Let $\Omega = \{(x,y) \in \mathbb{R}^2 \ : \ 0 < |x| < 1, \ 0 < y < 1\}$ and consider the function $u$ defined on $\Omega$ by (Sobolev Spaces by Adams, page 68, Example 3.10) $$ u(x,y) = \begin{cases} 1, \quad x > 0, \\ 0, \quad x < 0. \end{cases} $$

On page 80 (item (iv)) of this book Adams says that this function belongs to $C_B^1(\Omega)$ which consists of function in $C^1(\Omega)$ such that $D^\alpha u$ is bounded for $0 \le \alpha \le 1$.

But this function is discontinuous at $x=0$ so how can it be an element of any space of continuous functions? Is this a typo?


Its not a typo. The function is defined on two separate sets that are not path connected. On each set, it takes a constant value. Its maybe easier to see in 1D, this function is $$ f: [-1,0)\cup (0,1] \to \mathbb R, \quad f(x) = \frac{\operatorname{sgn(x)+1}}2$$ $f$ is continuous (even $C^\infty$) on its domain, but there is no continuous extension to $[-1,1]$ (and certainly no $C^1$ extension).

  • $\begingroup$ Ah right, I had overlooked the function was not defined on $x = 0$. Do you know why he goes on to say that although the function is in $C_B^1(\Omega)$, it is not in $C^1(\bar \Omega)$, which is the closed subspace of $C_1^j(\Omega)$ consisting of functions have uniformly continuous derivatives up to order $1$ on $\Omega$? ..how can $C^1(\bar \Omega)$ even be a subspace of $C_B^1(\Omega)$ as $C^1(\bar \Omega)$ is defined on $\bar \Omega$ which is bigger than $\Omega$? $\endgroup$ – eurocoder Nov 21 '18 at 11:23
  • $\begingroup$ Nevermind, he says on page that 10 that this is an abuse of notation. $\endgroup$ – eurocoder Nov 21 '18 at 11:26
  • $\begingroup$ If you are continuous on a bigger set, then the restrictions(which are invisible by the abuse of notation you mentioned) to smaller sets are naturally also continuous. That is, the inclusions are reversed; $A \subset B$ implies $C^1_B(B) \subset C^1_B(A)$. As a quick check, you may want to note that the extreme case of continuous functions defined only at a point $C^0(\{0\})$ contains every function $f:\mathbb R\to \mathbb R$. @eurocoder $\endgroup$ – Calvin Khor Nov 21 '18 at 11:26

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