Fundamental solution of a constant-coefficient operator

Define $E_y$ by the displayed pairing. By [F3], $T_{-y}\phi$ is a test function, and by [F1] composition with $E$ is a continuous complex-linear functional. Thus $E_y$ is a distribution; the formula is bilinear and uses no complex conjugation.

For every multi-index $\alpha$, use [F2] and then differentiate the translated test directly to obtain $⟨{\mathrm{\partial }}^{\alpha }{E}_y,\phi ⟩=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}⟨E,T_{-y}{\mathrm{\partial }}^{\alpha }\phi ⟩=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}⟨E,{\mathrm{\partial }}^{\alpha }{T}_{-y}\phi ⟩=⟨{\mathrm{\partial }}^{\alpha }E,T_{-y}\phi ⟩\mathrm{.}$ The equality $T_{-y}{\mathrm{\partial }}^{\alpha }\phi ={\mathrm{\partial }}^{\alpha }{T}_{-y}\phi$ follows because $T_{-y}\phi (x)=\phi (x+y)$ and $y$ is fixed.

By linearity of the distribution pairing, step 1.2 extends from each partial derivative to their finite constant-coefficient combination $L$. Hence $⟨L{E}_y,\phi ⟩=⟨LE,T_{-y}\phi ⟩$. If $LE={\delta }_0$, [F4] makes the right side $(T_{-y}\phi )(0)=\phi (y)={\delta }_y(\phi )$, so $L{E}_y={\delta }_y$. This proves the translated point-source assertion.

The calculation also covers the zero operator: its premise $LE={\delta }_0$ cannot hold since ${\delta }_0$ evaluates a test with value $1$ at zero as $1$. For $n=1$ the same multi-index computation applies unchanged; in the zero-dimensional formal case the only translation is by $0$ and the assertion is the premise itself. There are no spatial boundary endpoints on ${\mathbb{R}}^n$, and the proof uses only the fixed translation and finite algebra, not a choice axiom.