Local structure of distributions as derivatives of continuous functions

Take $\chi \in \mathcal{D}(\mathrm{\Omega })$ equal to one near $K$ using F3. The distribution $v=\chi u$ has compact support inside $\mathrm{supp}\chi$: it vanishes on the complement by its defining pairing. F2 therefore lets it act on restrictions of all smooth functions on ${\mathbb{R}}^n$; denote that action by $w(\psi )=u(\chi \psi {\mathrm{\mid }}_{\mathrm{\Omega }})$. F1 on $\mathrm{supp}\chi$ and the finite product rule give $\mathrm{\mid }w(\psi )\mathrm{\mid }\le A{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le m}{\sup }_{\mathrm{supp}\chi }\mathrm{\mid }{\mathrm{\partial }}^{\gamma }\psi \mathrm{\mid }$ for some $A,m$. If $u$ has global order at most $m$, its local estimate has that same exponent and multiplication does not raise it. Choose a nondegenerate open box $B={\prod }_i(a_i,b_i)$ containing $\mathrm{supp}\chi$ compactly in its interior. It need not lie in $\mathrm{\Omega }$.

Put $k=(m+1,\dots ,m+1)$ and $T={\mathrm{\partial }}^k$ on $\mathcal{D}(B)$. For $\mathrm{\mid }\gamma \mathrm{\mid }\le m$ let $q_i=k_i-{\gamma }_i\ge 1$. Repeatedly integrating in each coordinate from $a_i$ gives [step 1.1, F5, F8] ${\mathrm{\partial }}^{\gamma }\psi (x)={\int }_{a_1}^{x_1}\cdots {\int }_{a_n}^{x_n}{\prod }_i\frac{(x_i-t_i{)}^{q_i-1}}{(q_i-1)!}T\psi (t)dt\mathrm{.}$ All lower endpoint derivatives vanish because $\psi$ is compactly supported in $B$. The formula follows from the FTC in F8 by $q_i$ successive integrations; F5 changes the repeated integral over each simplex to the displayed polynomial kernel. Its kernel is bounded on the box by a finite constant depending on $B,\gamma ,m$. Hence $p_m(\psi )\le C_{B,m}\mathrm{\parallel }T\psi {\mathrm{\parallel }}_{L^1(B)}$. This also proves injectivity of $T$ as a map to $L^1$ classes: if $T\psi =0$ almost everywhere, all displayed integrals vanish, including the one for $\psi$. Thus on the complex subspace $E=T(\mathcal{D}(B))\subseteq L^1(B;\mathbb{C})$, $\mathrm{\ell }(T\psi )=w(\psi )$ is well-defined and has bound $\mathrm{\mid }\mathrm{\ell }(z)\mathrm{\mid }\le M\mathrm{\parallel }z{\mathrm{\parallel }}_1$, $M=A{C}_{B,m}$.

To extend this functional, regard $L^1(B;\mathbb{C})$ as a real normed space and set $p(z)=M\mathrm{\parallel }z{\mathrm{\parallel }}_1$. Consider all real-linear extensions of $\mathrm{Re}\mathrm{\ell }$ from the underlying real subspace $E$ to real subspaces, dominated above by $p$, ordered by extension. This is a set of graphs; it is nonempty, since the original real part is dominated. The union of a nonempty chain is a well-defined real-linear functional on the union subspace, still dominated by $p$; the empty chain has the original functional as an upper bound. F4 and F8 give a maximal such extension $h$ on a domain $D$.

If $z\mathrm{\notin }D$, define [step 3.1, algebra] $L_z={\sup }_{d\in D}(h(d)-p(d-z)),U_z={\inf }_{d\in D}(p(d+z)-h(d))\mathrm{.}$ For $d,e\in D$, $h(d)+h(e)=h(d+e)\le p(d+e)\le p(d-z)+p(e+z)$, so every lower candidate is at most every upper candidate. The choices $d=0$ show $-p(-z)\le L_z\le U_z\le p(z)$, so these are finite and $c=L_z$ belongs to the interval. Define $h^{\mathrm{\prime }}(d+tz)=h(d)+tc$. The decomposition is unique. For $t>0$ the upper bound on $c$ applied to $d\mathrm{/}t$ gives domination by $p(d+tz)$; for $t<0$ the lower bound applied to $d\mathrm{/}(-t)$ gives the same, and $t=0$ is the old bound. Thus $h^{\mathrm{\prime }}$ contradicts maximality. Consequently $D=L^1(B;\mathbb{C})$.

