[proofplan] We interpret $\Phi$ as a [distribution](/page/Distribution) via its action on [test functions](/page/Test%20Function) and compute its distributional Laplacian by duality: $(\Delta \Phi)(\varphi) = \Phi(\Delta \varphi)$. We split the integral over $B(0,R)$ into the singular ball $B(0,\varepsilon)$ and the annulus $B(0,R) \setminus \overline{B(0,\varepsilon)}$. The integral over the small ball vanishes by the [Dominated Convergence Theorem](/theorems/4) and local integrability of $\Phi$. On the annulus, where $\Phi$ is smooth and harmonic, [Green's identity](/theorems/5) reduces the volume integral to boundary integrals over $\partial B(0,\varepsilon)$. We evaluate the normal derivative term to extract $-\varphi(0)$ and show the remaining boundary term vanishes, yielding $(\Delta \Phi)(\varphi) = -\varphi(0) = -\delta_0(\varphi)$. [/proofplan] [step:Set up the distributional Laplacian via duality and split the domain near the singularity] Let $\varphi \in C_c^\infty(\mathbb{R}^n)$. The [distributional derivative](/page/Distributional%20Derivative) $\Delta \Phi$ acts on $\varphi$ by \begin{align*} (\Delta \Phi)(\varphi) := \Phi(\Delta \varphi) = \int_{\mathbb{R}^n} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x). \end{align*} Fix $R > 0$ such that $\operatorname{supp}(\varphi) \subseteq B(0,R)$. For $0 < \varepsilon < R$, we decompose the domain: \begin{align*} \int_{\mathbb{R}^n} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = \int_{B(0,\varepsilon)} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) + \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x). \end{align*} [guided] We want to compute the distributional Laplacian of $\Phi$. Since $\Phi$ is not smooth at the origin, we cannot compute $\Delta \Phi$ classically. Instead, we use the definition of distributional derivatives: for any [test function](/page/Test%20Function) $\varphi \in C_c^\infty(\mathbb{R}^n)$, the [distribution](/page/Distribution) $\Delta \Phi$ acts by \begin{align*} (\Delta \Phi)(\varphi) := \Phi(\Delta \varphi) = \int_{\mathbb{R}^n} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x). \end{align*} This integral is well-defined because $\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ and $\Delta \varphi$ is smooth with compact support. Since $\varphi$ has compact support, we fix $R > 0$ with $\operatorname{supp}(\varphi) \subseteq B(0,R)$, so the integral reduces to $B(0,R)$. The difficulty is that $\Phi$ is singular at the origin but harmonic everywhere else. The strategy is to excise a small ball $B(0,\varepsilon)$ around the singularity and work on the annulus where $\Phi$ is smooth. We decompose: \begin{align*} \int_{\mathbb{R}^n} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = \int_{B(0,\varepsilon)} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) + \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x). \end{align*} We will show the first integral vanishes as $\varepsilon \to 0$ (the singularity is too weak to contribute), while the second produces $-\varphi(0)$ via [Green's identity](/theorems/5). [/guided] [/step] [step:Show the integral over the singular ball $B(0,\varepsilon)$ vanishes by dominated convergence] Define $f_\varepsilon : B(0,R) \to \mathbb{R}$ by $f_\varepsilon(x) := \mathbf{1}_{B(0,\varepsilon)}(x) \cdot \Phi(x) \cdot \Delta \varphi(x)$. For $\mathcal{L}^n$-a.e. $x \in B(0,R)$, we have $\lim_{\varepsilon \to 0} f_\varepsilon(x) = 0$ since eventually $x \notin B(0,\varepsilon)$. Moreover, \begin{align*} |f_\varepsilon(x)| \leq \|\Delta \varphi\|_{L^\infty(B(0,R))} \cdot |\Phi(x)| \cdot \mathbf{1}_{B(0,R)}(x). \end{align*} The dominating function $\|\Delta \varphi\|_{L^\infty} \cdot |\Phi| \cdot \mathbf{1}_{B(0,R)}$ belongs to $L^1(\mathbb{R}^n, \mathcal{L}^n)$ because $\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. By the [Dominated Convergence Theorem](/theorems/4): \begin{align*} \lim_{\varepsilon \to 0} \int_{B(0,\varepsilon)} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = 0. \end{align*} [guided] The integral over $B(0,\varepsilon)$ captures the contribution of $\Phi$ near its singularity at the origin. We need to show this contribution vanishes as we shrink the ball. Define $f_\varepsilon : B(0,R) \to \mathbb{R}$ by $f_\varepsilon(x) := \mathbf{1}_{B(0,\varepsilon)}(x) \cdot \Phi(x) \cdot \Delta \varphi(x)$. Why does $f_\varepsilon \to 0$ pointwise $\mathcal{L}^n$-a.e.? For any fixed $x \neq 0$, once $\varepsilon < |x|$ the indicator $\mathbf{1}_{B(0,\varepsilon)}(x) = 0$, so $f_\varepsilon(x) = 0$ for all sufficiently small $\varepsilon$. The origin itself is a single point and has $\mathcal{L}^n$-measure zero. To apply the [Dominated Convergence Theorem](/theorems/4), we need an integrable dominator. We bound: \begin{align*} |f_\varepsilon(x)| \leq \|\Delta \varphi\|_{L^\infty(B(0,R))} \cdot |\Phi(x)| \cdot \mathbf{1}_{B(0,R)}(x). \end{align*} This bound is independent of $\varepsilon$. The constant $\|\Delta \varphi\|_{L^\infty(B(0,R))}$ is finite because $\varphi \in C_c^\infty(\mathbb{R}^n)$. The function $|\Phi| \cdot \mathbf{1}_{B(0,R)}$ is integrable because $\Phi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ — the singularity $|x|^{-(n-2)}$ (for $n \geq 3$) or $|\log|x||$ (for $n = 2$) is integrable near the origin in $\mathbb{R}^n$. Therefore the [Dominated Convergence Theorem](/theorems/4) gives \begin{align*} \lim_{\varepsilon \to 0} \int_{B(0,\varepsilon)} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = \int_{B(0,R)} \lim_{\varepsilon \to 0} f_\varepsilon(x) \, d\mathcal{L}^n(x) = 0. \end{align*} [/guided] [/step] [step:Apply Green's identity on the annulus to reduce to boundary integrals over $\partial B(0,\varepsilon)$] On the annular region $B(0,R) \setminus \overline{B(0,\varepsilon)}$, the function $\Phi$ is smooth and satisfies $\Delta \Phi = 0$ (it solves [Laplace's equation](/page/Laplace's%20Equation)). Applying [Green's second identity](/theorems/5) to $\Phi$ and $\varphi$ on this region with outward unit normal $\nu$: \begin{align*} \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} (\Phi \, \Delta \varphi - \varphi \, \Delta \Phi) \, d\mathcal{L}^n(x) = \int_{\partial(B(0,R) \setminus \overline{B(0,\varepsilon)})} \left( \Phi \, \frac{\partial \varphi}{\partial \nu} - \varphi \, \frac{\partial \Phi}{\partial \nu} \right) d\mathcal{H}^{n-1}(x). \end{align*} Since $\Delta \Phi = 0$ on the annulus and $\operatorname{supp}(\varphi) \subseteq B(0,R)$ implies $\varphi$ and $\nabla \varphi$ vanish on $\partial B(0,R)$, only the inner boundary $\partial B(0,\varepsilon)$ contributes. The outward normal to the annulus on $\partial B(0,\varepsilon)$ points inward, i.e., $\nu(x) = -x/|x|$. We obtain: \begin{align*} \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = \int_{\partial B(0,\varepsilon)} \left( \Phi \, \frac{\partial \varphi}{\partial \nu} - \varphi \, \frac{\partial \Phi}{\partial \nu} \right) d\mathcal{H}^{n-1}(x), \end{align*} where $\frac{\partial}{\partial \nu} = -\frac{\partial}{\partial r}$ denotes the derivative in the direction of the inward-pointing normal $\nu = -x/|x|$ on $\partial B(0,\varepsilon)$, and $r = |x|$. [guided] On the annulus $B(0,R) \setminus \overline{B(0,\varepsilon)}$, the function $\Phi$ is $C^\infty$ and harmonic: $\Delta \Phi = 0$. This is the key property that makes the excision strategy work — Green's identity will eliminate the volume integral of $\varphi \, \Delta \Phi$. We apply [Green's second identity](/theorems/5), which requires both functions to be $C^2$ on the closure of the domain. Here $\Phi \in C^\infty(\mathbb{R}^n \setminus \{0\})$ and $\varphi \in C_c^\infty(\mathbb{R}^n)$, so both are $C^2$ on $\overline{B(0,R) \setminus \overline{B(0,\varepsilon)}}$. The identity states: \begin{align*} \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} (\Phi \, \Delta \varphi - \varphi \, \Delta \Phi) \, d\mathcal{L}^n(x) = \int_{\partial(B(0,R) \setminus \overline{B(0,\varepsilon)})} \left( \Phi \, \frac{\partial \varphi}{\partial \nu} - \varphi \, \frac{\partial \Phi}{\partial \nu} \right) d\mathcal{H}^{n-1}(x). \end{align*} The boundary of the annulus consists of two pieces: the outer sphere $\partial B(0,R)$ and the inner sphere $\partial B(0,\varepsilon)$. On $\partial B(0,R)$, the boundary integral vanishes because $\operatorname{supp}(\varphi) \subseteq B(0,R)$, so $\varphi \equiv 0$ and $\nabla \varphi \equiv 0$ on $\partial B(0,R)$. On $\partial B(0,\varepsilon)$, the outward unit normal to the annulus points toward the origin: $\nu(x) = -x/|x|$, so $\frac{\partial}{\partial \nu} = -\frac{\partial}{\partial r}$ where $r = |x|$. Since $\Delta \Phi = 0$ on the annulus, the left-hand side simplifies to $\int \Phi \, \Delta \varphi \, d\mathcal{L}^n$, giving: \begin{align*} \int_{B(0,R) \setminus \overline{B(0,\varepsilon)}} \Phi(x) \, \Delta \varphi(x) \, d\mathcal{L}^n(x) = \int_{\partial B(0,\varepsilon)} \left( \Phi \, \frac{\partial \varphi}{\partial \nu} - \varphi \, \frac{\partial \Phi}{\partial \nu} \right) d\mathcal{H}^{n-1}(x). \end{align*} We now evaluate each boundary integral separately as $\varepsilon \to 0$. [/guided] [/step] [step:Evaluate the normal derivative of $\Phi$ on $\partial B(0,\varepsilon)$ to extract $-\varphi(0)$] On $\partial B(0,\varepsilon)$, parametrise by $x = \varepsilon \omega$ for $\omega \in \partial B(0,1)$. The gradient of $\Phi$ is radial: \begin{align*} \nabla \Phi(x) &= \begin{cases} -\dfrac{1}{2\pi} \cdot \dfrac{x}{|x|^2} & \text{if } n = 2, \\ -\dfrac{1}{n\alpha(n)} \cdot \dfrac{x}{|x|^n} & \text{if } n \geq 3. \end{cases} \end{align*} Since $\nu = -x/|x|$ on $\partial B(0,\varepsilon)$, the normal derivative is $\frac{\partial \Phi}{\partial \nu}(x) = \nabla \Phi(x) \cdot \nu(x)$. For $n = 2$: \begin{align*} \frac{\partial \Phi}{\partial \nu}(x) = -\frac{1}{2\pi} \cdot \frac{x}{|x|^2} \cdot \left(-\frac{x}{|x|}\right) = \frac{1}{2\pi\varepsilon}. \end{align*} For $n \geq 3$: \begin{align*} \frac{\partial \Phi}{\partial \nu}(x) = -\frac{1}{n\alpha(n)} \cdot \frac{x}{|x|^n} \cdot \left(-\frac{x}{|x|}\right) = \frac{1}{n\alpha(n)\varepsilon^{n-1}}. \end{align*} In both cases, $\frac{\partial \Phi}{\partial \nu}$ is constant on $\partial B(0,\varepsilon)$ with value $\frac{1}{n\alpha(n)\varepsilon^{n-1}}$ (using $n\alpha(n) = 2\pi$ when $n = 2$). Therefore: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, \frac{\partial \Phi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = \frac{1}{n\alpha(n)\varepsilon^{n-1}} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x). \end{align*} To evaluate the surface integral of $\varphi$, we expand $\varphi$ in a Taylor series about the origin. Since $\varphi \in C_c^\infty(\mathbb{R}^n)$: \begin{align*} \varphi(\varepsilon\omega) = \varphi(0) + \varepsilon \, \nabla\varphi(0) \cdot \omega + O(\varepsilon^2), \end{align*} where the $O(\varepsilon^2)$ remainder is uniform in $\omega \in \partial B(0,1)$. Integrating over $\partial B(0,\varepsilon)$ with the substitution $x = \varepsilon\omega$, $d\mathcal{H}^{n-1}(x) = \varepsilon^{n-1} \, d\mathcal{H}^{n-1}(\omega)$: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x) &= \varepsilon^{n-1} \int_{\partial B(0,1)} \varphi(\varepsilon\omega) \, d\mathcal{H}^{n-1}(\omega) \\ &= \varepsilon^{n-1} \left[ \varphi(0) \, \mathcal{H}^{n-1}(\partial B(0,1)) + \varepsilon \, \nabla\varphi(0) \cdot \int_{\partial B(0,1)} \omega \, d\mathcal{H}^{n-1}(\omega) + O(\varepsilon^2) \right]. \end{align*} The integral $\int_{\partial B(0,1)} \omega \, d\mathcal{H}^{n-1}(\omega) = 0$ by the symmetry of the unit sphere ($\omega \mapsto -\omega$ is a measure-preserving involution). Using $\mathcal{H}^{n-1}(\partial B(0,1)) = n\alpha(n)$: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x) = n\alpha(n)\varepsilon^{n-1} \, \varphi(0) + O(\varepsilon^{n+1}). \end{align*} Substituting: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, \frac{\partial \Phi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = \frac{1}{n\alpha(n)\varepsilon^{n-1}} \bigl( n\alpha(n)\varepsilon^{n-1}\varphi(0) + O(\varepsilon^{n+1}) \bigr) = \varphi(0) + O(\varepsilon^2). \end{align*} Therefore: \begin{align*} \lim_{\varepsilon \to 0} \int_{\partial B(0,\varepsilon)} \varphi(x) \, \frac{\partial \Phi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = \varphi(0). \end{align*} [guided] This is the central computation of the proof. We need to show that the boundary integral involving $\frac{\partial \Phi}{\partial \nu}$ extracts the value $\varphi(0)$ — this is the mechanism by which the Dirac delta emerges. On $\partial B(0,\varepsilon)$, parametrise by $x = \varepsilon\omega$ for $\omega \in \partial B(0,1)$. The gradient of $\Phi$ points radially. For $n \geq 3$, differentiating $\Phi(x) = \frac{1}{n(n-2)\alpha(n)} |x|^{-(n-2)}$: \begin{align*} \nabla \Phi(x) = \frac{1}{n(n-2)\alpha(n)} \cdot (-(n-2)) \cdot |x|^{-(n-2)-1} \cdot \frac{x}{|x|} = -\frac{1}{n\alpha(n)} \cdot \frac{x}{|x|^n}. \end{align*} For $n = 2$, differentiating $\Phi(x) = -\frac{1}{2\pi}\log|x|$: \begin{align*} \nabla \Phi(x) = -\frac{1}{2\pi} \cdot \frac{x}{|x|^2}. \end{align*} The outward normal to the annulus on $\partial B(0,\varepsilon)$ is $\nu = -x/|x|$ (pointing toward the origin). So the normal derivative is: \begin{align*} \frac{\partial \Phi}{\partial \nu}(x) = \nabla\Phi(x) \cdot \left(-\frac{x}{|x|}\right). \end{align*} For $n \geq 3$: \begin{align*} \frac{\partial \Phi}{\partial \nu}(x) = -\frac{1}{n\alpha(n)} \cdot \frac{x}{|x|^n} \cdot \left(-\frac{x}{|x|}\right) = \frac{1}{n\alpha(n)} \cdot \frac{|x|^2}{|x|^{n+1}} = \frac{1}{n\alpha(n)\varepsilon^{n-1}}. \end{align*} For $n = 2$, similarly $\frac{\partial \Phi}{\partial \nu} = \frac{1}{2\pi\varepsilon}$, which matches $\frac{1}{n\alpha(n)\varepsilon^{n-1}}$ since $n\alpha(n) = 2\alpha(2) = 2\pi$. The key observation is that $\frac{\partial \Phi}{\partial \nu}$ is constant on $\partial B(0,\varepsilon)$ and scales as $\varepsilon^{-(n-1)}$ — exactly the inverse of the surface area $\mathcal{H}^{n-1}(\partial B(0,\varepsilon)) = n\alpha(n)\varepsilon^{n-1}$. This scaling is built into the normalisation constant of $\Phi$ and is precisely why $\Phi$ produces a delta distribution. We factor out the constant normal derivative: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, \frac{\partial \Phi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = \frac{1}{n\alpha(n)\varepsilon^{n-1}} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x). \end{align*} Now we need to evaluate the surface integral of $\varphi$. Substituting $x = \varepsilon\omega$ with $d\mathcal{H}^{n-1}(x) = \varepsilon^{n-1} d\mathcal{H}^{n-1}(\omega)$ and expanding $\varphi$ in a Taylor series about the origin: \begin{align*} \varphi(\varepsilon\omega) = \varphi(0) + \varepsilon \, \nabla\varphi(0) \cdot \omega + O(\varepsilon^2), \end{align*} where the remainder is uniform in $\omega$ because $\varphi$ is smooth. Integrating term by term: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x) &= \varepsilon^{n-1} \int_{\partial B(0,1)} \varphi(\varepsilon\omega) \, d\mathcal{H}^{n-1}(\omega) \\ &= \varepsilon^{n-1} \Bigl[ \varphi(0) \cdot n\alpha(n) + \varepsilon \, \nabla\varphi(0) \cdot \underbrace{\int_{\partial B(0,1)} \omega \, d\mathcal{H}^{n-1}(\omega)}_{= \, 0} + O(\varepsilon^2) \Bigr]. \end{align*} Why does the linear term vanish? The map $\omega \mapsto -\omega$ is a measure-preserving involution of $\partial B(0,1)$, so $\int_{\partial B(0,1)} \omega_i \, d\mathcal{H}^{n-1}(\omega) = 0$ for each component $i$. This gives: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, d\mathcal{H}^{n-1}(x) = n\alpha(n)\varepsilon^{n-1}\varphi(0) + O(\varepsilon^{n+1}). \end{align*} Combining with the normal derivative factor: \begin{align*} \int_{\partial B(0,\varepsilon)} \varphi(x) \, \frac{\partial \Phi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = \frac{n\alpha(n)\varepsilon^{n-1}\varphi(0) + O(\varepsilon^{n+1})}{n\alpha(n)\varepsilon^{n-1}} = \varphi(0) + O(\varepsilon^2) \xrightarrow{\varepsilon \to 0} \varphi(0). \end{align*} The cancellation $\varepsilon^{n-1}/\varepsilon^{n-1} = 1$ is the heart of the argument: the surface area growth and the singularity strength of $\nabla\Phi$ are perfectly matched by the normalisation of $\Phi$. [/guided] [/step] [step:Show the boundary integral involving $\Phi \, \frac{\partial \varphi}{\partial \nu}$ vanishes as $\varepsilon \to 0$] We bound: \begin{align*} \left| \int_{\partial B(0,\varepsilon)} \Phi(x) \, \frac{\partial \varphi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) \right| \leq \|\nabla \varphi\|_{L^\infty(B(0,R))} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^{n-1}(x). \end{align*} Since $\Phi$ is constant on $\partial B(0,\varepsilon)$, we evaluate the surface integral of $|\Phi|$ directly. **Case $n = 2$:** $|\Phi(\varepsilon\omega)| = \frac{1}{2\pi}|\log\varepsilon|$, so \begin{align*} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^1(x) = \frac{|\log\varepsilon|}{2\pi} \cdot 2\pi\varepsilon = \varepsilon|\log\varepsilon| \to 0 \quad \text{as } \varepsilon \to 0. \end{align*} **Case $n \geq 3$:** $|\Phi(\varepsilon\omega)| = \frac{1}{n(n-2)\alpha(n)}\varepsilon^{-(n-2)}$, so \begin{align*} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^{n-1}(x) = \frac{\varepsilon^{-(n-2)}}{n(n-2)\alpha(n)} \cdot n\alpha(n)\varepsilon^{n-1} = \frac{\varepsilon}{n-2} \to 0 \quad \text{as } \varepsilon \to 0. \end{align*} In both cases: \begin{align*} \lim_{\varepsilon \to 0} \int_{\partial B(0,\varepsilon)} \Phi(x) \, \frac{\partial \varphi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) = 0. \end{align*} [guided] This integral involves $\Phi$ (not $\nabla\Phi$) on the sphere. The question is whether $\Phi$ grows fast enough to overcome the shrinking surface area. We use the bound $\left|\frac{\partial\varphi}{\partial\nu}\right| \leq |\nabla\varphi|$, so: \begin{align*} \left| \int_{\partial B(0,\varepsilon)} \Phi(x) \, \frac{\partial \varphi}{\partial \nu}(x) \, d\mathcal{H}^{n-1}(x) \right| \leq \|\nabla \varphi\|_{L^\infty(B(0,R))} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^{n-1}(x). \end{align*} Since $\Phi$ depends only on $|x|$, it is constant on $\partial B(0,\varepsilon)$ and we can factor it out. For $n = 2$: $|\Phi| = \frac{1}{2\pi}|\log\varepsilon|$ on $\partial B(0,\varepsilon)$, and $\mathcal{H}^1(\partial B(0,\varepsilon)) = 2\pi\varepsilon$, so \begin{align*} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^1(x) = \frac{|\log\varepsilon|}{2\pi} \cdot 2\pi\varepsilon = \varepsilon|\log\varepsilon| \to 0. \end{align*} The product $\varepsilon|\log\varepsilon|$ vanishes because the linear decay $\varepsilon \to 0$ beats the logarithmic blowup $|\log\varepsilon| \to \infty$. For $n \geq 3$: $|\Phi| = \frac{1}{n(n-2)\alpha(n)}\varepsilon^{-(n-2)}$ on $\partial B(0,\varepsilon)$, and $\mathcal{H}^{n-1}(\partial B(0,\varepsilon)) = n\alpha(n)\varepsilon^{n-1}$, so \begin{align*} \int_{\partial B(0,\varepsilon)} |\Phi(x)| \, d\mathcal{H}^{n-1}(x) = \frac{\varepsilon^{-(n-2)}}{n(n-2)\alpha(n)} \cdot n\alpha(n)\varepsilon^{n-1} = \frac{\varepsilon}{n-2} \to 0. \end{align*} The net power is $\varepsilon^{-(n-2)} \cdot \varepsilon^{n-1} = \varepsilon^1$, so this vanishes linearly. Contrast this with the previous step, where $|\nabla\Phi| \sim \varepsilon^{-(n-1)}$ exactly cancelled $\varepsilon^{n-1}$. Here, $|\Phi| \sim \varepsilon^{-(n-2)}$ is one power weaker, leaving $\varepsilon^1 \to 0$. [/guided] [/step] [step:Combine the limits to conclude $\Delta\Phi = -\delta_0$ in $\mathcal{D}'(\mathbb{R}^n)$] Collecting the results of the previous steps: \begin{align*} (\Delta\Phi)(\varphi) &= \lim_{\varepsilon \to 0} \left[ \int_{B(0,\varepsilon)} \Phi \, \Delta\varphi \, d\mathcal{L}^n + \int_{B(0,R)\setminus\overline{B(0,\varepsilon)}} \Phi \, \Delta\varphi \, d\mathcal{L}^n \right] \\ &= \lim_{\varepsilon \to 0} \left[ \int_{B(0,\varepsilon)} \Phi \, \Delta\varphi \, d\mathcal{L}^n + \int_{\partial B(0,\varepsilon)} \Phi \, \frac{\partial\varphi}{\partial\nu} \, d\mathcal{H}^{n-1} - \int_{\partial B(0,\varepsilon)} \varphi \, \frac{\partial\Phi}{\partial\nu} \, d\mathcal{H}^{n-1} \right] \\ &= 0 + 0 - \varphi(0) \\ &= -\varphi(0). \end{align*} Since $\delta_0(\varphi) = \varphi(0)$ for every $\varphi \in C_c^\infty(\mathbb{R}^n)$, we conclude: \begin{align*} \Delta\Phi = -\delta_0 \quad \text{in } \mathcal{D}'(\mathbb{R}^n). \end{align*} [/step]