Does multiplication by a test function stay in a Sobolev space? 
Let $u \in D^{1,\vec{p}}(\Omega)$ and $\phi \in C_c^{\infty}(\Omega)$.
  Then do we necessarily have $u\phi \in D^{1,\vec{p}}(\Omega)$?

My attempt
What we need to show is that $\partial_i (u \phi) \in L^{p_i}(\Omega)$ for each $i$. We can try to break up the integral as:
\begin{align*}
\int_{\Omega} |\partial_i (u \phi)|^{p_i}
&=
\int_{\Omega} |\partial_i u \cdot \phi + \partial_i \phi \cdot u|^{p_i}
\leq
\int_{\Omega} |\partial_i u \cdot \phi|^{p_i} + \int_{\Omega} |\partial_i \phi \cdot u|^{p_i}
\\
&=
\int_{\Omega} |\partial_i u|^{p_i} |\phi|^{p_i} + \int_{\Omega} |\partial_i \phi|^{p_i} |u|^{p_i}
\end{align*}
The first term is easily bounded given the nature of $\phi$ and $u$. But, the second term is more troublesome. We can bound the size of $|\partial_i \phi|$ and restrict the integral to a compact set. Then if $p_i \leq p^*$, we can easily bound our integral by $|\operatorname{supp} \partial_i \phi| + \|u\|_{p^*}$, which is finite since $D^{1,\vec{p}}(\Omega)$ is embedded in $L^{p^*}(\Omega)$. But we don't necessarily always have $p_i \leq p^*$; in this case, I don't know what to do.
Background
For $\vec{p} = (p_1, ..., p_N)$ and $\Omega \subseteq \mathbb{R}^N$, the Sobolev space $D^{1,\vec{p}}(\Omega)$ is defined as the completion of $C_c^{\infty}(\Omega)$ with respect to the norm:
\begin{align*}
\|u\|_{\vec{p}} = \sum \limits_{i=1}^N \|\partial_i u \|_{p_i}
\end{align*}
We think of this space as being continuously embedded into $L^{p^*}(\Omega)$ via the Sobolev inequality given here, where $p^* = Np/(N-p)$ and $1/p = \sum 1/p_i$.
 A: Just for convenience, let $K:=[a,b]^n$ for $a<b$ s.t. supp $\phi\subseteq K$. By the steps you already made in the question, we are left to bound
   $$\| u\|_{p_i}^{p_i}=\int_K dy~ |u(y)|^{p_i}\,,$$
for any $i$. Let us first assume that $u\in C_c^\infty(\Omega)\subseteq C_c^\infty(\mathbb{R}^n)$. With the neat shorthand notation $((x,y)):=(y_1,\ldots,y_{i-1},x,y_i,\ldots,y_{n-1}))$, we have
\begin{align}
    \int_K dy~ |u(y)|^{p_i}&=\int_a^b dx\int_{[a,b]^{n-1}}dy|u((x,y))|^{p_i}\\
    &\leq(b-a)\sup_{x'\in[a,b]}\int_{[a,b]^{n-1}}dy~|u((x',y))|^{p_i}\\
    &\leq(b-a)\sup_{x'\in[a,b]}\int_a^{x'} dx\int_{[a,b]^{n-1}}dy~ p_i|u((x,y))|^{p_i-1}|\partial_i u((x,y))|\\
    &\leq (b-a)p_i\int_K dy~|u(y)|^{p_i-1}|\partial_i u(y)|\,,
\end{align}
where, in the second to last step, we have used the fundamental theorem of calculus, together with the fact that $\partial_i|f|\leq|\partial_if|$ where $f$ stays away from zero.
Let us denote $\| f\|_{p,K}=(\int_K |f|^p)^{1/p}$. When we use Hölder's inequality on the above, we find
   $$\| u\|_{p_i,K}^{p_i}\leq \left\||u|^{p_i-1}\right\|_{q,K}\Vert\partial_iu\|_{p_i,K}$$
for $q$ with $\frac{1}{q}+\frac{1}{p_i}=1$. Since this implies $q(p_i-1)=p_i$ as well as $p_i/q=p_i-1$, the definitions give us
   $$\| |u|^{p_i-1}\|_{q,K}=\|u\|_{p_i,K}^{p_i-1}\,.$$
The following trick is a true classic! Note that we have obtained
   $$\|u\|^{p_i}_{p_i,K}\leq (b-a)p_i\|u\|_{p_i,K}^{p_i-1}\|\partial_iu\|_{p_i},$$
in which we can move $\| u\|_{p_i,K}^{p_i-1}$ to the left, and find
  $$\|u\|_{p_i,K}\leq (b-a)p_i\|\partial_iu\|_{p_i}\,.$$
Remember, we only proved this inequality for $u\in C_c^\infty(\Omega)$. Still, it implies that a Cauchy sequence $(u_n)$ in $D^{1,\vec{p}}$ is a Cauchy sequence w.r.t. $\|\cdot\|_{p_i,K}$ as well, implying that $\|\lim u_n\|_{p_i,K}<\infty$. Thus we have $\|u\|_{p_i,K}<\infty$ for any $u\in D^{1,\vec{p}}$, which means the answer to your question is... yes.
