The periodic Sobolev space H^k of the three-torus is a Banach algebra for k at least four

(Embedding control (a).) Since $k-(k-2)=2>3\mathrm{/}2$, [L1] with $(s,m)=(k,k-2)$ gives a $C^{k-2}$ representative of $u$ and the bound $\sup \mathrm{\mid }{D}^{\gamma }u\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{C^{k-2}}\le C_k\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}$ for every $\mathrm{\mid }\gamma \mathrm{\mid }\le k-2$.

(Each Leibniz term for smooth factors is bounded by $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^k}$.) First take smooth $u,v$ and fix $\mathrm{\mid }\alpha \mathrm{\mid }\le k$. Expand by [L3]. In a term $D^{\beta }u{D}^{\alpha -\beta }v$ two cases occur. If $\mathrm{\mid }\beta \mathrm{\mid }\ge k-1$, then $\mathrm{\mid }\alpha -\beta \mathrm{\mid }\le 1\le k-2$, so step 1.1 bounds the second factor in $L^{\mathrm{\infty }}$ and [L2] bounds the first in $L^2$. If $\mathrm{\mid }\beta \mathrm{\mid }\le k-2$, step 1.1 bounds the first factor in $L^{\mathrm{\infty }}$, while [L2] bounds the second in $L^2$. Both cases give an $L^2$ bound $C_k\mathrm{\parallel }u{\mathrm{\parallel }}_{H^k}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^k}$. They exhaust the possibilities because both orders cannot be at least $k-1$ when $\mathrm{\mid }\alpha \mathrm{\mid }\le k$ and $k\ge 4$.

(Conclusion of (b).) Summing the finitely many terms of [L3] and using [L4] gives the stated product bound for smooth $u,v$. For general $u,v\in H^k$, put $u_N=P_{N}u$ and $v_N=P_{N}v$. By [L5], these converge in $H^k$ and, by [L1], uniformly. The smooth estimate applied to $u_N{v}_N-u_M{v}_M=(u_N-u_M)v_N+u_M(v_N-v_M)$ shows that $(u_N{v}_N)$ is Cauchy in $H^k$. Completeness [L5] gives a limit, and uniform convergence identifies it with $uv$. For every smooth test function $\eta$, pass to the limit in $\int u_N{v}_N{D}^{\alpha }\eta =(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\int D^{\alpha }(u_N{v}_N)\eta$; uniform convergence on the left and $L^2$ convergence of the derivatives on the right show that the limits are the weak derivatives of $uv$. The product estimate passes to the limit by norm continuity and $\mathrm{\parallel }{P}_{N}w{\mathrm{\parallel }}_{H^k}\le \mathrm{\parallel }w{\mathrm{\parallel }}_{H^k}$.

((c): smoothing by a classical factor.) First take $w_N=P_{N}w$, which is smooth. The $C^k$ Leibniz rule [L3] and [L6] give its weak derivatives by the displayed finite sum. Since $\mathrm{\mid }{D}^{\beta }\phi \mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_{C^k}$ and $\mathrm{\parallel }{D}^{\alpha -\beta }{w}_N{\mathrm{\parallel }}_2\le \mathrm{\parallel }{w}_N{\mathrm{\parallel }}_{H^k}$, summing the binomial coefficients gives $\mathrm{\parallel }\phi w_N{\mathrm{\parallel }}_{H^k}\le C_k\mathrm{\parallel }\phi {\mathrm{\parallel }}_{C^k}\mathrm{\parallel }{w}_N{\mathrm{\parallel }}_{H^k}$. Applying this estimate to $w_N-w_M$ and using completeness [L5] shows that $\phi w_N$ converges in $H^k$; uniform convergence identifies its limit as $\phi w$. Passing the weak derivative identities against smooth test functions gives $\phi w\in H^k$, and the bound follows from $\mathrm{\parallel }{w}_N{\mathrm{\parallel }}_{H^k}\le \mathrm{\parallel }w{\mathrm{\parallel }}_{H^k}$.