Set $\mathrm{\Lambda }(z)=h(z)-ih(iz)$. Real linearity and $\mathrm{\Lambda }(iz)=i\mathrm{\Lambda }(z)$ make it complex-linear. On $E$ it equals $\mathrm{\ell }$ since $\mathrm{Re}\mathrm{\ell }(iz)=-\mathrm{Im}\mathrm{\ell }(z)$. If $\mathrm{\Lambda }(z)\mathrm{\ne }0$, put $\omega =\overline{\mathrm{\Lambda }(z)}\mathrm{/}\mathrm{\mid }\mathrm{\Lambda }(z)\mathrm{\mid }$. Then $\mathrm{\mid }\mathrm{\Lambda }(z)\mathrm{\mid }=\mathrm{Re}\mathrm{\Lambda }(\omega z)=h(\omega z)\le M\mathrm{\parallel }z{\mathrm{\parallel }}_1$; the zero case has the same bound. F6 on the finite Borel Lebesgue measure of $B$ gives a bounded Borel representative $b$, after setting its exceptional null values to zero, with $w(\psi )={\int }_{B}bT\psi$.

Extend $b$ by zero outside $B$ and define $G(x)={\int }_{\{t_i\le x_i\mathrm{\forall }i\}}b(t)dt$. It is continuous on all of ${\mathbb{R}}^n$: for $x,x^{\mathrm{\prime }}$, the symmetric difference of the two lower orthants intersected with $B$ lies in the union of coordinate slabs of widths at most $\mathrm{\mid }{x}_i-x_i^{\mathrm{\prime }}\mathrm{\mid }$. Thus $\mathrm{\mid }G(x)-G(x^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}{\sum }_i\mathrm{\mid }{x}_i-x_i^{\mathrm{\prime }}\mathrm{\mid }{\prod }_{j\mathrm{\ne }i}(b_j-a_j)$. For a test $\zeta$ on ${\mathbb{R}}^n$, F5 swaps $t,x$ in the absolutely integrable function $b(t)1_{t_i\le x_i\mathrm{\forall }i}{\mathrm{\partial }}_1\cdots {\mathrm{\partial }}_n\zeta (x)$: $t$ lies in bounded $B$ and $x$ in the compact derivative support. F8 in each $x_i$ gives the inner integral $(-1{)}^n\zeta (t)$, because all upper endpoint values vanish. Therefore $\int G{\mathrm{\partial }}_1\cdots {\mathrm{\partial }}_n\zeta =(-1{)}^n\int b\zeta$.

Apply step 6.1 to $\zeta =T\psi$, $\psi \in \mathcal{D}(B)$. Then $w(\psi )=(-1{)}^n\int G{\mathrm{\partial }}^{k+\mathbf{1}}\psi$. Put $f=(-1{)}^{\mathrm{\mid }k\mathrm{\mid }}G{\mathrm{\mid }}_{\mathrm{\Omega }}$ and $\alpha =k+\mathbf{1}$. Since $\mathrm{\mid }\alpha \mathrm{\mid }=\mathrm{\mid }k\mathrm{\mid }+n$, F7 gives ${\mathrm{\partial }}^{\alpha }{u}_f(\phi )=(-1{)}^n\int G{\mathrm{\partial }}^{\alpha }\phi =w(\phi )=u(\phi )$ for every $\phi \in {\mathcal{D}}_K$. The function $f$ is continuous and locally integrable. Empty $K$ has only the zero test, so $f=0,\alpha =0$ suffices there; zero bounds in the construction give zero functionals and cause no division by a norm. This proves the claim and the stated exponent control. $\mathrm{\square }$