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.

  • $\begingroup$ 2) is false. Not all derivatives are continuous functions. $\endgroup$ Apr 16, 2019 at 23:38
  • $\begingroup$ @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). $\endgroup$
    – WoodWorker
    Apr 16, 2019 at 23:47
  • $\begingroup$ I know -- I mentioned Darboux theorem. $\endgroup$
    – WoodWorker
    Apr 16, 2019 at 23:51

1 Answer 1


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 $[\frac{a+b}{2}, b]$.)

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

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