# right hand limit of convex function at boundary

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.

• 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. – Paul Sinclair Feb 21 at 1:28
• Thank you for the counter example! Now I have to figure out where I can go from here... – user3456032 Feb 22 at 14:37