I've been stuck with prove the following problem. Can you help me?
Suppose that $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ is a function such that $\log f(x)$ is concave. Then, it has derivative $ \frac{f^\prime_{-}(x) + f^\prime_{+}(x) }{2f(x)}$ except, possibly, on a countable set,
where $f^\prime_{+}(x)$ and $f^\prime_{-}(x)$ are right and left derivatives, respectively.
Thanks in advance.
As mentioned in the comments, convex functions are differentiable a.e. (see here). Note that the negation of a convex function is concave and vice versa, so we can use this differentiability result for concave functions. With this, we can define $$ h(x) = e^{(\log f(x))} = f(x) $$ the composition of a function with is differentiable a.e. ( $\log f(x)$ in this case) and a function which is differentiable everywhere ($e^x$ in this case), will be differentiable a.e., and so $f(x)$ will be differentiable a.e. Therefore, when $f'(x)$ exists, we have $f'(x) = f'(x+) = f'(x-)$ and so $$ f'(x) = \frac{f'(x+) + f'(x-)}{2} $$ when the derivative exists. The same composition argument can now be applied to $f(x)$ and $\log(x)$ to give $$ \frac{d}{dx} \log (f(x)) = \frac{f'(x)}{f(x)} = \frac{f'(x+) + f'(x-)}{2f(x)} $$ where the first inequality follows from the chain rule. Again, this is only when $f'(x)$ exists