Calderón–Zygmund Decomposition and Singular Integrals — Examples

By [F1], a dyadic interval meeting $(0,1]$ is either contained in it or contains it. The former have average $1$. A containing interval of length $2^a$, $a\ge 0$, has average $2^{-a}$, exceeding $\lambda =1\mathrm{/}4$ exactly when $a=0$ or $a=1$. The unique ancestors of these lengths are $(0,1]$ and $(0,2]$, while $(0,4]$ has average exactly $\lambda$ and every coarser ancestor has smaller average. Intervals disjoint from $(0,1]$ have average zero. Thus all bad intervals are contained in $(0,2]$, which is itself bad and is the unique maximal bad interval, with average $1\mathrm{/}2=2\lambda$.

Reading off the formulas of the decomposition [F2] for the single maximal bad interval $Q=(0,2]$: $\mathrm{\mid }Q\mathrm{\mid }=2$, $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_{Q}f=1\mathrm{/}2$, so $b=f-\frac{1}{2}{\mathbf{1}}_{(0,2]}=\frac{1}{2}{\mathbf{1}}_{(0,1]}-\frac{1}{2}{\mathbf{1}}_{(1,2]}$ and $g=\frac{1}{2}{\mathbf{1}}_{(0,2]}$.

The checks: $f={\mathbf{1}}_{(0,1]}$ has $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$ and ${\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1=4$; $\sum \mathrm{\mid }Q\mathrm{\mid }=\mathrm{\mid }(0,2]\mathrm{\mid }=2\le 4$; $\mathrm{\mid }g\mathrm{\mid }=1\mathrm{/}2=2\lambda$ on its support; and $\int b=\frac{1}{2}\lambda ((0,1])-\frac{1}{2}\lambda ((1,2])=\frac{1}{2}-\frac{1}{2}=0$. This is the asserted finite verification of maximality, of the $2\lambda$ average bound, and of the parent-good property.

The published estimates give exactly the size bound $\mathrm{\mid }{K}_j(x)\mathrm{\mid }\le c_n\mathrm{\mid }x{\mathrm{\mid }}^{-n}$, the $\delta =1$ first-difference bound in the regime $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }>0$ with constant $A_2^{\mathrm{\prime }}=C_n$, and the vanishing of all spherical means of $K_j$. The formula for $K_j$ makes it smooth, hence measurable and locally integrable, off the origin.

$R_j$ is $L^2$-bounded with norm at most one, since its symbol $m_j$ is measurable with $\mathrm{\mid }{m}_j\mathrm{\mid }\le 1$ on $\{\xi \mathrm{\ne }0\}$ and the Plancherel multiplier bound gives $\mathrm{\parallel }{R}_j\mathrm{\parallel }\le \mathrm{\parallel }{m}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$; thus $B=1$ is an admissible $L^2$ norm bound.

$R_j$ is skew-adjoint: for $f,g\in L^2$, $⟨R_{j}f,g⟩=\int m_j\widehat{f}\bar{\widehat{g}}=\int \widehat{f}\overline{(-m_j)\widehat{g}}=⟨f,-R_{j}g⟩$, because $\overline{m_j}=-m_j$ for the purely imaginary symbol; hence $R_j^{\ast }=-R_j$.

The pointwise size bound gives the annular condition with $A_1=c_n\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log 2$. For every $h\mathrm{\ne }0$, integrating the raw difference bound in [F1] and using polar coordinates yields ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }{K}_j(x-h)-K_j(x)\mathrm{\mid }dx\le C_n\mathrm{\mid }h\mathrm{\mid }{\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }x{\mathrm{\mid }}^{-n-1}dx=C_n\mathrm{\mid }h\mathrm{\mid }\frac{\mathrm{\mid }{S}^{n-1}\mathrm{\mid }}{2\mathrm{\mid }h\mathrm{\mid }}=\frac{1}{2}{C}_n\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\mathrm{.}$ Thus Hörmander's condition holds with $A_2=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{2}^{-1}{C}_n$. Together with step 1.1 this proves that $K_j$ is a Calderón–Zygmund kernel, and its raw first-difference bound now makes it standard $1$-Hölder with constant $C_n$.

