How can a discontinuous function belong to $C_B^1(\Omega)$, the space of continuous functions $u$ with bounded derivatives?

Question

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?

Answer 1

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).