[proofplan] The proof has two parts. First, we establish that $u \in C^{2}(\mathbb{R}^n)$ by passing derivatives through the convolution integral. The key device is to write $u = \Phi * f$ and transfer all derivatives onto the smooth, compactly supported factor $f$. The [Dominated Convergence Theorem](/theorems/4), together with the [local integrability of $\Phi$](/theorems/5), justifies the limit exchange. Second, we show $-\Delta u = f$ by combining Part 1 with the [distributional identity $\Delta \Phi = -\delta_0$](/theorems/27) and the convolution rule for distributions. [/proofplan] [step:Rewrite $u$ as a convolution with derivatives transferred to $f$] Since $f \in C_c^{2}(\mathbb{R}^n)$, there exists $R > 0$ such that $\operatorname{supp} f \subset B(0, R)$. By the definition of $\Phi$ in the [Fundamental Solution of Laplace's Equation](/theorems/566), the function $u$ takes the form \begin{align*} u(x) = \int_{\mathbb{R}^n} \Phi(x - y) f(y) \, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} \Phi(y) f(x - y) \, d\mathcal{L}^n(y), \end{align*} where the second equality follows from the substitution $y \mapsto x - y$, which preserves $\mathcal{L}^n$. Because $\operatorname{supp} f \subset B(0, R)$, the function $y \mapsto f(x - y)$ is supported in $B(x, R)$, and the integral reduces to \begin{align*} u(x) = \int_{B(x, R)} \Phi(y) f(x - y) \, d\mathcal{L}^n(y). \end{align*} This representation is the starting point: all derivatives will be placed on $f(x - y)$, which is $C^{2}$ in $x$ for each fixed $y$. [guided] Why transfer derivatives onto $f$ rather than differentiating $\Phi$ directly? The function $\Phi$ has a singularity at the origin — in dimension $n \geq 3$, $\Phi(y) \sim |y|^{2-n}$ as $y \to 0$, and its gradient $\nabla \Phi(y) \sim |y|^{1-n}$ is even more singular. Differentiating under the integral sign with respect to $\Phi$ would require controlling these singularities, which is the subject of the much harder Part 2. By contrast, $f$ is $C^{2}$ with compact support, so $\partial_{x_i} f(x - y) = -\partial_{y_i} f(x - y)$ is bounded and compactly supported uniformly in $x$. The only singularity in the integrand comes from $\Phi(y)$, and by the [local integrability of $\Phi$](/theorems/5), $\Phi \in L^1_{\text{loc}}(\mathbb{R}^n)$, so the integral is well-defined. The substitution $y \mapsto x - y$ is a reflection composed with a translation. Both operations preserve $\mathcal{L}^n$ (the reflection has Jacobian determinant $(-1)^n$, so $|\det J| = 1$), giving \begin{align*} u(x) = \int_{\mathbb{R}^n} \Phi(x - y) f(y) \, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} \Phi(y) f(x - y) \, d\mathcal{L}^n(y). \end{align*} Since $f(x - y) = 0$ whenever $|y - x| > R$, the effective integration domain is $B(x, R)$. [/guided] [/step] [step:Pass first-order derivatives through the integral via the Dominated Convergence Theorem] Fix $i \in \{1, \ldots, n\}$ and $x \in \mathbb{R}^n$. For $h \neq 0$, define the difference quotient \begin{align*} D_h : \mathbb{R}^n & \to \mathbb{R} \\ z & \mapsto \frac{f(z + h