When is $\exp(f(x))$ concave?

Question

I'm interested in the set of twice-differentiable functions $f$ such that $\exp(f(x))$ is concave for $x \in [0,1]$. This is equivalent to asking for the set of functions $f$ such that $$ \left(\frac{df}{dx}(x)\right)^2 + \frac{d^2 f}{dx^2}(x) < 0 $$ for every $x \in [0,1]$. The reasoning for this equivalence is below. I'm purposefully avoiding a non-strict inequality to avoid the trivial solution $f(x) = c$ for some constant $c \in \mathbb R$. If the entire set can’t be characterized, I would be satisfied with a single example of $f$.

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Because \begin{equation} \frac{d \exp \circ f}{dx}(x) = \exp(f(x)) \cdot \frac{df}{dx}(x) \end{equation} and so \begin{align} \frac{d^2 \exp \circ f}{dx^2}(x) &= \exp(f(x)) \cdot \frac{df}{dx}(x) \cdot \frac{df}{dx}(x) + \exp(f(x)) \cdot \frac{d^2f}{dx^2}(x) \\ &= \exp(f(x)) \cdot \left[\left(\frac{df}{dx}(x)\right)^2 + \frac{d^2f}{dx^2}(x)\right] \end{align} Because $\exp(f(x)) \geq 0$ for every $x \in [0,1]$, then we want to determine the functions $f$ such that $$ \left(\frac{df}{dx}(x)\right)^2 + \frac{d^2f}{dx^2}(x) < 0 $$ for every $x \in [0,1]$.

Answer 1

In general, if $u:[0, 1]\to (-\infty, 0)$ is a continuous function, the function $f$ that fullfils $$(\exp(f(t)))''=u(t)$$ is concave. We know that such an ODE has a solution, for each $u$, and solving it gives us $$f(x)=\log\left(k_2+k_1x+\int_{0}^{x}\int_{0}^{y}u(t)dt\right)$$

You can just plug in your favourite negative function $u$ and you have many examples :)

Answer 2

Here is an example: $$ F(x)=\ln(\ln(x+2)). $$ It is well defined on $[0,1]$ and $$ e^{F(x)}=\ln(x+2) $$ Is clearly concave(even strictly).

Answer 3

A possibility is to consider $f(x)=\cos(ax)$.

This quickly reduces to a quadratic $a^2(\sin^2(ax)-\cos(ax))<0$ and then one can check that it is a solution when $a<\arccos(\frac{-1+\sqrt{5}}{2})\approx0.90456$.

This isn't a full solution, as $f$ isn't characterised completely, but it is indeed an example, and a very simple one at that.

Answer 4

Let $f(x):=\ln u(x)$, where $u$ is any positive, twice-differentiable function that is concave in $[0,1]$. Then $f$ is also twice-differentiable and $\exp(f(x))$ is concave in $[0,1]$.