Schwartz derivatives are integrable

Put $W(x)={\prod }_{j<n}(1+x_j^2)$ and $A={\sum }_{ϵ\in \{0,1{\}}^n}{p}_{\alpha +2ϵ,\beta }(f)$. Expanding the product and taking absolute values gives $W(x)\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }\le A$. Also $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }\le B:=p_{\alpha \beta }(f)\le A$. Each factor $(1+t^2{)}^{-1}$ has integral at most $6$: its integral on $[-1,1]$ is at most $2$, while on $2^k\le \mathrm{\mid }t\mathrm{\mid }<2^{k+1}$ it contributes at most $2\cdot 2^k{2}^{-2k}$; the sum for $k\ge 0$ is $4$. All functions are continuous and hence Borel measurable. For $n=1$ this proves $\int W^{-1}\le 6$. For $n>1$, [F2] identifies integration of this Borel function against ${\lambda }_n$ with integration against ${\lambda }_{n-1}\times {\lambda }_1$; induction on $n$ and [F1] therefore give $\int W^{-1}\le 6^n$.

For $g=x^{\alpha }{\mathrm{\partial }}^{\beta }f$, step 1.1 gives $\mathrm{\parallel }g{\mathrm{\parallel }}_1\le 6^{n}A$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le B$. If $1<p<\mathrm{\infty }$, integrating $\mathrm{\mid }g{\mathrm{\mid }}^p\le B^{p-1}\mathrm{\mid }g\mathrm{\mid }$ gives $\mathrm{\parallel }g{\mathrm{\parallel }}_p\le B^{1-1\mathrm{/}p}(6^{n}A{)}^{1\mathrm{/}p}\le 6^{n}A$. When $B=0$, $g=0$ and this conclusion holds directly. This treats both endpoint spaces and every intermediate exponent.