Lipschitz continuity for the function $f(x)=x\ln x$

Question

For which $L$, Lipschitz continuity is fulfilled for the function $f(x)=x\ln x$

We have: $$|f(x_1)-f(x_2)|=|x_1\ln x_1 - x_2 \ln x_2|\le L|x_1-x_2| $$ However I don't have idea how to find this $L$.

What's the domain of $f$? – Clement Yung Jun 13 '20 at 10:05 — Clement Yung, Jun 13 '20 at 10:05

score 2 · Answer 1 · answered Jun 13 '20 at 10:15

Assuming $x_1,x_2\in (0,\infty)$, for no $L$. Let $x_2\to x_1$, then $$\left|{x_1\ln x_1-x_2\ln x_2\over x_1-x_2}\right|<L$$since $f(x)=x\ln x$ is differentiable over $(0,\infty)$, we can write $$|\ln x_1+1|<L$$ which is impossible for small enough $x_1$, hence $f(x)$ is not Lipschitz continuous over $(0,\infty)$.

score 1 · Answer 2 · answered Jun 13 '20 at 20:25

Suppose $f(a,b)\to \mathbb R$ is differentiable. Then $f$ is Lipschitz on $(a,b)$ iff $f'$ is bounded on $(a,b).$

In your problem we can think about $x\ln x$ on $(0,1).$ Its derivative is $\ln x +1,$ which is unbounded as $x\to 0^+.$ Therefore $x\ln x$ is not Lipschitz on $(0,1).$

Lipschitz continuity for the function $f(x)=x\ln x$

2 Answers2