Discrete Young inequality, the Sobolev product bound, and the transport product bound on the torus

Let $A,B:{\mathbb{Z}}^3\to [0,+\mathrm{\infty })$ be nonnegative with $A\in {\mathrm{\ell }}^1$, $B\in {\mathrm{\ell }}^2$. For each $k$, Cauchy-Schwarz [F2] gives $({\sum }_{l}A(k-l)B(l){)}^2\le ({\sum }_{l}A(k-l)\left)\right({\sum }_{l}A(k-l)B(l{)}^2)\le \mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}{\sum }_{l}A(k-l)B(l{)}^2$. Summing over $k$ and reindexing by [F1] (all terms nonnegative) yields ${\sum }_k({\sum }_{l}A(k-l)B(l){)}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}{\sum }_{l}B(l{)}^2{\sum }_{k}A(k-l)=\mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}^2\mathrm{\parallel }B{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}^2$, because ${\sum }_{k}A(k-l)={\sum }_{m}A(m)=\mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}$ for each fixed $l$. Hence $\mathrm{\parallel }A\ast B{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }B{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}$.

For $f,g\in H^{\sigma }$ with $\sigma >3\mathrm{/}2$, define the nonnegative families $A_l:=w_l^{\sigma }\mathrm{\mid }\widehat{f}(l)\mathrm{\mid }$, $B_m:=\mathrm{\mid }\widehat{g}(m)\mathrm{\mid }$, $A_l^{\mathrm{\prime }}:=\mathrm{\mid }\widehat{f}(l)\mathrm{\mid }$, $B_m^{\mathrm{\prime }}:=w_m^{\sigma }\mathrm{\mid }\widehat{g}(m)\mathrm{\mid }$. By the subadditivity $w_k\le w_{k-l}+w_l$ and convexity of $t\mapsto t^{\sigma }$ for $\sigma \ge 1$, one has $w_k^{\sigma }\le 2^{\sigma -1}(w_{k-l}^{\sigma }+w_l^{\sigma })$; multiplying the coefficient identity of [F5] by $w_k^{\sigma }$ and applying the triangle inequality therefore gives, for every $k$, the coefficient bound $w_k^{\sigma }\mid \widehat{fg}(k)\mid \le 2^{\sigma -1}((A\ast B)(k)+(A^{\mathrm{\prime }}\ast B^{\mathrm{\prime }})(k))$.

For complex $a,b$, ${\sum }_l\mathrm{\mid }a(k-l)b(l)\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_2<\mathrm{\infty }$ because $\mathrm{\mid }b(l)\mathrm{\mid }\le \mathrm{\parallel }b{\mathrm{\parallel }}_2$, so the convolution is absolutely convergent. The same bound holds: pointwise $\mathrm{\mid }(a\ast b)(k)\mathrm{\mid }\le (\mathrm{\mid }a\mathrm{\mid }\ast \mathrm{\mid }b\mathrm{\mid })(k)$, so $\mathrm{\parallel }a\ast b{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }\mathrm{\mid }a\mathrm{\mid }\ast \mathrm{\mid }b\mathrm{\mid }{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le \mathrm{\parallel }\mathrm{\mid }a\mathrm{\mid }{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }\mathrm{\mid }b\mathrm{\mid }{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}=\mathrm{\parallel }a{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}$ by step 1.1 and the triangle inequality in ${\mathrm{\ell }}^2$. This is claim 1.

Applying Young to the two convolutions of step 1.2, using the change of variable $l\mapsto k-l$ to commute the factors in the first, gives $\mathrm{\parallel }fg{\mathrm{\parallel }}_{H^{\sigma }}\le 2^{\sigma -1}(\mathrm{\parallel }A{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\mathrm{\parallel }B{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}+\mathrm{\parallel }{A}^{\mathrm{\prime }}{\mathrm{\parallel }}_{{\mathrm{\ell }}^1}\mathrm{\parallel }{B}^{\mathrm{\prime }}{\mathrm{\parallel }}_{{\mathrm{\ell }}^2})$. Thus $\mathrm{\parallel }A{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\sigma }}$, $\mathrm{\parallel }{B}^{\mathrm{\prime }}{\mathrm{\parallel }}_2=\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{\sigma }}$, while [F3] gives $\mathrm{\parallel }{A}^{\mathrm{\prime }}{\mathrm{\parallel }}_1\le W_{\sigma }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\sigma }}$ and $\mathrm{\parallel }B{\mathrm{\parallel }}_1\le W_{\sigma }\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{\sigma }}$. Therefore $\mathrm{\parallel }fg{\mathrm{\parallel }}_{H^{\sigma }}\le 2^{\sigma }{W}_{\sigma }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\sigma }}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{\sigma }}$; the finite weighted coefficient norm also proves $fg\in H^{\sigma }$. This is claim 2.

The coefficient estimate and steps 1.2–3.1 apply to each of the typed products, since their coefficient norm is at most the product of the input norms; thus claim 2 covers scalar-scalar, scalar-vector and componentwise vector-vector products with the same constant. For claim 3 write $(f\cdot \mathrm{\nabla })g={\sum }_{j=1}^3{f}_j{\mathrm{\partial }}_{j}g$. Since $g\in H^{\sigma +1}$, each ${\mathrm{\partial }}_{j}g\in H^{\sigma }$ with $\mathrm{\parallel }{\mathrm{\partial }}_{j}g{\mathrm{\parallel }}_{H^{\sigma }}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{H^{\sigma +1}}$, and each component $f_j$ of $f$ lies in $H^{\sigma }$ with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{H^{\sigma }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\sigma }}$. Claim 2 applied to the pair $(f_j,{\mathrm{\partial }}_{j}g)$ gives $\mathrm{\parallel }{f}_j{\mathrm{\partial }}_{j}g{\mathrm{\parallel }}_{H^{\sigma }}\le C_{\sigma }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\sigma }}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{\sigma +1}}$; summing the three components with the triangle inequality gives claim 3.