# 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 $$1/p = \sum 1/p_i$$.

• Huh? $\phi\in C_c^\infty$, so $\lVert\partial_i\phi\rVert_\infty$ are bounded. – user10354138 Jun 25 at 21:21
• @user10354138 Sure, so we bound $|\partial_i \phi|^{p_i}$ by a constant $C$, and we get $C \int_{\operatorname{supp} \partial_i \phi} |u|^{p_i}$. But how do we know this integral is finite? – Sambo Jun 25 at 21:24
• Use $u\in D^{1,p}$. – user10354138 Jun 25 at 21:36
• @user10354138 How so? – Sambo Jun 25 at 21:37

Just for convenience, let $$K:=[a,b]^n$$ for $$a s.t. supp $$\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.

• Fantastic! Did you come up with this yourself, or is it from a reference? It seems that you show something stronger, namely that $u \in L^{p_i}_{loc}(\Omega)$. This seems like an important result! – Sambo Jul 8 at 18:00
• I'm also curious about the "pass to the completion" step. My idea is to a sequence $u_n$ converging to $u$ in $D^{1,\vec{p}}$ and also pointwise a.e. (extracting a subsequence). Fatou's lemma tells us that the norm of $u$ is less or equal to the limit of the $u_n$ norms, and we apply your inequality to get less or equal to the limit of the $\partial_i u_n$ norms. This is equal to the norm of $\partial_i u$ since $D^{1,\vec{p}}$ convergence implies convergence of the partials in $L^{p_i}$. – Sambo Jul 8 at 19:21
• Yes, that's the right approach if you see $D^{1,\vec{p}}(\Omega)$ as a subset of $L^{p^*}(\Omega)$. If you see $D^{1,\vec{p}}(\Omega)$ as the abstract completion of $C_c^\infty(\Omega)$ (and assume wlog that $\Omega\subseteq K$) then you can canonically embed it in $L^{p_i}(K)$, and you will automatically get this bound. – Teun Jul 9 at 10:31
• I came up with this myself. It was a nice problem for the weekend. :) Now that I've looked in the literature I read that, at least for some different variant of $D^{1,p}$ this is apparently referred to as `common knowledge': pdfs.semanticscholar.org/9abc/…. – Teun Jul 9 at 10:33
• Indeed, I was considering $D^{1,\vec{p}}(\Omega)$ as embedded into $L^{p^*}(\Omega)$. By the "canonical embedding" into $L^{p_i}(K)$, do you mean identifying the Cauchy sequence $(u_n)_n$ with the limit of $(\partial_i u_n)_n$? If so, I don't see how to even make sense of the inequality in that case. – Sambo Jul 9 at 15:45