Lipschitz function $f(x)=x \log|x|$

Question

Define the function $f:[-b,b]\to\mathbb{R}$, $f(x)=x\log(|x|), x\neq 0$ and $f(0)=0$ and prove that this function is lipschitz.

My attempt: I wrote $f(x)=\frac{1}{2}x\log(x^2)$ and prove that the derivative is continuous, however this function is not differentiable at $x=0$.

Answer 1

Lipschitz continuity requires a fixed $K$ such that $|f(x)-f(y)|\le K|x-y|$ for all $x$ and $y$ in the interval. In particular, since $y=0\in[-b,b]$, we would need

$$|x\ln|x||=|f(x)|=|f(x)-0|=|f(x)-f(0)|\le K|x-0|=K|x|$$

for all $x\not=0$, which is to say, we would need $|\log x|\le K$ for all $0\lt x\le b$. But that's clearly not the case, so the function is not Lipschitz.

Answer 2

In fact, it is not Lipschitz-continuous, because $\lim_{x\to 0} f'(x)=-\infty$. If $S$ is the set where a Lipschitz-continuous function is differentiable and $f':S\to\Bbb R$ is its derivative (calculated pointwise), then by definition $f'$ must be bounded on $S$.