Let $f\in L^2$ have compact support and let $\phi \in C_c^{\mathrm{\infty }}$ be supported off it. If either is zero the formula is immediate; otherwise the supports have distance $d>0$. By skew-adjointness and [F2], $\int (R_{j}f)\phi =⟨R_{j}f,\bar{\phi }⟩=-\int f(y)\overline{R_j\stackrel{‾}{\phi }(y)}dy=-\iint K_j(y-x)f(y)\phi (x)dxdy=\iint K_j(x-y)f(y)\phi (x)dxdy$. The kernel is real and odd, and the positive separation and bounded supports make the double integral absolutely convergent. Fubini and injectivity of locally integrable distribution pairings identify $R_{j}f$ almost everywhere off its support with $\int K_j(x-y)f(y)dy$.

Steps 2.1, 1.2 and 2.2 verify the annular and Hörmander bounds for $K_j$, the $L^2$ bound for $R_j$ and the off-support representation for compactly supported $L^2$ inputs; by steps 1.1 and 2.1 the kernel is standard $1$-Hölder. Therefore $R_j$ is a standard-kernel Calderón–Zygmund operator in the sense of the base definition and the standard-kernel definition.

Explicit values of the truncations. For $x\mathrm{\notin }\{0,1\}$ and $0<\epsilon <d(x):=\min \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }x-1\mathrm{\mid }\}$ the truncation is a sum of ordinary integrals with no singular point in the domain, and direct antiderivatives give: for $x<0$ or $x>1$, $H_{\epsilon }f(x)=\frac{1}{\pi }{\int }_0^1\frac{dy}{x-y}=\frac{1}{\pi }\log \frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}=q(x)$; for $0<x<1$, $H_{\epsilon }f(x)=\frac{1}{\pi }({\int }_0^{x-\epsilon }+{\int }_{x+\epsilon }^1)\frac{dy}{x-y}=\frac{1}{\pi }(\log x-\log \epsilon +\log \epsilon -\log (1-x))=\frac{1}{\pi }\log \frac{x}{1-x}=q(x)$, since $\mathrm{\mid }x\mathrm{/}(x-1)\mathrm{\mid }=x\mathrm{/}(1-x)$ there. Hence $H_{\epsilon }f(x)=q(x)$ for every $\epsilon <d(x)$, so ${\lim }_{\epsilon \downarrow 0}{H}_{\epsilon }f(x)=q(x)$ at every $x\mathrm{\notin }\{0,1\}$. Moreover, for $0<x<1$ direct integration gives $\pi H_{\epsilon }f(x)=(\log (x\mathrm{/}\epsilon ){)}_{+}-(\log ((1-x)\mathrm{/}\epsilon ){)}_{+}$. Since $u\mapsto u_{+}$ is $1$-Lipschitz, $\mathrm{\mid }{H}_{\epsilon }f(x)\mathrm{\mid }\le \mathrm{\mid }q(x)\mathrm{\mid }$. Outside $[0,1]$ the integrand has one sign, so deleting part of the integration domain also gives $\mathrm{\mid }{H}_{\epsilon }f(x)\mathrm{\mid }\le \mathrm{\mid }q(x)\mathrm{\mid }$. Thus $G$ dominates the truncations almost everywhere. It is locally integrable because $\mathrm{\mid }q(x)\mathrm{\mid }\le {\pi }^{-1}(\mathrm{\mid }\log \mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }+\mathrm{\mid }\log \mathrm{\mid }x-1\mathrm{\mid }\mathrm{\mid })$ and ${\int }_0^a\mathrm{\mid }\log t\mathrm{\mid }dt<\mathrm{\infty }$ for finite $a>0$.