e_i) - f(z)}{h}. \end{align*} Since $f \in C^{2}(\mathbb{R}^n)$, Taylor's theorem with integral remainder gives \begin{align*} \left| D_h(z) - \partial_{x_i} f(z) \right| \leq \frac{|h|}{2} \sup_{w \in \mathbb{R}^n} \left| \partial_{x_i}^{2} f(w) \right| =: \frac{|h|}{2} M_{ii} \end{align*} for all $z \in \mathbb{R}^n$, where $M_{ii} < \infty$ because $\partial_{x_i}^{2} f$ is continuous with compact support. In particular, $D_h(z) \to \partial_{x_i} f(z)$ pointwise as $h \to 0$. Now consider the difference quotient of $u$: \begin{align*} \frac{u(x + h e_i) - u(x)}{h} = \int_{\mathbb{R}^n} \Phi(y) \, D_h(x - y) \, d\mathcal{L}^n(y). \end{align*} We apply the [Dominated Convergence Theorem](/theorems/4) to pass the limit $h \to 0$ inside the integral. The hypotheses are verified as follows: - **Pointwise convergence.** For each $y \in \mathbb{R}^n$, $\Phi(y) D_h(x - y) \to \Phi(y) \partial_{x_i} f(x - y)$ as $h \to 0$. - **Integrable dominating function.** For all $h$ with $|h| \leq 1$, both $f(x - y + h e_i)$ and $f(x - y)$ vanish unless $y \in B(x, R + 1)$. The bound $|D_h(z)| \leq \|\nabla f\|_{L^\infty}$ (by the Mean Value Theorem) gives \begin{align*} |\Phi(y) D_h(x - y)| \leq \|\nabla f\|_{L^\infty} \, |\Phi(y)| \, \mathbb{1}_{B(x, R+1)}(y). \end{align*} By the [local integrability of $\Phi$](/theorems/5), the function $|\Phi| \, \mathbb{1}_{B(x, R+1)} \in L^1(\mathbb{R}^n, \mathcal{L}^n)$, so the dominating function is integrable. The Dominated Convergence Theorem yields \begin{align*} \partial_{x_i} u(x) = \int_{\mathbb{R}^n} \Phi(y) \, \partial_{x_i} f(x - y) \, d\mathcal{L}^n(y). \end{align*} The right-hand side is continuous in $x$ because $\partial_{x_i} f$ is continuous and compactly supported, and $\Phi$ is locally integrable. Hence $u \in C^{1}(\mathbb{R}^n)$. [guided] The idea is to write the difference quotient of $u$ as an integral involving the difference quotient of $f$, and then exchange limit and integral. The exchange is not automatic because $\Phi$ has a singularity at the origin, so we must produce an integrable dominating function. Fix $i \in \{1, \ldots, n\}$. For $h \neq 0$, define \begin{align*} D_h : \mathbb{R}^n & \to \mathbb{R} \\ z & \mapsto \frac{f(z + h e_i) - f(z)}{h}. \end{align*} Since $f \in C^{2}(\mathbb{R}^n)$, the function $t \mapsto f(z + t e_i)$ is $C^{2}$ in $t$, and Taylor's theorem gives \begin{align*} f(z + h e_i) = f(z) + h \, \partial_{x_i} f(z) + \int_0^h (h - t) \, \partial_{x_i}^{2} f(z + t e_i) \, d\mathcal{L}^1(t). \end{align*} Dividing by $h$ and subtracting $\partial_{x_i} f(z)$: \begin{align*} \left| D_h(z) - \partial_{x_i} f(z) \right| = \left| \frac{1}{h} \int_0^h (h - t) \, \partial_{x_i}^{2} f(z + t e_i) \, d\mathcal{L}^1(t) \right| \leq \frac{|h|}{2} M_{ii}, \end{align*} where $M_{ii} := \sup_{w \in \mathbb{R}^n} |\partial_{x_i}^{2} f(w)| < \infty$ since $\partial_{x_i}^{2} f$ is continuous with compact support. For the dominating function, note that by the Mean Value Theorem applied to $t \mapsto f(z + t e_i)$, \begin{align*} |D_h(z)| = \left| \frac{f(z + h e_i) - f(z)}{h} \right| \leq \sup_{w \in \mathbb{R}^n} |\partial_{x_i} f(w)| = \|\nabla f\|_{L^\infty}. \end{align*} Moreover, $D_h(x - y) = 0$ unless $x - y \in \operatorname{supp} f + [-1,1] e_i$ (for $|h| \leq 1$), which forces $y \in B(x, R + 1)$. Combining: \begin{align*} |\Phi(y) D_h(x - y)| \leq \|\nabla f\|_{L^\infty} \, |\Phi(y)| \, \mathbb{1}_{B(x, R+1)}(y). \end{align*} By the [local integrability of $\Phi$](/theorems/5), $\Phi \in L^1_{\text{loc}}(\mathbb{R}^n)$, so $\int_{B(x,R+1)} |\Phi(y)| \, d\mathcal{L}^n(y) < \infty$. The dominating function is therefore in $L^1(\mathbb{R}^n, \mathcal{L}^n)$. The [Dominated Convergence Theorem](/theorems/4) now applies: the integrands converge pointwise and are dominated by an integrable function, so \begin{align*} \partial_{x_i} u(x) = \lim_{h \to 0} \int_{\mathbb{R}^n} \Phi(y) D_h(x - y) \, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} \Phi(y) \, \partial_{x_i} f(x - y) \, d\mathcal{L}^n(y). \end{align*} To confirm continuity of $\partial_{x_i} u$, note that as $x \to x_0$, the integrand $\Phi(y) \partial_{x_i} f(x - y) \to \Phi(y) \partial_{x_i} f(x_0 - y)$ pointwise and is dominated by $\|\partial_{x_i} f\|_{L^\infty} |\Phi(y)| \mathbb{1}_{B(x_0, R+1)}(y)$ (for $x$ near $x_0$). By the Dominated Convergence Theorem again, $\partial_{x_i} u(x) \to \partial_{x_i} u(x_0)$. [/guided] [/step] [step:Pass second-order derivatives through the integral to establish $u \in C^{2}(\mathbb{R}^n)$] The same argument applies verbatim to the second derivatives. Fix $i, j \in \{1, \ldots, n\}$. Since $f \in C_c^{2}(\mathbb{R}^n)$, the partial derivative $\partial_{x_i} f$ is $C^{1}$ with compact support. Define the difference quotient of $\partial_{x_i} f$ in the $e_j$-direction: \begin{align*} \widetilde{D}_h(z) := \frac{\partial_{x_i} f(z + h e_j) - \partial_{x_i} f(z)}{h}. \end{align*} By Taylor's theorem, $|\widetilde{D}_h(z) - \partial_{x_i x_j}^{2} f(z)| \leq \frac{|h|}{2} \sup_{w} |\partial_{x_j}(\partial_{x_i x_j}^{2} f)(w)|$. However, $f$ is only assumed $C^{2}$, so we instead use the uniform continuity of $\partial_{x_i x_j}^{2} f$ (continuous with compact support implies uniformly continuous) and the Mean Value Theorem to obtain pointwise convergence $\widetilde{D}_h(z) \to \partial_{x_i x_j}^{2} f(z)$ with the uniform bound $|\widetilde{D}_h(z)| \leq \|\partial_{x_i x_j}^{2} f\|_{L^\infty}$. Applying the [Dominated Convergence Theorem](/theorems/4) with the same dominating function $\|\partial_{x_i x_j}^{2} f\|_{L^\infty} |\Phi(y)| \mathbb{1}_{B(x, R+1)}(y)$ (integrable by the [local integrability of $\Phi$](/theorems/5)): \begin{align*} \partial_{x_i x_j}^{2} u(x) = \int_{\mathbb{R}^n} \Phi(y) \, \partial_{x_i x_j}^{2} f(x - y) \, d\mathcal{L}^n(y). \end{align*} The right-hand side is continuous in $x$ by the same dominated convergence argument used for $\partial_{x_i} u$. Since all second-order partial derivatives of $u$ exist and are continuous, $u \in C^{2}(\mathbb{R}^n)$. [/step] [step:Establish $-\Delta u = f$ using the distributional identity $\Delta \Phi = -\delta_0$] Having shown $u \in C^{2}(\mathbb{R}^n)$, we