Fundamental Solutions Newtonian Potentials and Green Functions

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.

By [F1] and [F2], ${\omega }_{n-1}=n\mathrm{\mid }{B}_1\mathrm{\mid }$ is finite and strictly positive. In dimension two, [F3] gives $\mathrm{\mid }{B}_1\mathrm{\mid }=\pi$, so ${\omega }_1=2\pi$ and the logarithmic coefficient in the Statement is exactly $1\mathrm{/}{\omega }_1$. The operator sign is $L=-\mathrm{\Delta }$ using the definition [F9]. This uses the Countable Choice assumption [A1] through the measure convention.

In dimension one, ${\int }_{-R}^R\mathrm{\mid }{\mathrm{\Phi }}_1(x)\mathrm{\mid }dx=R^2\mathrm{/}2$, so ${\mathrm{\Phi }}_1$ is locally integrable. For a test $\phi$, [F7] gives $⟨-{\mathrm{\Phi }}_1^{\mathrm{\prime }\mathrm{\prime }},\phi ⟩=-{\int }_{\mathbb{R}}{\mathrm{\Phi }}_1(x){\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx$. The slopes of ${\mathrm{\Phi }}_1$ are $+1\mathrm{/}2$ to the left of zero and $-1\mathrm{/}2$ to the right; integrating by parts separately on the two half-lines, the compact-support endpoint terms vanish and ${\int }_{\mathbb{R}}{\mathrm{\Phi }}_1{\phi }^{\mathrm{\prime }\mathrm{\prime }}=-\phi (0)$, so the negative integral equals $\phi (0)={\delta }_0(\phi )$ by [F8]. Thus $-{\mathrm{\Phi }}_1^{\mathrm{\prime }\mathrm{\prime }}={\delta }_0$ and [F5] identifies it as the one-dimensional fundamental-solution analogue for $L=-d^2\mathrm{/}d{x}^2$.

Let $0<R\le 1$. For $n\ge 3$, [F4] gives ${\int }_{B_R}\mathrm{\mid }{\mathrm{\Phi }}_n(x)\mathrm{\mid }dx=\frac{1}{n-2}{\int }_0^{R}rdr=\frac{R^2}{2(n-2)}<\mathrm{\infty }$. For $n=2$, using ${\omega }_1=2\pi$ from step 1.1, it gives ${\int }_{B_R}\mathrm{\mid }{\mathrm{\Phi }}_2(x)\mathrm{\mid }dx={\int }_0^{R}r(-\log r)dr=-\frac{R^2}{2}\log R+\frac{R^2}{4}<\mathrm{\infty }$. The polar integration use [F4] also assumes [A1].

The formulas are continuous away from zero, so they are integrable on every compact set avoiding the pole. Step 2.1 proves integrability in a neighborhood of the pole; hence ${\mathrm{\Phi }}_n\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. By [F10], changing its value at the single point zero does not change its almost-everywhere class, and [F6] defines the corresponding regular functional.

The zero-dimensional case is outside the stated $n\ge 2$ kernel and has no unit-sphere convention used here. The polar endpoint $r=0$ is included as an improper integral in step 2.1; away from zero the kernel is smooth. Countable Choice is used only through [F1], [F2], and [F4]; no full Axiom of Choice is used.

For $\mathrm{\Omega }={\mathbb{R}}^n$, every translated compact support $x-\mathrm{supp}\phi$ lies in $\mathrm{\Omega }$, so the safe domain is all of ${\mathbb{R}}^n$. Also ${\mathrm{\partial }}^{\alpha }\phi$ is a test function with support contained in $\mathrm{supp}\phi$. Thus all three convolutions are defined there by [F2].

By the smoothness and parameter-differentiation conclusion in [F1], differentiating the pairing in [F2] gives ${\mathrm{\partial }}_x^{\alpha }(T\ast \phi )(x)=⟨T(y),({\mathrm{\partial }}^{\alpha }\phi )(x-y)⟩=T\ast ({\mathrm{\partial }}^{\alpha }\phi )(x)$.

