function which is not Lipschitz but in $W^{1,\infty}$ on a domain with a slit

Solution 1:

The following graphical solution IMO says a million more words than I could write. All you need to do is cut your square $(-1,1)^2$ along the line $[0,1)\times \{0\} = \{(x_1,0)\in\mathbb R^2: x_1\in[0,1)\}$, and then smoothly raise the flap in the upper right quadrant:

Plotted on math3d. In symbols,

$$u(x_1,x_2) = \begin{cases} \hat u(x_1) & x_2>0, \\ 0 & \text{otherwise}\end{cases}$$ where $\hat u$ is a smooth function such that $\hat u(s)=0$ when $s\le 0$ and positive otherwise. Such functions exist; I chose to suitably modify the bump function from Wikipedia by setting $$\hat u(s) = \begin{cases} e^{-1/s^2} & s>0, \\ 0 & s\le 0.\end{cases}$$ Not only is the function in $W^{1,\infty}$, it is smooth everywhere it is defined. But it fails to satisfy the Lipschitz condition $$\sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|} < \infty$$ as you allow $x$ to approach the slit from above and $y$ from below.

(By the way I recall this question from either chapter 1 or 2 of Alinhac's Hyperbolic PDEs.)

Solution 2:

See "Counterexample and quasiconvexity" here.

Let $g$ be the appropriately defined inverse to the function $h: (0, \sqrt{2}) \times (0, 2 \pi) \to \Omega$ given by $ h(y_1,y_2):=(y_1 \cos(y_2), y_1 \sin(y_2))$. Let $f:=u \circ g$.

$f$ is not Lipschitz on $\Omega$:

Let $k \in (1, \infty)$. Consider the points $(\sqrt{(k-1)/k},1/\sqrt{k}), (\sqrt{(k-1)/k},-1/\sqrt{k}) \in \Omega$. Notice we have $$|f(x_1,x_2)-f(x_1,-x_2)|=2\pi \; \text{ BUT } \; \sqrt{0+4/k}=2/\sqrt{k}<2\pi$$

Hopefully, this helps as I am venturing out on some limbs here.