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

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).
• 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 $\Omega$? – eurocoder Nov 21 '18 at 11:23
• 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 – Calvin Khor Nov 21 '18 at 11:26