Distributional convergence to the $L^2$ transform. Let $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ and put $b_{\epsilon }(y):=\frac{1}{\pi }{\int }_{\mathrm{\mid }x-y\mathrm{\mid }>\epsilon }\frac{\overline{\phi (x)}}{x-y}dx$; on the domain $\mathrm{\mid }x-y\mathrm{\mid }\ge \epsilon$ the double integral ${\iint }_{\mathrm{\mid }x-y\mathrm{\mid }>\epsilon }\frac{\mathrm{\mid }f(y)\phi (x)\mathrm{\mid }}{\mathrm{\mid }x-y\mathrm{\mid }}dydx$ is finite because $\mathrm{\mid }x-y{\mathrm{\mid }}^{-1}\le {\epsilon }^{-1}$ and $f$ and $\phi$ are bounded with bounded support, so Fubini applies and $⟨H_{\epsilon }f,\phi ⟩={\int }_0^{1}f(y)b_{\epsilon }(y)dy$. As $\epsilon \downarrow 0$ one has $b_{\epsilon }(y)=-\frac{1}{\pi }{\int }_{\mathrm{\mid }x-y\mathrm{\mid }>\epsilon }\frac{\overline{\phi (x)}}{y-x}dx\to -H\bar{\phi }(y)=-\overline{H\phi (y)}$, using that the kernel is real and [F2]; the convergence is dominated by a constant depending on $\phi$, because applying the estimate $\mathrm{\mid }{\int }_{\mathrm{\mid }t\mathrm{\mid }>\epsilon }\psi (y+t)t^{-1}dt\mathrm{\mid }\le {\int }_{\mathrm{\mid }t\mathrm{\mid }\le 1}\mathrm{\mid }\psi (y+t)-\psi (y)\mathrm{\mid }\mathrm{\mid }t{\mathrm{\mid }}^{-1}dt+{\int }_{\mathrm{\mid }t\mathrm{\mid }>1}\mathrm{\mid }\psi (y+t)\mathrm{\mid }dt\le 2\mathrm{\parallel }{\psi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}+2\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$ to $\psi =\bar{\phi }$ bounds all $b_{\epsilon }(y)$ uniformly. Hence dominated convergence on the finite-measure set $(0,1]$ gives $⟨H_{\epsilon }f,\phi ⟩\to -{\int }_0^{1}f(y)\overline{H\phi (y)}dy=-⟨f,H\phi ⟩=⟨f,H^{\ast }\phi ⟩=⟨Hf,\phi ⟩$.

$q\mathrm{\notin }{L}^1(\mathbb{R})$: for $x\ge 2$ one has $\frac{x}{x-1}=1+\frac{1}{x-1}$ with $\frac{1}{x-1}\le 1$, and $\log (1+u)\ge \frac{u}{2}$ for $0\le u\le 1$, so $q(x)=\frac{1}{\pi }\log (1+\frac{1}{x-1})\ge \frac{1}{2\pi (x-1)}\ge \frac{1}{2\pi x}$. Therefore ${\int }_2^{\mathrm{\infty }}\mathrm{\mid }q\mathrm{\mid }\ge \frac{1}{2\pi }{\int }_2^{\mathrm{\infty }}\frac{dx}{x}=+\mathrm{\infty }$.

Identification of the limit. On each compact $K$ the domination $\mathrm{\mid }{H}_{\epsilon }f\mathrm{\mid }\le G$ of step 1.1 with $G\in L^1(K)$ and the pointwise convergence $H_{\epsilon }f\to q$ off the null set $\{0,1\}$ let dominated convergence pass the limit inside the pairing: $⟨H_{\epsilon }f,\phi ⟩\to \int q\bar{\phi }$ for every test function $\phi$ supported in $K$, hence for every test function. Comparing with step 1.2, $\int (q-Hf)\bar{\phi }=0$ for every $\phi \in C_c^{\mathrm{\infty }}$, and both $q$ and the $L^2$ class $Hf$ are locally integrable, so $q=Hf$ almost everywhere; in particular $f\in L^1\cap L^2$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$ and $Hf=q$.

