The Sobolev space $W^{k,p}$ is an algebra above the critical index

The product estimate for derivatives. Let $w_1,w_2\in W^{k,p}(\mathrm{\Omega };\mathbb{K})$ and let ${\gamma }_1,{\gamma }_2$ be multi-indices with $\mathrm{\mid }{\gamma }_1\mathrm{\mid }+\mathrm{\mid }{\gamma }_2\mathrm{\mid }\le k$; put $s_i:=k-\mathrm{\mid }{\gamma }_i\mathrm{\mid }\ge 0$, so that $s_1+s_2=2k-\mathrm{\mid }{\gamma }_1\mathrm{\mid }-\mathrm{\mid }{\gamma }_2\mathrm{\mid }\ge k$. By [F1] choose exponents $q_i\in [1,\mathrm{\infty }]$ with $D^{{\gamma }_i}{w}_i\in L^{q_i}(\mathrm{\Omega })$ and $\mathrm{\parallel }{D}^{{\gamma }_i}{w}_i{\mathrm{\parallel }}_{L^{q_i}}\le C\mathrm{\parallel }{w}_i{\mathrm{\parallel }}_{W^{k,p}(\mathrm{\Omega })}$ as follows: $\frac{1}{q_i}=0$ when $s_{i}p>n$; $\frac{1}{q_i}=\frac{1}{p}-\frac{s_i}{n}$ when $s_{i}p<n$; and $\frac{1}{q_i}=\frac{1}{2n}$ when $s_{i}p=n$, which is available because the critical embedding supplies every finite exponent. In every case $\frac{1}{q_1}+\frac{1}{q_2}\le \frac{1}{p}$: for two subcritical exponents this is $\frac{2}{p}-\frac{s_1+s_2}{n}\le \frac{2}{p}-\frac{k}{n}<\frac{1}{p}$ because $kp>n$; for one critical and one subcritical exponent it is $\frac{1}{2n}+\frac{1}{p}-\frac{s}{n}\le \frac{1}{p}$ because $s\ge 1$; for two critical exponents it is $\frac{1}{n}\le \frac{1}{p}$ because $p\le n$ (recall $s_{i}p=n$ with $s_i\ge 1$ forces $p\le n$); and a supercritical exponent contributes $0$. Hence by generalized Holder [F3], $D^{{\gamma }_1}{w}_1{D}^{{\gamma }_2}{w}_2\in L^p(\mathrm{\Omega })$ with $\mathrm{\parallel }{D}^{{\gamma }_1}{w}_1{D}^{{\gamma }_2}{w}_2{\mathrm{\parallel }}_{L^p}\le C(n,p,k,\mathrm{\Omega })\mathrm{\parallel }{w}_1{\mathrm{\parallel }}_{W^{k,p}}\mathrm{\parallel }{w}_2{\mathrm{\parallel }}_{W^{k,p}}$.

The Leibniz identity and the algebra bound. By [F2] choose $u_j,v_j\in C^{\mathrm{\infty }}(\mathrm{\Omega })\cap W^{k,p}(\mathrm{\Omega })$ with $u_j\to u$ and $v_j\to v$ in $W^{k,p}(\mathrm{\Omega })$. Fix $\mathrm{\mid }\alpha \mathrm{\mid }\le k$; for the smooth factors, repeated classical differentiation gives the finite Leibniz formula, and [F4] identifies its classical derivatives with weak derivatives: $D^{\alpha }(u_j{v}_j)={\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right)D^{\beta }{u}_j{D}^{\alpha -\beta }{v}_j$. Applying step 1.1 to the pairs $(u_j-u,v_j)$ and $(u,v_j-v)$ with $({\gamma }_1,{\gamma }_2)=(\beta ,\alpha -\beta )$ shows that each summand converges in $L^p(\mathrm{\Omega })$ to $D^{\beta }u{D}^{\alpha -\beta }v$, and the case $\alpha =0$ gives $u_j{v}_j\to uv$ in $L^p(\mathrm{\Omega })$; therefore, for every test function $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, ${\int }_{\mathrm{\Omega }}uv{\mathrm{\partial }}^{\alpha }\phi ={\lim }_j{\int }_{\mathrm{\Omega }}{u}_j{v}_j{\mathrm{\partial }}^{\alpha }\phi =(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\lim }_j{\int }_{\mathrm{\Omega }}{D}^{\alpha }(u_j{v}_j)\phi =(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\int }_{\mathrm{\Omega }}{g}_{\alpha }\phi$ with $g_{\alpha }:={\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right)D^{\beta }u{D}^{\alpha -\beta }v\in L^p(\mathrm{\Omega })$. By the characterization [F4], $g_{\alpha }$ is the weak $\alpha$-derivative of $uv$ for every $\mathrm{\mid }\alpha \mathrm{\mid }\le k$, so $uv\in W^{k,p}(\mathrm{\Omega })$, and $\mathrm{\parallel }uv{\mathrm{\parallel }}_{W^{k,p}(\mathrm{\Omega })}\le {\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le k}\mathrm{\parallel }{g}_{\alpha }{\mathrm{\parallel }}_{L^p}\le C(n,p,k,\mathrm{\Omega })\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{k,p}}\mathrm{\parallel }v{\mathrm{\parallel }}_{W^{k,p}}$ by step 1.1, which is the asserted algebra inequality. Completeness [F5] makes it a Banach algebra with continuous multiplication (after an equivalent norm rescaling if a submultiplicative norm is required).