Are all smooth functions Lipschitz?

Question

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!

Consider $f(x) = x^2$ there is no $L$ such that $\forall x,y \in \mathbb R, |f(x) - f(y)| \le L |x-y|$ — Doug M, Dec 05 '17 at 21:19
It is true, however, that smooth functions are locally Lipschitz. — carmichael561, Dec 05 '17 at 21:20

score 3 · Accepted Answer · answered Dec 05 '17 at 22:11

3

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.

answered Dec 05 '17 at 22:11

Thanks for answering. What about the converse? Is every lipschitz function whose derivative exist $\beta$-smooth? – SMA.D Dec 06 '17 at 10:43
Didn't find anything. :( – SMA.D Dec 06 '17 at 13:24
Lipschitz implies bounded derivative? – Dec 06 '17 at 15:33

Are all smooth functions Lipschitz?

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