# If $g$ is continous on $[a,b]$ with bounded upper and lower derivatives on $(a,b)$, will $g$ be Lipschitz?

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.

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.

• +1 Would you mind explaining why $S$ is closed? (I suspect it has something to do with Intermediate and/or Extreme Value Theorem?) Commented Jan 29, 2020 at 13:49
• @Pascal'sWager I add this detail.
– Slup
Commented Jan 29, 2020 at 13:54
• Thank you. Your answer was very helpful and it makes sense now! Commented Jan 29, 2020 at 19:49
• 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/… Commented Jan 29, 2020 at 19:51
• How do you know Fact implies f is Lipschitz on [a,b]? Commented Nov 27, 2022 at 6:32

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 $$0 \le \overline Dg(\xi) = \overline Df(\xi) - \frac{f(d)-f(c)}{d-c} \le L - \frac{f(d)-f(c)}{d-c} \\ \implies f(d) - f(c) \le L(d-c) \, .$$