Does Lipschitz continuity on $[a,b]$ and differentiability on $(a,b)$ imply differentiability at $b$

Question

Suppose $f: [a, b] \rightarrow \mathbb R^n$ is Lipschitz continuous. Suppose also that $f$ is differentiable on $(a,b)$ and that there is a continuous function $g: [a, b] \rightarrow \mathbb R^n$ which has the property that $g(x)$ is the derivative of $f$ at $x$ for all $x \in (a,b)$. Is it necessarily the case that $f$ is differentiable at $b$ (in the sense of left-side limits) and, moreover that the derivative of $f$ at $b$ is $g(b)$.

Answer 1

We don't need all of those assumptions. Assume $n=1.$ Suppose $f:(a,b]\to \mathbb R$ is continuous, $f$ is differentiable in $(a,b),$ and $\lim_{x\to b^-} f'(x)$ exists. Then $f'(b)$ exists (as a left sided derivative).

Proof: Choose any $c\in (a,b).$ Then $f$ is continuous on $[c,b]$ and differentiable in $(c,b).$ For any $x\in (c,b),$ the mean value theorem shows there exists $c_x\in (x,b)$ such that

$$\tag 1 \frac{f(b)-f(x)}{b-x} = f'(c_x)$$

As $x\to b^-,$ $c_x\to b^-.$ By our assumption on $f,$ $\lim_{x\to b^-} f'(c_x)$ exists. Thus by $(1),$ $f'(b)$ exists.

So we're done for $n=1,$ and the result for higher dimensions follows easily.

Answer 2

Let $L$ be the Lipschitz-constant, i.e. $|f(x)-f(y)| \leq L |x-y|$. Then $|f'(x)| \leq L$ for all $x \in (0,b)$. Thus, the derivative is bounded. However, $f$ do not have to be differentiable in $0$ or $b$.

Counterexample: Let $\rm{Si}(x) := \int_0^x \frac{\sin(t)}{t} \, \rm{d}t$, then $f(x) = \rm{Si}(1/x) +x \cos(1/x)$ has the derivate $f'(x)=\sin(1/x)$. Since $f'$ is bounded, $f$ is Lipschitz-continuous on $[0,1]$. (Note that we can define $f$ also in $x=0$.) However, $f$ is not differentiable in $0$.