Is there a strictly increasing and strictly concave real-valued function defined over the entire real line?
I am looking for a real-valued function $f: \mathbb{R} \to \mathbb{R}$ that satisfies the following properties:
Strictly increasing: $f(x_1) < f(x_2)$ for all $x_1 < x_2$.
Strictly concave:$$f(\lambda x_1 + (1 - \lambda) x_2) > \lambda f(x_1) + (1 - \lambda) f(x_2)$$for all distinct $x_1, x_2 \in \mathbb{R}$ and $\lambda \in (0, 1)$.
unbounded i.e. for $x \to \infty$, also $f(x) \to \infty$.
In other words, I’m searching for a function that is strictly increasing and strictly concave across the entire real line $\mathbb{R}$, with no domain restrictions (i.e., the function should be well-defined for all real numbers, both positive and negative).
To clarify the context:
- The square root function $\sqrt{x}$ is strictly increasing and strictly concave but is not defined for negative real numbers.
- The function $f(x) = -1/x$ is strictly increasing and strictly concave for $x > 0$ but fails these properties for $x < 0$.
- The function $f(x) = -\exp(-x)$ is well-defined, strictly increasing, and strictly concave for all $x \in \mathbb{R}$, but it is bounded as $x \to \infty$.
Is there any function that meets all the criteria of being strictly increasing, strictly concave, and defined over the entire real line, while remaining unbounded? If no such function exists, I would appreciate a brief explanation of why that is the case.
Any help or insights would be greatly appreciated. Thanks!