone have some thoughts about doing this or prove my intuition is wrong?

• Can you prove it for $n=2$? – D.W. Nov 11 '19 at 20:30
• For $n = 2$, we need to prove $f(1,1) \geq f(0,1) + f(1,0)$. I can't think out a proof for it either... But from some examples like $f(x) = \sqrt{x \cdot y}$ which is a concave function, we can easily verify this is true. – Minbiao Nov 12 '19 at 16:35
• I think I forgot to mention one more constraint on the function $f(x)$, this concave function is constrained to be increasing. – Minbiao Nov 13 '19 at 17:15

## 1 Answer

Consider the function $$f(x, y) = 1 - e^{-(x + y)}$$. Now $$f(0, 0) = 0$$, $$f$$ is increasing and concave, since $$g(t) = -e^{-t}$$ is concave.

But $$f(1, 0) + f(0, 1) = 2(1 - e^{-1}) > 1 - e^{-2} = f(1, 1)$$, hence the claim doesn't hold even when $$n = 2$$.

The claim doesn't hold for convex functions either. $$f(x, y) = 2(x + y) + |x - y|$$ is one counterexample: $$f(0, 0) = 0$$, $$f$$ is increasing and convex, since

$$\alpha f(x_{1}, y_{1}) + \beta f(x_{2}, y_{2}) = \ldots + \alpha |x_{1} - y_{1}| + \beta |x_{2} - y_{2}| \geq \ldots + |\alpha(x_{1} - y_{1}) + \beta(x_{2} - y_{2})| = f(\alpha x_{1} + \beta x_{2}, \alpha y_{1} + \beta y_{2})$$

By the triangle inequality, but $$f(0, 1) + f(1, 0) = 6 > 4 = f(1, 1)$$.