Suppose $T:L^1(\mathbb{R};\mathbb{C})\to L^1(\mathbb{R};\mathbb{C})$ were bounded and agreed with the $L^2$ Hilbert transform on $L^1\cap L^2$. Since $f\in L^1\cap L^2$ by step 2.1, the $L^1$ class $Tf$ would equal the $L^1$ class $Hf$, which step 2.1 identifies with $q$; but $q\mathrm{\notin }{L}^1$ by step 1.3, whereas $Tf\in L^1$ by definition of $T$. This contradiction shows that no such $T$ exists, which is exactly the failure of strong type $(1,1)$; the statement says nothing about weak type $(1,1)$.

The function $q$ is unbounded above and below on every punctured neighbourhood of the endpoints: for $0<x<1$, $\frac{\mathrm{\mid }x\mathrm{\mid }}{\mathrm{\mid }x-1\mathrm{\mid }}=\frac{x}{1-x}$, which tends to $0$ as $x\to 0^{+}$ and to $+\mathrm{\infty }$ as $x\to 1^{-}$; since $\log$ is continuous, strictly increasing, with ${\lim }_{u\downarrow 0}\log u=-\mathrm{\infty }$ and ${\lim }_{u\to \mathrm{\infty }}\log u=+\mathrm{\infty }$, one has $q(x)\to -\mathrm{\infty }$ as $x\to 0^{+}$ and $q(x)\to +\mathrm{\infty }$ as $x\to 1^{-}$. Hence for every $M>0$ the sets $\{q>M\}$ and $\{q<-M\}$ contain nondegenerate intervals, so both have positive Lebesgue measure and $q$ is not essentially bounded.

Suppose $T:L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})\to L^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$ were bounded and agreed with the $L^2$ Hilbert transform on $L^{\mathrm{\infty }}\cap L^2$. Since $f\in L^{\mathrm{\infty }}\cap L^2$ by [F1], the class $Tf$ would equal the class $Hf=q$; but $q$ is not essentially bounded by step 1.1, whereas every class in $L^{\mathrm{\infty }}$ is essentially bounded. This contradiction shows that no bounded $L^{\mathrm{\infty }}\to L^{\mathrm{\infty }}$ extension compatible on $L^{\mathrm{\infty }}\cap L^2$ exists, which is the asserted failure; nothing here concerns a BMO-valued estimate.

Since $k$ obeys the size bound with $A_1=1$ and $f={\mathbf{1}}_{B(0,1)}\in L^1\cap L^2$, [F2] applies, and polar coordinates [F1] give, for every $0<\epsilon <1$, $T_{\epsilon }f(0)={\int }_{\epsilon <\mathrm{\mid }y\mathrm{\mid }<1}\mathrm{\mid }y{\mathrm{\mid }}^{-n}dy={\int }_{\epsilon }^1{r}^{-n}\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{r}^{n-1}dr=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{\int }_{\epsilon }^1\frac{dr}{r}=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log \frac{1}{\epsilon },$ with $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }>0$; the doubly truncated integral likewise equals $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log (\min \{1,N\}\mathrm{/}\epsilon )$ for $0<\epsilon <\min \{1,N\}$, and it vanishes if $\epsilon \ge 1$; fixing $N=2$ and sending $\epsilon \downarrow 0$ proves $T^{\ast \ast }f(0)=+\mathrm{\infty }$.

Consequently ${\lim }_{\epsilon \downarrow 0}{T}_{\epsilon }f(0)=+\mathrm{\infty }$: the symmetric truncations do not converge at the origin, and the maximal truncated operator is infinite there, although every individual truncation is finite.

