If we suppose that a function is differentiable at every point on its domain (but place no more restrictions than that), does it follow that the function is Lipschitz on its domain? I think that it does not, and I suspect that it is because we do not require that the function is continuously differentiable. But I'm not sure how to prove my thought - specifically, I can't come up with a counterexample.
If I am right that it is not Lipschitz, can we say if it will be locally Lipschitz at every point?
If $f\colon\Bbb R\longrightarrow\Bbb R$ is the function defined by $f(x)=x^2$, then $f$ is differentiable and even a $C^\infty$ function, but it is not Lipschitz continuous.
On the other hand, if you define$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}\end{array}$$then $f$ is not Lipschitz continuous on any neighborhood of $0$. This has to do with the fact that $f'$ is unbounded (which is stronger than being discontinuous).