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By the Mean Value Theorem from ordinary calculus one knows that, if $f$ is continous on $[a,b]$ and differentiable on $(a,b)$ with bounded derivatives, then $f$ has to be Lipschitz on $[a,b]$.

Now I ask a more general question: If $g$ is continous on $[a,b]$ with bounded upper and lower derivatives on $(a,b)$, will $g$ be Lipschitz? If not, what would be a counterexample?

I really don't know how to proceed. One idea I had was approximating $g$ with a piecewise linear function $\phi$, but I don't know if this gets us anywhere.

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2 Answers 2

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Suppose that $$M = 1 + \sup_{x\in (a,b)}\{|\overline{D}f(x)|,|\underline{D}f(x)|\}$$ and $f$ is continuous.

Fact. Pick $[c,d]\subseteq (a,b)$. Then $$|f(d) - f(c)|\leq M(d-c)$$ Proof. Define $$S = \big\{x\in [c,d]\,\big|\,|f(x) - f(c)|\leq M(x-c)\big\}$$ Clearly $S$ is closed as $f$ is continuous and $c\in S$. Note that for every $x_0\in [c,d]$ there exists $\delta_{x_0}>0$ such that $$\bigg|\frac{f(x)-f(x_0)}{x-x_0}\bigg|< M$$ for all $x\in (x_0-\delta,x_0+\delta)$. We can rewrite it to $$\big|f(x)-f(x_0)\big| \leq M\cdot |x-x_0|$$ Pick $u = \sup S$. If $u < d$, then $$\bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(c)\bigg| \leq \bigg|f\left(u+\frac{\delta_{u}}{2}\right) - f(u)\bigg| + \bigg|f(u) - f(c)\bigg|\leq M\cdot \frac{\delta_{u}}{2} + M\cdot (u - c) = M\left(u+\frac{\delta_{u}}{2} - c\right)$$ Then $u+\frac{\delta_{u}}{2}\in S$. This is a contradiction with $u = \sup S$. This implies $u = d$ and hence $$|f(d) - f(c)|\leq M\cdot (d-c)$$

Now Fact implies that $f$ is Lipschitz.

Remark.

Pick a sequence $\{x_n\}_{n\in \mathbb{N}}$ of elements of $S$ such that $\lim x_n = x \in [c,d]$. Then $$|f(x_n) - f(c)| \leq M\cdot (x_n-c)$$ By taking limits of both sides of these inequality and by continuity of $f$ we deduce that $$|f(x) - f(c)|\leq M\cdot (x-c)$$
Hence $x\in S$ and this shows that $S$ is closed.

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  • $\begingroup$ +1 Would you mind explaining why $S$ is closed? (I suspect it has something to do with Intermediate and/or Extreme Value Theorem?) $\endgroup$ Commented Jan 29, 2020 at 13:49
  • $\begingroup$ @Pascal'sWager I add this detail. $\endgroup$
    – Slup
    Commented Jan 29, 2020 at 13:54
  • $\begingroup$ Thank you. Your answer was very helpful and it makes sense now! $\endgroup$ Commented Jan 29, 2020 at 19:49
  • $\begingroup$ If you have time, I have another question which is similar in that it involves generalizing calculus derivative properties to upper/lower derivatives. I would greatly appreciate your guidance :) math.stackexchange.com/questions/3527390/… $\endgroup$ Commented Jan 29, 2020 at 19:51
  • $\begingroup$ How do you know Fact implies f is Lipschitz on [a,b]? $\endgroup$
    – FactorY
    Commented Nov 27, 2022 at 6:32
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It suffices to prove the following:

Let $f:[c, d] \to \Bbb R$ be a continuous function with an upper derivative that is bounded above: $$ \forall x \in [c, d]: \overline Df(x) \le L \, . $$ Then $$ f(d) -f(c) \le L(d-c) \, . $$

If both the upper derivative and the lower derivative are bounded $$ -L \le \underline D f(x) \le \overline D f(x) \le L $$ then the above can be applied to $f$ and $-f$, and it follows that $$ |f(d) - f(c) | \le L (d-c) $$ on any subinterval $[c, d] \subset (a,b)$, so that $f$ is Lipschitz continuous.

The proof of the above claim resembles that of the mean-value theorem (and Rolle's theorem): Consider the function $$ g(x) = f(x) - (x-c)\frac{f(d)-f(c)}{d-c} \, . $$ Then $g(c) = g(d)$, so that $g$ attains its maximum at a point $\xi \in (c, d]$. It follows that $$ \overline Dg(\xi) \ge \lim_{\delta \to 0} \sup \bigl\{ \frac{g(\xi +h)-g(\xi)}{h} \bigm\vert -\delta < h < 0 \bigr\} \ge 0 $$ and therefore $$ 0 \le \overline Dg(\xi) = \overline Df(\xi) - \frac{f(d)-f(c)}{d-c} \le L - \frac{f(d)-f(c)}{d-c} \\ \implies f(d) - f(c) \le L(d-c) \, . $$

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