No principal-value distribution for $k$ exists along any sequence ${\delta }_j\downarrow 0$. Indeed, take the nonnegative Schwartz test $\phi (x)=e^{-\mathrm{\mid }x{\mathrm{\mid }}^2}$, with $\phi (0)=1$; by continuity of $\phi$ there is $\rho >0$ with $\phi \ge \phi (0)\mathrm{/}2$ on $B(0,\rho )$, so for every $\delta \in (0,\rho )$ polar coordinates give ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge \delta }k(x)\phi (x)dx\ge {\int }_{\delta <\mathrm{\mid }x\mathrm{\mid }<\rho }\mathrm{\mid }x{\mathrm{\mid }}^{-n}\frac{\phi (0)}{2}dx=\frac{\phi (0)}{2}\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log \frac{\rho }{\delta },$ which tends to $+\mathrm{\infty }$ as $\delta \downarrow 0$; hence the limit in [F3] fails for this $\phi$ and every sequence ${\delta }_j\downarrow 0$. Together with step 2.1 and [F2] this shows that the pointwise size condition by itself yields neither a principal-value distribution nor a finite maximal truncated operator, so principal-value cancellation is an additional requirement. In fact $k$ also satisfies Hörmander's condition: $\mathrm{\mid }\mathrm{\nabla }k(z)\mathrm{\mid }=n\mathrm{\mid }z{\mathrm{\mid }}^{-n-1}$ gives $\mathrm{\mid }k(x-y)-k(x)\mathrm{\mid }\le n{2}^{n+1}\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^{-n-1}$ on $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }$, whose integral is bounded uniformly in $y$ by polar coordinates. Thus these kernel bounds alone do not assert a principal value or an associated $L^2$-bounded operator.

Differentiating $\mathrm{\Gamma }(x)=\mathrm{\mid }x{\mathrm{\mid }}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})$ twice off the origin gives $k_{ij}(x)={\mathrm{\partial }}_{ij}\mathrm{\Gamma }(x)={\omega }_{n-1}^{-1}(n{x}_i{x}_j\mathrm{\mid }x{\mathrm{\mid }}^{-n-2}-{\delta }_{ij}\mathrm{\mid }x{\mathrm{\mid }}^{-n})$ for $x\mathrm{\ne }0$: the first derivative is $(2-n)(n-2{)}^{-1}{\omega }_{n-1}^{-1}{x}_i\mathrm{\mid }x{\mathrm{\mid }}^{-n}=-{\omega }_{n-1}^{-1}{x}_i\mathrm{\mid }x{\mathrm{\mid }}^{-n}$, and differentiating once more with ${\mathrm{\partial }}_j(x_i\mathrm{\mid }x{\mathrm{\mid }}^{-n})={\delta }_{ij}\mathrm{\mid }x{\mathrm{\mid }}^{-n}-n{x}_i{x}_j\mathrm{\mid }x{\mathrm{\mid }}^{-n-2}$ gives the displayed expression. This function is smooth on ${\mathbb{R}}^n\setminus \{0\}$ and homogeneous of degree $-n$.

Size and gradient bounds: $\mathrm{\mid }{k}_{ij}(x)\mathrm{\mid }\le {\omega }_{n-1}^{-1}(n\mathrm{\mid }{x}_i{x}_j\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^{-n-2}+\mathrm{\mid }x{\mathrm{\mid }}^{-n})\le (1+n){\omega }_{n-1}^{-1}\mathrm{\mid }x{\mathrm{\mid }}^{-n}$ by $\mathrm{\mid }{x}_i{x}_j\mathrm{\mid }\le \mathrm{\mid }x{\mathrm{\mid }}^2$; and since $k_{ij}$ is $C^{\mathrm{\infty }}$ off the origin and homogeneous of degree $-n$, each partial derivative is homogeneous of degree $-n-1$, continuous on the compact unit sphere, and therefore satisfies $\mathrm{\mid }\mathrm{\nabla }{k}_{ij}(x)\mathrm{\mid }\le C_n\mathrm{\mid }x{\mathrm{\mid }}^{-n-1}$ for all $x\mathrm{\ne }0$ with $C_n:=n{\sup }_{\mathrm{\mid }z\mathrm{\mid }=1}\mathrm{\mid }\mathrm{\nabla }{k}_{ij}(z)\mathrm{\mid }<\mathrm{\infty }$.

