Is a piecewise linear function with uniformly bounded derivative uniformly continuous

continuityreal-analysisuniform-continuity

Let $(a_k)_{k \geq 0} \subset \Bbb R$ be any sequence.

Let $f : \Bbb R \rightarrow \Bbb R$ be the continuous function defined by

$$f(x) = \begin{cases}
a_0 \text{ if } x \leq 0 \\
a_k + (a_{k+1}-a_k)(x-k) \text{ if } x \in [k, k+1]
\end{cases}$$

i.e. $f$ is a continuous, piecewise linear function.

Assume that $f$ has a uniformly bounded derivative on $\Bbb R \setminus \Bbb N$ (i.e. the sequence $(a_{k+1}-a_k)_{k \geq 0}$ is bounded). Can we conclude that $f$ is uniformly continuous ?

If not, what if we further assume some regularity for $(a_k)_{k \geq 0}$, e.g. that $(a_k)_{k\geq 0}$ is a decreasing sequence converging to $0$ ?

Best Answer

Hint: verify that $f(x)=f(0)+\int_0^{x}f'(t)\, dt$ and conclude that $f$ is actually a Lipschtz function.