The signed derivative definition [F3] gives $({\mathrm{\partial }}^{\alpha }T)\ast \phi (x)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}⟨T(y),{\mathrm{\partial }}_y^{\alpha }\phi (x-y)⟩$. Since ${\mathrm{\partial }}_y^{\alpha }\phi (x-y)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}({\mathrm{\partial }}^{\alpha }\phi )(x-y)$, the two signs cancel and this equals the expression in step 1.2. Hence all three smooth functions agree on the safe domain.

The same calculation includes $\alpha =0$; if $\phi =0$ or $T=0$ each side is zero; for $n=1$ it is the same one-variable derivative calculation, and formally for $n=0$ the only multi-index is zero and the identity is tautological. There are no spatial boundary endpoints on ${\mathbb{R}}^n$, and no choice is used.

For real $u$, the gradient field $F=\mathrm{\nabla }u$ belongs to $C^1(\bar{\mathrm{\Omega }};{\mathbb{R}}^n)$ by the stated $C^2$ closure convention. By [F2], $\mathrm{div}F=\mathrm{\Delta }u$, and by [F3], $F\cdot \nu =g$ at each boundary point.

Apply the divergence theorem [F1] to the field in step 1.1. It gives ${\int }_{\mathrm{\Omega }}\mathrm{\Delta }udx={\int }_{\mathrm{\partial }\mathrm{\Omega }}gdS$. This use of [F1] requires exactly the Countable Choice assumption [A1].

Since $f=-\mathrm{\Delta }u$, negating the identity in step 2.1 yields ${\int }_{\mathrm{\Omega }}fdx=-{\int }_{\mathrm{\partial }\mathrm{\Omega }}gdS$. For complex-valued $u,f,g$, apply this real calculation separately to real and imaginary parts.

If $f=0$, the identity says the total outward Neumann flux is zero; if also $g=0$, both sides vanish. The theorem applies on every boundary component with the outward orientation specified in [F3]. The domain class in [F4] excludes the empty set and dimensions zero or one. No converse or sufficiency for existence is asserted.

Fix $x,z\in \mathrm{\Omega }\setminus \{y\}$. If $x=z$, the constant path at $x$ lies in the punctured set. Otherwise [F1] gives a polygonal path $\gamma :[0,1]\to \mathrm{\Omega }$ from $x$ to $z$. Choose $r>0$ small enough that ${\bar{B}}_2(y,r)\subset \mathrm{\Omega }$ and $r<\min \{\mathrm{\mid }x-y\mathrm{\mid },\mathrm{\mid }z-y\mathrm{\mid }\}$. If $\gamma$ avoids $y$, it already gives the required path.

Suppose $\gamma$ meets $y$. The preimage $A={\gamma }^{-1}[{\bar{B}}_2(y,r)]$ is a finite union of closed intervals: on each of the finitely many affine pieces in [F1], the preimage of the convex closed ball is a closed interval, possibly empty or a point. Its connected components are therefore finitely many closed intervals $[a_j,b_j]$. Since the endpoints $x,z$ lie outside the closed ball, each component is contained in $(0,1)$, and continuity and maximality give $\gamma (a_j),\gamma (b_j)\in S_2(y,r)$. If $a_j=b_j$, that component is a single sphere point and cannot contain $y$.

For each component with $a_j<b_j$, use [F2] to choose a continuous path on $S_2(y,r)$ from $\gamma (a_j)$ to $\gamma (b_j)$, reparameterized on $[a_j,b_j]$. Replace $\gamma$ on those finitely many intervals by these sphere paths and retain it on the intervening closed intervals. The pieces agree at every endpoint, so [F3] gives a continuous path $\tilde{\gamma }:[0,1]\to \mathrm{\Omega }$. Every replacement lies on a sphere of positive radius and hence misses $y$; the retained portions lie outside the closed ball, apart from singleton components already on the sphere. Some component has positive length because $\gamma$ meets the interior point $y$. Thus $\tilde{\gamma }$ avoids $y$ and joins $x$ to $z$. This also covers any zero-length polygonal pieces and tangencies to the sphere.