The spherical mean vanishes: for $r>0$, substituting $x=r\omega$ and using the homogeneity, ${\int }_{S^{n-1}}{k}_{ij}(r\omega )d{\sigma }_{n-1}(\omega )=r^{-n}{\omega }_{n-1}^{-1}{\int }_{S^{n-1}}(n{\omega }_i{\omega }_j-{\delta }_{ij})d{\sigma }_{n-1}(\omega )$. The surface measure is invariant under coordinate permutations and under the sign change ${\omega }_i\mapsto -{\omega }_i$, so ${\int }_{S^{n-1}}{\omega }_i{\omega }_{j}d\sigma =0$ for $i\mathrm{\ne }j$ and all $n$ integrals $\int {\omega }_i^{2}d\sigma$ are equal to ${\omega }_{n-1}\mathrm{/}n$ since their sum is $\int \mathrm{\mid }\omega {\mathrm{\mid }}^{2}d\sigma ={\omega }_{n-1}$; hence $\int (n{\omega }_i{\omega }_j-{\delta }_{ij})d\sigma =n{\delta }_{ij}{\omega }_{n-1}\mathrm{/}n-{\delta }_{ij}{\omega }_{n-1}=0$, and the spherical mean of $k_{ij}$ vanishes.

Principal value and the local delta term. By step 2.2, ${\int }_{\epsilon <\mathrm{\mid }x\mathrm{\mid }<1}{k}_{ij}(x)dx=0$. Thus for every Schwartz test $\phi$, the limit $⟨V_{ij},\phi ⟩={\lim }_{\epsilon \downarrow 0}{\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }{k}_{ij}(x)\phi (x)dx$ exists: near zero subtract $\phi (0)$, giving a majorant $C\mathrm{\mid }x{\mathrm{\mid }}^{1-n}{\sup }_{\mathrm{\mid }z\mathrm{\mid }\le 1}\mathrm{\mid }\mathrm{\nabla }\phi (z)\mathrm{\mid }$, and the tail is integrable by Schwartz decay. These bounds also prove $V_{ij}\in {\mathcal{S}}^{\mathrm{\prime }}$. Integration by parts on $\{\epsilon <\mathrm{\mid }x\mathrm{\mid }<R\}$ using [F2] first shows ${\mathrm{\partial }}_i\mathrm{\Gamma }$ is the regular distribution of $-{\omega }_{n-1}^{-1}{x}_i\mathrm{\mid }x{\mathrm{\mid }}^{-n}$: the inner boundary term from $\mathrm{\Gamma }$ is $O(\epsilon )$, and outer terms vanish as $R\to \mathrm{\infty }$. Applying integration by parts once more gives $⟨{\mathrm{\partial }}_{ij}\mathrm{\Gamma },\phi ⟩={\lim }_{\epsilon \downarrow 0}[{\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }{k}_{ij}\phi -{\omega }_{n-1}^{-1}{\int }_{S^{n-1}}{\omega }_i{\omega }_j\phi (\epsilon \omega )d\sigma ]$. The inward normal of the exterior region at radius $\epsilon$ is $-\omega$, which fixes the minus sign. Step 2.2 evaluates the boundary limit as ${\delta }_{ij}\phi (0)\mathrm{/}n$. Hence ${\mathrm{\partial }}_{ij}\mathrm{\Gamma }=V_{ij}-({\delta }_{ij}\mathrm{/}n){\delta }_0$; summing the diagonal identities, whose off-origin kernels have zero trace, proves $-\mathrm{\Delta }\mathrm{\Gamma }={\delta }_0$.