$f(x-y+z) > f(x) - f(y) + f(z)$ for $x<y<z$ if $f$ is concave?

Question

While studying the original literature of the Hausdorff-Measure "Dimension und äußeres Maß" (1918), on page 168 there is given the inequality $$f(x-y+z) > f(x) - f(y) + f(z) \text{ for } x < y < z.$$ Since the publication is pretty long ago and it's written in a strange style, I'm not quite sure which conditions $f$ needs to fulfill. I guess it needs to be a concave (convex upwards) function, at least for simple examples it seems to work. But I can't find a proof for this statement.

I would be very happy if someone could say whether this statement is correct and in the best case could provide proof for it.

Answer 1

It is correct that the inequality holds for concave function. It is a special case of Karamata's inequality.

• If $y \ge x+z-y$ then $(z, x)$ majorizes $(y, x+z-y)$ because $z \ge y$ and $z+x = y + (x+z-y)$.
• Otherwise $(z, x)$ majorizes $(x+z-y, y)$.

In both cases, for a concave function $f$ it follows that $$ f(z) + f(x) \le f(y) + f(x+z-y) \, . $$ If $f$ is strictly concave then the inequality is strict.

---

You can also derive it directly by writing down the concavity condition for $x < y < z$ $$ f(y) \ge \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(z) $$ and for $x < x+z-y < z$ $$ f(x+z-y) \ge \frac{y-x}{z-x}f(x) + \frac{z-y}{z-x}f(z) $$ and add these two inequalities. Again, the inequalities are strict if $f$ is strictly concave.