The construction proves path-connectedness, including equal endpoints by the constant path in step 1.1. The set is open: for each $x\mathrm{\ne }y$ in $\mathrm{\Omega }$, openness of $\mathrm{\Omega }$ gives a ball about $x$ contained in $\mathrm{\Omega }$, and shrinking its radius below $\mathrm{\mid }x-y\mathrm{\mid }$ makes it avoid $y$. It is nonempty because a ball about $y$ contains $y+t{e}_1\mathrm{\ne }y$ for some sufficiently small $t>0$, where $e_1=(1,0,\dots ,0)$. By [F4] it is connected. The proof covers zero-length polygonal pieces, tangent singleton components and the case $x=z$; no infinite family of detours or iff claim is used. [F3, F4, given, step 1.1, step 3.1, algebra, cases, discharge-construct] \square

First take real-valued functions and put $w=u-v$. By [F6], $D(\mathrm{\nabla }w)=D(\mathrm{\nabla }u)-D(\mathrm{\nabla }v)$; taking traces and using [F7] gives $\mathrm{\Delta }w=0$ on every component. By [F2] and the same derivative linearity, ${\mathrm{\partial }}_{\nu }w={\mathrm{\partial }}_{\nu }u-{\mathrm{\partial }}_{\nu }v=0$ on its boundary. For complex-valued functions, apply this real argument separately to their real and imaginary parts.

Suppose $n\ge 2$ and fix a connected component $U$. Apply [F1] with $a=b=w$ on $U$. The volume integrand is $w\mathrm{\Delta }w+\mathrm{\mid }Dw{\mathrm{\mid }}^2=\mathrm{\mid }Dw{\mathrm{\mid }}^2$ and the boundary integrand is $w{\mathrm{\partial }}_{\nu }w=0$, so ${\int }_U\mathrm{\mid }Dw{\mathrm{\mid }}^{2}dx=0$. Continuity of $Dw$ forces $Dw=0$ throughout $U$: if it were nonzero at one point, it would be bounded away from zero on a small ball of positive measure. By [F3], $w$ is constant on $U$. The invocation of [F1] uses precisely the Countable Choice assumption [A1].

Suppose $n=1$. By [F5], each connected component $U$ is an interval; boundedness makes it an interval with finite endpoints $a<b$. The equation in step 1.1 is $w^{\mathrm{\prime }\mathrm{\prime }}=0$. For any $x<y$ in $U$, the mean value theorem [F4] applied to $w^{\mathrm{\prime }}$ on $[x,y]$ gives $w^{\mathrm{\prime }}(y)-w^{\mathrm{\prime }}(x)=w^{\mathrm{\prime }\mathrm{\prime }}(c)(y-x)=0$, so $w^{\mathrm{\prime }}$ is constant on $U$. Its continuous trace at the right endpoint is zero because the outward normal there is $+1$ and ${\mathrm{\partial }}_{\nu }w=0$. Thus $w^{\mathrm{\prime }}=0$ on $U$, and [F3] gives that $w$ is constant there.

The components are handled independently, so their constants need not agree. If there is only one component, the conclusion is one constant on all of $\mathrm{\Omega }$; zero difference is included. In dimensions at least two, the only use of ${\mathrm{A}\mathrm{C}}_{\omega }$ is through [F1] under [A1]; the interval proof in dimension one uses no choice.

By [F3] and [F4], ${\omega }_{n-1}=n\mathrm{\mid }{B}_1\mathrm{\mid }$ is positive and finite in every stated dimension.

For $n\ge 3$ and any $R>0$, assign the finite value $0$ to the kernel at the pole as permitted by [F11]; then $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }{\mathbf{1}}_{B_R}$ is Borel. Apply [F2] and use [F1] and $\sigma (S^{n-1})={\omega }_{n-1}$ from [F3] to obtain ${\int }_{B_R}\mathrm{\mid }\mathrm{\Phi }(x)\mathrm{\mid }dx=\frac{1}{(n-2){\omega }_{n-1}}{\int }_0^R{r}^{n-1}{r}^{2-n}{\omega }_{n-1}dr=\frac{R^2}{2(n-2)}<\mathrm{\infty }\mathrm{.}$

