Is $U^{-1}(U(x)+U(y))$ a convex function in general?

Question

Let $U(x)$ be a positive, strictly increasing, strictly convex $C^2$ function in $x$, is it generally true that $U^{-1}(U(x)+U(y))$ is a convex function in $x,y$ ? For $U(x)=e^x$, it is well known that the log-sum-exponential function is convex. The statement is also true for power functions $U(x)=x^{1+\beta}$. What about the general case?

Answer 1

Here is a counterexample.

Let $U(x) := \mathrm{e}^{\sqrt{x}}$. We have $$U'(x) = \frac{1}{2\sqrt{x}}\mathrm{e}^{\sqrt{x}}, \quad U''(x) = \frac{1}{4x^{5/2}}\mathrm{e}^{\sqrt{x}}(x^{3/2} - x).$$ Thus, $U(x)$ is positive, strictly increasing, strictly convex on $x > 1$.

We have $U^{-1}(x) = \ln^2 x$. We have $$f(x, y) = U^{-1}(U(x) + U(y)) = \ln^2 \left(\mathrm{e}^{\sqrt{x}} + \mathrm{e}^{\sqrt{y}}\right).$$We have $$\frac{\partial^2 f(x, 3/2)}{\partial x^2}\Big\vert_{x=3/2} < 0.$$

Note that $U(x) := \mathrm{e}^{\sqrt{x}}$ satisfies $U'''(x) U'(x) - 2[U''(x)]^2 > 0$ on $(1, 3)$. See below.

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Some thoughts.

Assume that $U'''(x)$ exists. If $U'''(x) U'(x) - 2[U''(x)]^2 > 0$, then $U^{-1}(U(x) + U(y))$ is not convex.

For $U(x) = \mathrm{e}^x, x^2$ etc, we have $U'''(x) U'(x) - 2[U''(x)]^2 \le 0$.

Reasoning. Let $z := U^{-1}(U(x) + U(y))$. We have $U(z) = U(x) + U(y)$. Taking derivative with respect to $x$ on both sides, we have $$U'(z) \cdot \frac{\partial z}{\partial x} = U'(x)$$ which results in $$\frac{\partial z}{\partial x} = \frac{U'(x)}{U'(z)}.$$ Similarly, we have $$\frac{\partial z}{\partial y} = \frac{U'(y)}{U'(z)}.$$ Then we have $$\frac{\partial}{\partial x}\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \frac{U'(x)}{U'(z)} = \frac{U''(x) U'(z)^2 - U'(x)^2 U''(z)}{U'(z)^3}, $$ etc.

If $U^{-1}(U(x) + U(y))$ is convex, then $$U''(x) U'(z)^2 - U'(x)^2 U''(z) \ge 0, $$ or $$\frac{U''(x)}{U'(x)^2} \ge \frac{U''(z)}{U'(z)^2}. \tag{1}$$ Note that $U(z) \ge U(x)$ which results in $z > x$. Thus, if $\left(\frac{U''(x)}{U'(x)^2}\right)' > 0$, then (1) does not hold. We have $$\left(\frac{U''(x)}{U'(x)^2}\right)' = \frac{U'''(x)U'(x) - 2U''(x)^2}{U'(x)^3}.$$