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Let $\Omega =\left(-1,1 \right)^2\setminus\left( \left[0,1 \right)\times \{ 0\}\right)$

and $u(x_1,x_2)=\hat u(x_1)$ for $x_1,x_2 >0$ and $u(x_1,x_2)=0$ else.

I am looking for a function $\hat u$ such that $u \in W^{1,\infty}(\Omega)$ and $u$ is not Lipschitz continous on $\Omega$.

I have failed to come up with a suitable function $\hat u$. Any hints what I may try ?

Would appreciate any help.

Calvin Khor
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XPenguen
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2 Answers2

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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:

enter image description here

enter image description here

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.)

Calvin Khor
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  • (There are much simpler $W^{1,\infty}$ examples which I will omit in the spirit of hinting) – Calvin Khor Dec 19 '21 at 04:09
  • Could I take the function which is $1$ on the upper semi-ball, $0$ on the lower and $U$ being $B(0,1)\setminus(-1,1)\times{0}$? It will have weak derivative equal to zero and is not Lipschitz continuous... – Kadmos Jul 06 '24 at 22:36
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    @Kadmos Yes. Your function's domain isn't connected, so its kind of different :) – Calvin Khor Jul 07 '24 at 09:39
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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.

M A Pelto
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