For $n=2$ and any $R>0$, again assign $\mathrm{\Phi }(0)=0$ as permitted by [F11]. By [F1]–[F3] and step 1.1, ${\int }_{B_R}\mathrm{\mid }\mathrm{\Phi }(x)\mathrm{\mid }dx=\frac{{\omega }_1}{2\pi }{\int }_0^{R}r\mathrm{\mid }\log r\mathrm{\mid }dr\mathrm{.}$ If $0<R\le 1$, the radial integral is $-\frac{R^2}{2}\log R+\frac{R^2}{4}$; if $R\ge 1$, splitting at $1$ gives $\frac{R^2}{2}\log R-\frac{R^2}{4}+\frac{1}{2}$. Both values are finite, including at $R=1$, and the prefactor is finite by step 1.1.

Let $K\subset {\mathbb{R}}^n$ be compact. By [F6], $K$ is closed and bounded, hence Borel and Lebesgue measurable by [F7]; boundedness and the Euclidean triangle inequality give a centered ball $B_R$ containing $K$. By [F5] and steps 2.1–2.2, ${\int }_K\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }\le {\int }_{B_R}\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }<\mathrm{\infty }$ (and the empty $K$ has integral zero). Since [F1] and [F11] make $\mathrm{\Phi }$ measurable, this is $\mathrm{\Phi }\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$ by [F8]'s definition.

The regular functional in [F8] is therefore well-defined; [F9], under [A1], proves it is a distribution. The pole value changes only a singleton, which is null by [F10], and the radial integrals prove finiteness at the improper endpoint $r=0$ and every finite outer radius $R>0$. The claim assumes $n\ge 2$; it makes no global-integrability assertion at $R=\mathrm{\infty }$, and Countable Choice is used only through the named polar, surface-measure, ball-measure, Borel-measurability, singleton-null, and embedding interfaces, not full AC.

For any real $\alpha$, induction on $j$ using [F12] and [F7] gives $(t^{\alpha }{)}^{(j)}=c_j{t}^{\alpha -j}$ on $(0,\mathrm{\infty })$, where $c_0=1$ and $c_{j+1}=c_j(\alpha -j)$. Each derivative is continuous by the real-power continuity in [F7], so $t^{\alpha }$ is smooth under [F2]. Also ${\log }^{\mathrm{\prime }}(t)=t^{-1}$ by [F8]; applying the same derivative calculation to $t^{-1}$ shows every higher derivative of $\log t$ exists and is continuous. Thus both radial profiles used in [F1] are smooth for $t>0$.

On $U={\mathbb{R}}^n\setminus \{0\}$, put $s(x)={\sum }_{i=1}^n{x}_i^2$. Its coordinate functions and their finite sums and products are smooth by direct coordinate differentiation and [F2]–[F4]. Since $s(x)>0$ on $U$, the radius $r(x)=s(x{)}^{1\mathrm{/}2}$ is smooth there by [F7], step 1.1, and closure under composition [F4]. Composing $r$ with the power profile for $n\ge 3$ or the logarithm profile for $n=2$ proves that $\mathrm{\Phi }$ is smooth on $U$.

For a smooth radial profile $q(r)$ and $r=\mathrm{\mid }x\mathrm{\mid }>0$, the chain rule [F5] and partial-derivative formula [F6] give ${\mathrm{\partial }}_{i}q(r)=q^{\mathrm{\prime }}(r)x_i\mathrm{/}r$. Differentiating again by the chain and product rules [F5], [F7] and [F9] gives ${\mathrm{\partial }}_i^{2}q(r)=q^{\mathrm{\prime }\mathrm{\prime }}(r)x_i^2\mathrm{/}{r}^2+q^{\mathrm{\prime }}(r)(1\mathrm{/}r-x_i^2\mathrm{/}{r}^3)$. Summing over $i$ and using ${\sum }_i{x}_i^2=r^2$ and the Laplacian definition [F10] yields $\mathrm{\Delta }q(r)=q^{\mathrm{\prime }\mathrm{\prime }}(r)+(n-1)q^{\mathrm{\prime }}(r)\mathrm{/}r$. The calculation is on $r>0$, where all derivatives used exist by steps 1.1 and 2.1.

