1

‎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.

koohyar eslami
  • 417
  • Is the statement true? For $f(x)=x$ ($x \geq 0$), the statement boils down to $f'(x) = 1/x$, right? – LinAlg Feb 06 '20 at 02:49
  • @LinAlg It would boil down to $\frac{d}{dx}\log x = \frac{1}{x}$, so it's correct in this case. It's also very easy to prove in the differentiable case. You can just apply chain-rule. – forgottenarrow Feb 06 '20 at 04:11
  • 1
    https://math.stackexchange.com/questions/946311/a-convex-function-is-differentiable-at-all-but-countably-many-points – LinAlg Feb 07 '20 at 13:07

1 Answers1

2

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

JMK
  • 795