# If $f$ is differentiable on $[a, b]$, is $f$ Lipschitz of order $1$ on $[a, b]$?

If $$f$$ is differentiable on $$[a, b]$$, is $$f$$ Lipschitz of order $$1$$ on $$[a, b]$$?

The author puts the following function as a counterexample, but I don't understand why. $$f(x)= \begin{cases} x^2 \sin\left(\frac{1}{x^2}\right), & \text{if } x\neq 0,\\ 0, & \text{if } x=0, \end{cases}$$ on $$[0,1]$$. I would appreciate any help.

• But could be done $$f(x)= \begin{cases} x^2 \sin\frac{1}{x^2}\qquad x\neq 0\\ 0\qquad\quad\qquad x=0 \end{cases}$$ It is differentiable on $[0,1]$ and $$f'(x)= \begin{cases} 2 x \sin\frac{1}{x^2}-\frac{2}{x} \cos\frac{1}{x^2}, &x\neq 0\\ 0, &x=0 \end{cases}$$ Jun 23, 2022 at 3:44
• If the function was Lipschitz, its derivative would be bounded on $[0,1]$, but it is not. The derivative should be continuous on $[0,1]$ in order to satisfy the Lipschitz condition.
– Gary
Jun 23, 2022 at 3:47
• If I understand that, it was an error of the author, but then I don't know how to continue to conclude that it is not of order 1 Jun 23, 2022 at 3:56
• Show that if $f$ was Lipschitz with Lipschitz constant $L$, then $|f'(x)|\leq L$ for all $x\in [0,1]$. (Hint re-order the Lipschitz condition and let $y\to x$)
– Gary
Jun 23, 2022 at 4:13
• math.stackexchange.com/questions/3043139/… Jun 23, 2022 at 4:19