If $n\ge 3$, set $q(r)=r^{2-n}\mathrm{/}((n-2){\omega }_{n-1})$. Then $q^{\mathrm{\prime }}(r)=-r^{1-n}\mathrm{/}{\omega }_{n-1}$ and $q^{\mathrm{\prime }\mathrm{\prime }}(r)=(n-1)r^{-n}\mathrm{/}{\omega }_{n-1}$ by [F7] and [F9], so step 3.1 gives $\mathrm{\Delta }\mathrm{\Phi }=0$. If $n=2$, set $q(r)=-(2\pi {)}^{-1}\log r$. Then $q^{\mathrm{\prime }}(r)=-(2\pi r{)}^{-1}$ by [F8]–[F9] and $q^{\mathrm{\prime }\mathrm{\prime }}(r)=(2\pi r^2{)}^{-1}$ by applying [F7] to $r^{-1}$; hence $q^{\mathrm{\prime }\mathrm{\prime }}+q^{\mathrm{\prime }}\mathrm{/}r=0$. These cases exhaust $n\ge 2$.

For fixed $y$, translation $x\mapsto x-y$ has affine coordinate functions, so direct differentiation gives its identity derivative and zero higher derivatives. The chain rule [F5] therefore gives ${\mathrm{\Delta }}_x(\mathrm{\Phi }(x-y))=(\mathrm{\Delta }\mathrm{\Phi })(x-y)=0$ whenever $x\mathrm{\ne }y$. The translated function is smooth there by [F4], hence is harmonic by [F11]. The assumption ${\mathrm{A}\mathrm{C}}_{\omega }$ in [A1] is carried from the kernel convention but is not used in these pointwise derivative calculations.

On ${\mathbb{R}}^n\setminus \{0\}$ the formulas in [F1] are continuous, so assigning the finite value $0$ at the closed singleton $\{0\}$ gives a Borel representative $\tilde{\mathrm{\Phi }}$ of the kernel.

By [F3] and [F4], using [A1], choose a Borel representative $\tilde{f}$ equal to $f$ almost everywhere; in the complex-valued case apply [F4] to the real and imaginary parts. Since $f$ is finite-valued, the Borel set where the resulting extended-real representative is infinite is contained in its null exceptional set; reset it to zero there, which keeps it Borel and equal to $f$ almost everywhere and gives a finite-valued representative. For each fixed $x$, changing $f$ to $\tilde{f}$ changes the integrand only on the fixed null exceptional set. Changing the assigned kernel value at zero, arbitrary by [F2], changes the integrand only at $y=x$, a null singleton by [F7]. Thus both choices preserve the Lebesgue integral, including whether its absolute value is finite.

The function $H(x,y)=\tilde{\mathrm{\Phi }}(x-y)\tilde{f}(y)$ is Borel on ${\mathbb{R}}^n\times {\mathbb{R}}^n$ by [F8]. By [F6] both Lebesgue measure spaces are $\sigma$-finite, so [F5] shows that $J(x):={\int }_{{\mathbb{R}}^n}\mathrm{\mid }H(x,y)\mathrm{\mid }dy$ is measurable. Applying [F5] to the positive and negative parts of the real and imaginary parts of $H$ also makes their section integrals measurable.

The set $A_f=\{x:J(x)<\mathrm{\infty }\}$ is measurable. On $A_f$, all four section integrals from step 2.1 are finite, and their signed combination is $Nf(x)$, so $Nf$ is measurable on its pointwise domain. If $A_f$ has full measure, extending by zero on $A_f^c$ gives a measurable representative on ${\mathbb{R}}^n$; any other extension is equal to it almost everywhere and hence represents the same class. No finiteness claim is made at points outside $A_f$.

If $f=0$, then $A_f={\mathbb{R}}^n$ and $Nf=0$. The pole assignment and the finite/infinite boundary of the defining integral are handled in steps 1.2–3.1. This definition assumes $n\ge 2$; it uses Countable Choice only for the Borel representative and the stated measure interfaces, and no full Axiom of Choice.