Find the solution to $f(x)=-f(1/x)$.

Question

Let $f:\mathbb{R}_{++}\rightarrow \mathbb{R}$ be monotone increasing, concave and twice differentiable, such that $f(1)=0$. What is the class of such $f$s that satisfy the property $f(x)=-f\left( \frac{1}{x} \right)$ ?

I know that, for example, the logarithm function does the job since it is increasing, concave, twice differentiable and $\log \left( x^{-1} \right)= - \log (x) $.

EDIT Here is an attempt. Do you think it is correct? By combining $f(1)=0$ with $f(x)=-f\left( \frac{1}{x} \right)$ we have

$$f(x)+f\left( \frac{1}{x} \right)=f(1)$$

so, for all $x\in \mathbb{R}_{++}$,

$$f(x)+f\left( \frac{1}{x} \right)=f \left( x \cdot \frac{1}{x} \right)$$

Let $g(x)=f(e^{x})$, then

$$g\left(x + \frac{1}{x} \right)= f \left( e^{x+ \frac{1}{x} } \right)=f \left( e^{x} e^{\frac{1}{x}} \right) = f\left( e^{x} \right) + f\left( e^{\frac{1}{x}} \right)= g(x) + g\left(\frac{1}{x}\right)$$ so that it must be $$g\left(x + \frac{1}{x} \right)= g\left( x \right) + g\left(\frac{1}{x}\right)$$ This is an instance of the Cauchy functional equation whose only continuous solution is given by $g(x)=cx$ for some constant $c$. The corresponding solution to my case is: $$f(x)=c \log x$$

Any feedback would be much appreciated.

Answer 1

Setting $x=e^t$, your functional equation becomes :

$$f(e^t)=-f(e^{-t})\tag{1}$$

Let $F=f \circ \exp$ ($\circ$ = function's composition).

(therefore $F$ is a function $\mathbb{R} \to \mathbb{R}$).

(1) becomes the equivalent functional equation

$$F(t)=-F(-t)\tag{2}$$

which amounts to say that $F$ is an odd function $\mathbb{R} \to \mathbb{R}$.

Can you take it from here, taking into account the different constraints you have on $f$ ?