# Proving some derivative properties

I want to prove the following for $$f:[a,b] \to \mathbb{R}$$.

(1) If $$f$$ differentiable in $$(a,b)$$, and $$\lim_{x \to a+} f'(x)$$ exists, then $$f'(a)$$ exists.

(2) If $$f'(z)$$ exists for $$z \in (a,b)$$ and one-sided limits $$\lim_{x \to z+} f'(x)$$ and $$\lim_{x\to z-}f'(x)$$ exist, then both are equal to $$f'(z)$$.

(3) The upper Dini derivative $$D^+f = \limsup_{h \to 0+} \frac{f(x+h) - f(x)}{h}$$ exists although possibly with value $$+\infty$$. If $$f$$ is continuous but not differentiable it is not possible that $$D^+f(x) = + \infty$$ at every point in $$(a,b)$$.

I proved (1) with mean value theorem since $$\lim_{h \to 0+} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0+} f'(c) = f'(a)$$ where $$a < c < a+h$$. I also know that (2) can be proved with Darboux theorem implying derivative cannot have a jump discontinuity.

Supposedly using (2) we can answer (3), but I am unable to proceed.

• 2) is false. Not all derivatives are continuous functions. Apr 16, 2019 at 23:38
• @KaviRamaMurthy: Sorry. I left out a condition that both the left and right limits of the derivative also exist.My problem is really with (3). Apr 16, 2019 at 23:47
• I know -- I mentioned Darboux theorem. Apr 16, 2019 at 23:51

Here is an outline of a possible proof for (3), though it doesn't really use (2):

Suppose, to the contrary, that we did have $$D^+ f(x) = +\infty$$ for every $$x \in [a, b)$$.

• Fix a positive real number $$M$$, and define $$A_M := \{ x \in [a, b] \mid f(x) - f(a) \ge M (x - a) \}$$. Then $$a \in A_M$$ and $$A_M$$ is bounded above by $$b$$, so $$\sup A_M$$ exists.
• Show that $$A_M$$ is closed, so $$\sup A_M \in A_M$$.
• Show that if $$\sup A_M < b$$, then using the assumption $$D^+ f(\sup A_M) = \infty$$, we can get a contradiction.

Now, from the fact that $$b \in A_M$$ no matter how large $$M$$ is, derive a contradiction.

(And then, to show it's also impossible to have $$D^+ f(x) = +\infty$$ for every $$x \in (a, b)$$, apply the result you prove above to the restriction of $$f$$ to $$[\frac{a+b}{2}, b]$$.)

• Thank you -- this clears it up. I see no connection with (2) either. Apr 19, 2019 at 16:15