Does bounded and continuous implies Lipschitz? If a function $f : \mathbb{R} \rightarrow \mathbb{R}$ is integrable, bounded and continuous, is it also Lipschitz continuous?

No. Let 
$$
f(x)=\left\{ \begin{array}{ll}
0&, x\leq 0\\
\sqrt{x} &,x\in [0,1]\\
-x^2+2x &, x\in [1,2]\\
0&, x\geq 2.
\end{array}\right.
$$
$f$ is continuous, bounded and integrable, but it is not Lipschitz, since $f|_{[0,1]}$ is not Lipschitz.
