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Denote by $f$ a monotonically decreasing, convex function defined on $[0,\infty)$ that has a derivative $f'$ on $(0,\infty)$.

I would like to show that if $f(0)$ exists and is finite (and $\lim_{x \to 0} f(x) = f(0)$), then the right hand limit $f_+'(x) = \lim_{h \searrow 0} \frac{f(x+h)-f(x)}{h}$ exists and is finite at $0$, and that $\lim_{x\to 0} f'(x) = f_+'(0)$ (in my setting it would be fine to assume that $f'$ is continuous (or even differentiable) on $(0,\infty)$).

I have so far tried to follow (and then modify) https://proofwiki.org/wiki/Convex_Real_Function_is_Left-Hand_and_Right-Hand_Differentiable and the cited reference (1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach). In this case it is shown (on the interior of an interval) that $F_x(h) = \frac{f(x+h)-f(x)}{h}$ is an increasing function in $h$ and hence $\lim_{h\to 0} F_x(h) = f_+'(x)$ has to exists. This existence is already unclear to me, and the closest answer to this I could find was mentioned in the question Proof that Right hand and Left hand derivatives always exist for convex functions. where an inequality involving limits based on $h' < 0$ is used. In my case I can not reflect around the boundary point and I am hence searching for another way to show the statement.

Any hints or comments are greatly appreciated.

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    $\begingroup$ Consider the function $$f(x) = \begin{cases} -\sqrt{16 - (x - 4)^2} & x\in[0,1]\\1-\sqrt 7 - x& x \in (1,\infty)\end{cases}$$ It is tangent to the $y$-axis. $\endgroup$ – Paul Sinclair Feb 21 at 1:28
  • $\begingroup$ Thank you for the counter example! Now I have to figure out where I can go from here... $\endgroup$ – user3456032 Feb 22 at 14:37

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