Type of convex function?

Question

I want a convex function $f:\mathbb{R} \to \mathbb{R}$ with the following property: given points $x,d \in \mathbb{R}$, and $\alpha \in (0,1)$, we have $$f(x + \alpha d) \geq \alpha f(x + d).$$

Is this possible? If so, what are some assumptions on $f$ that can be made to ensure this (the more general the better)?

Thanks!

Answer 1

The condition essentially says:

For any $x$ and $y$, the line from $(x,0)$ to $(y,f(y))$ is strictly below the function.

or, more elegantly

Any line segment from the $x$-axis to the graph of the function is strictly below the function.

Suppose we fixed some point $(x,f(x))$ on the graph. We need that the line from this to anywhere on the $x$ axis is strictly below the graph - however, by moving the point on the $x$ axis towards $\infty$, we can give the line arbitrarily small slope in either direction - implying that $f(x)$ must be bounded below by lines of arbitrarily small slope and therefore be constant.

To be rigorous, suppose $f(x_1)=c_1$ and $f(x_2)=c_2$ for $c_1> c_2\geq 0$ (since we can easily show $f$ is non-negative by letting $\alpha$ go towards $0$). We are going to draw a line through $(x_1,c_1)$ and $(x_2,\frac{c_1+c_2}2)$ and show that this must be a lower bound, but obviously isn't. In particular, to ensure rigor, we choose $x$, $\alpha$, and $d$ and use your property. In particular: $$d=\frac{2x_1-2x_2}{1-\frac{c_2}{c_1}}$$ $$\alpha = 1-\frac{1-\frac{c_2}{c_1}}{2}$$ $$x=x_1-d$$ where, it is clear that $1>\frac{c_2}{c_1}\geq 0$ and hence that $\alpha\in(0,1)$. Then, when we plug things in, we get $$f(x+\alpha d)\geq \alpha f(x+d)$$ $$f(x_2)\geq \alpha f(x_1)$$ $$c_2 \geq \alpha c_1$$ $$c_2 \geq c_1-\frac{c_1-c_2}2$$ $$c_2 \geq \frac{c_1+c_2}2$$ and multiplying by $2$ and shuffling terms gives $$c_2 \geq c_1$$ contradicting hypothesis. Therefore no pair of points $x_1,x_2$ where $f(x_1)>f(x_2)$ may exist, implying that $f$ is constant and non-negative.