Consider the function
$$f(x) = \begin{bmatrix}x_2\\-x_1+\varepsilon(1-x_1^2)x_2\end{bmatrix}$$
Show that this function is locally lipschitz on the set $D=\{x\in\mathbb{R}^2\,\colon \Vert x\Vert<r\}$ for any finite $r>0$.
So we know that $f$ is Lipschitz if the norm of its jacobian is bounded. Here:
$$\frac{\partial f}{\partial x} = \begin{bmatrix}0&1\\-1-2\varepsilon x_1x_2 & \varepsilon(1-x_1^2)\end{bmatrix}$$
Then
$$\left\Vert\frac{\partial f}{\partial x}\right\Vert = \max\left\{1, \left\vert-1-2\varepsilon x_1x_2\right\vert + \left\vert\varepsilon(1-x_1^2)\right\vert\right\}$$
Now consider
\begin{align}
\left\vert-1-2\varepsilon x_1x_2\right\vert + \left\vert\varepsilon(1-x_1^2)\right\vert &\le 1 + 2\vert\varepsilon\vert\vert x_1x_2\vert + \vert\varepsilon\vert(1+x_1^2)\\
&\le 1 + \vert\varepsilon\vert\left(x_1^2+x_2^2 + (1+x_1^2+x_2^2)\right)\\
&< 1+\vert\varepsilon\vert(2r^2+1)
\end{align}
Hence the Jacobian of $f$ is fixed is bounded above hence $f$ is locally lipschitz continuous.
Is this correct? Also is this globally lipschitz continuous?
Best Answer
I see no problem with your work showing that it is locally Lipschitz. It is NOT globally Lipschitz since the derivative of $f_2$ with respect to $x_1$ gets arbitrarily large. Indeed, for fixed $x_2 \neq 0$, consider $$g(x) = -x + \epsilon x_2 (1-x^2).$$ If we can show that this map is not globally Lipschitz, I hope you will agree that it follows immediately that $f$ is not globally Lipschitz. We see $$\lvert g'(x) \rvert = \lvert-1 - 2\epsilon x_2x \rvert \to \infty \,\,\,\,\, \text{ as } x \to \infty.$$ In particular, for any $K > 0$, we can find $L >0$ such that $\lvert x \rvert > L$ gives $\lvert g'(x) \rvert > K$. Choose any $x,y > L, x \neq y$. Then by the mean value theorem, there is $c$ between $x,y$ such that $$\lvert g(x) - g(y) \rvert = \lvert g'(c) \rvert \lvert x-y \rvert > K \lvert x - y \rvert.$$ Thus $g$ is not Lipschitz with constant $K$. But $K$ was arbitrary, so $g$ is not Lipschitz.