If function $u$, is $ C^1$ function on two almost disjoint compact sets, then is $u \in W^{1,\infty}$ of union of two sets?

39 Views Asked by At

Let B denote the open unit ball in $R^n$, $B_+ = \{x \in B : x_n > 0\}$ and $B_- = \{x \in B : x_n < 0\}$. Also Suppose u ∈ $C^1(\overline B_+)\bigcap C^1(\overline B_-)$.

I was trying to figure out is it true that u ∈ $W^{1,∞}(B)$? Is there any counterexample? If yes, then what kind of modification or addition of the assumptions is needed to make the statement correct?

Any kind of help is greatly appreciated. Thanks.

1

There are 1 best solutions below

1
On BEST ANSWER

This is not true in general. Recall that $u\in W^{1,\infty}(B)$ implies that $u$ is Lipschitz continuous. We define $u\equiv 1$ on $B_+$ and $u\equiv -1$ on $B_-$, then $u$ is not Lipschitz continuous. I think if your impose the condition that $u$ is continuous on $\{x\in B:\,x_n=0\}$, then it is true.