Distributions supported at one point

Suppose its support is contained in $\{a\}$. Obtain an order $m$ and constant from F2. Take a fixed smooth $\chi$ equal to one near zero with support in a ball $B(0,R)$, using F3 in Euclidean space. For all sufficiently small $\epsilon >0$, ${\chi }_{\epsilon }(x)=\chi ((x-a)\mathrm{/}\epsilon )$ is supported inside $\mathrm{\Omega }$ and equals one near $a$. Hence $u(\phi )=u({\chi }_{\epsilon }\phi )$ for every test.

Suppose ${\mathrm{\partial }}^{\gamma }\phi (a)=0$ for every $\mathrm{\mid }\gamma \mathrm{\mid }\le m$. For fixed $\gamma$, Taylor's formula F4 on a ball about $a$ applied to ${\mathrm{\partial }}^{\gamma }\phi$ gives ${\mathrm{\partial }}^{\gamma }\phi (a+h)=o(\mathrm{\mid }h{\mathrm{\mid }}^{m-\mathrm{\mid }\gamma \mathrm{\mid }})$; when $\mathrm{\mid }\gamma \mathrm{\mid }=m$ this is continuity with value zero. The little-o bounds are uniform over $\mathrm{\mid }h\mathrm{\mid }\le R\epsilon$ by their definition: their suprema are $o({\epsilon }^{m-\mathrm{\mid }\gamma \mathrm{\mid }})$. The product rule expands a derivative of order $\beta$, $\mathrm{\mid }\beta \mathrm{\mid }\le m$, of ${\chi }_{\epsilon }\phi$ into finitely many terms [step 1.1, F2, F4, algebra] $\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right){\epsilon }^{-\mathrm{\mid }\beta -\gamma \mathrm{\mid }}({\mathrm{\partial }}^{\beta -\gamma }\chi )((x-a)\mathrm{/}\epsilon ){\mathrm{\partial }}^{\gamma }\phi (x),\gamma \le \beta \mathrm{.}$ Each has supremum $o({\epsilon }^{m-\mathrm{\mid }\beta \mathrm{\mid }})$, hence tends to zero, even when $\mathrm{\mid }\beta \mathrm{\mid }=m$. All derivatives vanish outside the shrinking support. Thus F2 and step 1.1 give $\mathrm{\mid }u(\phi )\mathrm{\mid }\le C{\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le m}\sup \mathrm{\mid }{\mathrm{\partial }}^{\beta }({\chi }_{\epsilon }\phi )\mathrm{\mid }\to 0$, so $u(\phi )=0$.

Take a fixed cutoff $\eta$ equal to one near $a$ and compactly supported in $\mathrm{\Omega }$. For $\mathrm{\mid }\alpha \mathrm{\mid }\le m$ put $q_{\alpha }(x)=\eta (x)(x-a{)}^{\alpha }\mathrm{/}\alpha !$. Direct monomial differentiation gives ${\mathrm{\partial }}^{\beta }{q}_{\alpha }(a)=1$ for $\beta =\alpha$ and zero for every other $\mathrm{\mid }\beta \mathrm{\mid }\le m$. Therefore $\phi -{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\mathrm{\partial }}^{\alpha }\phi (a)q_{\alpha }$ has zero jet through degree $m$. Step 2.1 makes its pairing zero, giving $u(\phi )=\sum u(q_{\alpha }){\mathrm{\partial }}^{\alpha }\phi (a)$. By F1 this is the claimed representation with $c_{\alpha }=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}u(q_{\alpha })$.

Conversely, every test supported in $\mathrm{\Omega }\setminus \{a\}$ has all derivatives zero at $a$, so F1 makes every displayed combination vanish there. Its support is therefore contained in $\{a\}$. Evaluate a zero combination on the tests $q_{\alpha }$ constructed with the largest order occurring to see each coefficient is zero. This proves uniqueness. Finally a nonzero distribution cannot have empty support by locality (as used in F2), so a nonzero combination has support exactly $\{a\}$. The zero combination is allowed with $m=0,c_0=0$. All jets and sums are finite and no choice is used. $\mathrm{\square }$