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I have derived the following theorems :

  1. If $\displaystyle \lim_{x \to a} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a} f'(x)=f'(a)$

  2. If $\displaystyle \lim_{x \to a^+} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a^+} f'(x)=f'(a)$

  3. If $\displaystyle \lim_{x \to a^-} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a^-} f'(x)=f'(a)$

However, I recently saw an argument that had a step using the following claim:

If $\displaystyle \lim_{x \to a^-} f_-'(x)$ exists and $f$ is continuous in some interval around $a$, then $\displaystyle \lim_{x \to a^-} f'_-(x)=f'_-(a)$, where $f'_-$ denotes the left derivative $(*)$.

I looked back at my proofs for 1,2, and 3 and saw that the theorems I used in these proofs (e.g. the mean value theorem) are not applicable for $(*)$. Under what general circumstances is $(*)$ valid, and how does the proof proceed?

S.C.
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  • @ParamanandSingh The mean value theorem requires me to know that there is an open neighborhood of differentiability. The best I can do with $\displaystyle \lim_{x \to a^-}f'_-(x)=L$ is to claim that there is an open neighborhood of left differentiability. – S.C. Feb 20 '22 at 09:05
  • Yes that's why I deleted my comments. Sometimes one reads too fast and misses the details. – Paramanand Singh Feb 20 '22 at 09:20

1 Answers1

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Let $f:[a,b] \to \Bbb R$ be a continuous function. An immediate consequence of the mean-value theorem is that if $f$ is differentiable on $(a, b)$ with non-negative (non-positive) derivative then $f$ is increasing (decreasing). Here and in the following, increasing/decreasing is meant in the weak sense.

This can be generalized to functions having left or right derivatives in $(a, b)$ only:

Let $f:[a,b] \to \Bbb R$ be a continuous function. If $f$ has a right derivative at every point in $(a, b)$ then $$ \tag{$1$} \left ( \forall x \in (a, b): f_+'(x) \ge 0 \right) \implies \text{$f$ is increasing on $[a, b]$.} $$ $$ \tag{$2$} \left ( \forall x \in (a, b): f_+'(x) \le 0 \right) \implies \text{$f$ is decreasing on $[a, b]$.} $$ If $f$ has a left derivative at every point in $(a, b)$ then $$\tag{$3$} \left ( \forall x \in (a, b): f_-'(x) \ge 0 \right) \implies \text{$f$ is increasing on $[a, b]$.} $$ $$ \tag{$4$} \left ( \forall x \in (a, b): f_-'(x) \le 0 \right) \implies \text{$f$ is decreasing on $[a, b]$.} $$

For a proof of $(1)$ see for example $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing, the other variants follow easily.

Using the above we can prove the following:

If $f:[a, b] \to \Bbb R$ is continuous and has a left derivative at each point in $(a, b)$ then $$ \tag{$5$} f_+'(a) = \lim_{x \to a^+} f_-'(x) $$ $$ \tag{$6$} f_-'(b) = \lim_{x \to b^-} f_-'(x) $$ if the limit on the right exists. Similarly, if $f$ has a right derivative at each point in $(a, b)$ then $$ \tag{$7$} f_+'(a) = \lim_{x \to a^+} f_+'(x) $$ $$ \tag{$8$} f_-'(b) = \lim_{x \to b^-} f_+'(x) $$ if the limit on the right exists.

$(6)$ is what you are asking for, so we will prove that. The other variants can be proved in the same way or can be reduced to $(6)$.

Let us assume that the limit $$ \lim_{x \to b^-} f_-'(x) = A $$ exists. Given $\epsilon > 0$ there is a $x_0 \in (a, b)$ such that $$ \tag{$*$} A - \epsilon < f_-'(x) < A + \epsilon $$ for $x_0 < x < b$. So the left derivative of $f(x) - (A+\epsilon)x$ is $\le 0$ on $(x_0, b)$. Using $(4)$ above, one can conclude that $f(x) - (A+\epsilon)x$ is decreasing on $[x_0, b]$, so that $$ f(x) - (A+\epsilon)x \ge f(b) - (A+\epsilon)b \implies \frac{f(x)-f(b)}{x-b} \le A + \epsilon \, . $$ In the same way one uses the left inequality in $(*)$ and $(3)$ to show that $$ \frac{f(x)-f(b)}{x-b} \ge A - \epsilon $$ for $x_0 < x < b$.

So we have shown that for every $\epsilon > 0$ there is a $x_0 \in (a, b)$ such that $$ A - \epsilon \le \frac{f(x)-f(b)}{x-b} \le A + \epsilon $$ and that proves that $f$ has a left derivative at $x=b$ with $$ f_-'(b) = A = \lim_{x \to b^-} f_-'(x) \, . $$

Remark: As Paramanand Singh pointed out, requiring continuity at the endpoints of the interval is important. Otherwise counterexamples for all of the above can be found easily, e.g. by defining $f(a)= f(b) = 1$, $f(x) = 0$ otherwise.

The underlying reason is that the mean-value theorem, Rolle's theorem and the above generalizations $(1)-(4)$ all go back to the fact that a continuous function on a compact interval attains its maximum and its minimum. And if the left (right) derivative exists at a maximum then it must be non-negative (non-positive). At a minimum it is the other way around.

Martin R
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  • Cheers! Looking forward to reading the linked post and this. Much appreciated. – S.C. Feb 20 '22 at 07:21
  • @MartinR Thanks a lot from me as well. For convex $f$ the proof should be quite direct. – Kurt G. Feb 20 '22 at 08:10
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    Nice proof. When the function is continuous then even one sided monotone nature at each point is sufficient to conclude monotone nature in the interval. Just want to add that continuity is essential here. – Paramanand Singh Feb 20 '22 at 09:22
  • @ParamanandSingh: Thanks for your feedback. I have tried to incorporate that into the answer. – Martin R Feb 21 '22 at 08:39
  • @S.Cramer: Please let me know if anything is unclear. – Martin R Feb 24 '22 at 08:56
  • Everything is clear, but I think there is just one minor typo: You write - "So the left derivative of $f(x) - (A+\epsilon)x$ is $\le 0$ on $(x_0, 1)$". I'm guessing that open interval should be $(x_0,b)$ – S.C. Feb 25 '22 at 08:43
  • @S.Cramer: Yes, indeed. I had initially written this for the interval $(0, 1)$ and later decided to change that to $(a, b)$. I'll fix it right now, thanks! – Martin R Feb 25 '22 at 08:45
  • Maybe a dumb question - Is it always the case that if, for an arbitrary $\varepsilon \gt 0$ and $\delta \gt 0$, I have $-\varepsilon \leq F(x)-L \leq \varepsilon$, then I must have $-\varepsilon \lt F(x)-L \lt \varepsilon$ for some smaller $\delta'$?. I just worked it out - it would seem so. – S.C. Feb 25 '22 at 08:49
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    @S.Cramer: Sure, you can choose a smaller $\delta$ such that $\epsilon/2 \le F(x) - L \le \epsilon/2$. But in this proof it actually makes no difference whether you choose the limit definition with $\le$ or with $<$. – Martin R Feb 25 '22 at 08:51