# Sobolev space, continuous function

I have a question about Sobolev space.

Let $\Omega$ be an open subset of $\mathbb{R}^{d}$,

we consider the Sobolev space

$H^{1}(\Omega):=\left\{ u \in L^{2}(\Omega) : D_{j}u \in L^{2}(\Omega), j=1,\ldots,n \right\}$

with norm

$\|u\|^{2}_{H^{1}(\Omega)}:=\|u\|^{2}_{L^{2}(\Omega)}+\sum_{j=1}^{n}\|D_{j}u\|_{L^{2}(\Omega)}^{2}$,

where $D_{j}u=\partial u/ \partial x_{j}$ is the distributional deriavtive. Moreover, we let

$X:=H^{1}(\Omega) \cap C_{c}(\bar{\Omega})$,

where $C_{c}(\bar{\Omega})$ denotes the space of all continuous real valued functions on $\bar{\Omega}$ with compact support and

$H^{1}(\Omega) \cap C_{c}(\bar{\Omega})=\left\{ u|{}_{\Omega} \in H^{1}(\Omega): u \in C_{c}(\bar{\Omega})\right\}.$ (I think this is usual definition)

My question

Let $u \in X$ and $\varphi \in C(\bar{\Omega})$ then can we show $u \varphi \in X$?

I think $u \varphi \in C_{c}(\bar{\Omega})$ is clear. But I don't know $u \varphi \in H^{1}(\Omega)$. $u \varphi$ is differentiable (in distribution sence)?

• It is OK if $\phi\in C^1$. If not, I think that $\partial_j\phi$ can be irregular and therefore the term $u\partial_j\phi$ could not belong to $L^2$. – Giuseppe Negro Apr 19 '16 at 9:49
For instance, the function $\varphi := (t \mapsto |t|^\alpha)$ on $\Omega=(-1,1)$ belong to $C(\bar \Omega)$, but not to $H^1(\Omega)$ if $\alpha > 0$ is sufficiently small. Now, for $u \in X$ with $u(t) = 1$ for all $t \in (-1/2, 1/2)$ you will not have $u \, \varphi \in H^1(\Omega)$.