Dirac comb

For $n\ge 1$, the unit-lattice Dirac comb is

${\mathrm{III}}_{{\mathbb{Z}}^n}={\sum }_{k\in {\mathbb{Z}}^n}{\delta }_k,⟨{\mathrm{III}}_{{\mathbb{Z}}^n},\phi ⟩={\sum }_{k\in {\mathbb{Z}}^n}\phi (k)\mathrm{.}$

This defines a tempered distribution, not merely a formal series. Indeed, choose an integer $L>n+1$. The shell $\{k\in {\mathbb{Z}}^n:m\le \mathrm{\mid }k\mathrm{\mid }<m+1\}$ has at most $(2m+3{)}^n$ points, and the Schwartz-seminorm definition gives a constant $A_{n,L}$ such that

$\mathrm{\mid }\phi (k)\mathrm{\mid }\le A_{n,L}(1+\mathrm{\mid }k\mathrm{\mid }{)}^{-L}{\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le L}{p}_{\alpha ,0}(\phi )\mathrm{.}$

The resulting shell series is bounded by a constant times ${\sum }_{m\ge 1}{m}^{n-L}$, which converges by The p-series for a real exponent p converges exactly when p is greater than one. Thus the lattice sum is absolutely convergent and obeys one finite seminorm estimate, so Finite seminorm bound characterizes tempered distributions applies. On a compactly supported test only finitely many terms remain, and the restriction agrees with the locally finite sum of the Dirac distributions Dirac delta and its derivatives. The zero test gives zero. The shell decomposition is canonical, so no choice axiom is used.