Can we prove the following statement:
$$\|\triangledown f(x)-\triangledown f(y)\|\leq\beta\|x-y\|\xrightarrow{?} \| f(x)-f(y)\|\leq L\|x-y\|$$
i.e. every smooth function is Lipschitz? If it is not correct please tell me under what conditions it can be correct.
In the comments of this question an example is given that exponential function is smooth and is not Lipschitz. However I can't find any $\beta$ such that $\|e^x-e^y\|\leq \beta\|x-y\|$ and derivative of exponential equals to itself!
Best Answer
Smooth functions are locally Lipschitz. In general, they are not globally Lipschitz: for example, $$ e^{x+1}-e^x = e^x(e-1) \to \infty\ \text{ as }\ x\to\infty $$ which shows there is no bound $|e^y-e^x|\le C|y-x|$. A similar computation can be made with $f(x) = x^2$.
Under the additional assumption that the derivative is bounded, the function is Lipschitz. This is a direct consequence of the Mean Value Theorem: see Bounded derivative implies Lipschitz, for example.