now prove $-\Delta u = f$. By Part 1, we may compute $\Delta u$ by differentiating under the integral: \begin{align*} \Delta u(x) &= \sum_{i=1}^{n} \partial_{x_i}^{2} u(x) = \sum_{i=1}^{n} \int_{\mathbb{R}^n} \Phi(y) \, \partial_{x_i}^{2} f(x - y) \, d\mathcal{L}^n(y) \\ &= \int_{\mathbb{R}^n} \Phi(y) \, \Delta_x f(x - y) \, d\mathcal{L}^n(y), \end{align*} where we exchanged the finite sum and the integral, and $\Delta_x$ denotes the Laplacian with respect to $x$ (which, in the integrand, acts on $f(x - y)$ with $y$ fixed). The key identity is: for any $\varphi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} \int_{\mathbb{R}^n} \Phi(y) \, \Delta \varphi(y) \, d\mathcal{L}^n(y) = -\varphi(0). \end{align*} This is precisely the statement that $\Delta \Phi = -\delta_0$ in the sense of [distributions](/theorems/27). While our $f$ is only $C^{2}$ rather than $C^\infty$, the distributional identity $\Delta \Phi = -\delta_0$ holds when tested against any $\varphi \in C_c^{2}(\mathbb{R}^n)$, since the integration-by-parts argument in the [proof of the distributional Laplacian identity](/theorems/27) requires only two derivatives of the test function. For each fixed $x \in \mathbb{R}^n$, define \begin{align*} \varphi_x : \mathbb{R}^n & \to \mathbb{R} \\ y & \mapsto f(x - y). \end{align*} Then $\varphi_x \in C_c^{2}(\mathbb{R}^n)$ (since $f \in C_c^{2}(\mathbb{R}^n)$ and the map $y \mapsto x - y$ is a diffeomorphism). Moreover, $\Delta_y \varphi_x(y) = \Delta_y f(x - y) = \Delta_x f(x - y)$, because \begin{align*} \partial_{y_i} f(x - y) = -\partial_{x_i} f(x - y) \quad \Longrightarrow \quad \partial_{y_i}^{2} f(x - y) = \partial_{x_i}^{2} f(x - y). \end{align*} Applying the distributional identity to $\varphi_x$: \begin{align*} \int_{\mathbb{R}^n} \Phi(y) \, \Delta_x f(x - y) \, d\mathcal{L}^n(y) &= \int_{\mathbb{R}^n} \Phi(y) \, \Delta_y \varphi_x(y) \, d\mathcal{L}^n(y) = -\varphi_x(0) = -f(x). \end{align*} Therefore $\Delta u(x) = -f(x)$, which gives $-\Delta u(x) = f(x)$ for all $x \in \mathbb{R}^n$. [guided] The strategy for Part 2 is to exploit the fact that $\Phi$ is a fundamental solution — meaning $\Delta \Phi = -\delta_0$ in the distributional sense — to convert the convolution $\Phi * (\Delta f)$ into an evaluation of $f$. From the regularity result in the previous steps, we have an explicit formula for $\Delta u$: we may move the Laplacian inside the integral because we proved that each second derivative passes through. The finite sum $\Delta = \sum_{i=1}^n \partial_{x_i}^{2}$ commutes with the integral since the sum is finite, giving \begin{align*} \Delta u(x) = \int_{\mathbb{R}^n} \Phi(y) \, \Delta_x f(x - y) \, d\mathcal{L}^n(y). \end{align*} Now comes the central observation. The [distributional identity $\Delta \Phi = -\delta_0$](/theorems/27) states that for every test function $\varphi \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} \int_{\mathbb{R}^n} \Phi(y) \, \Delta \varphi(y) \, d\mathcal{L}^n(y) = -\varphi(0). \end{align*} The proof of this identity (see [theorem 27](/theorems/27)) uses integration by parts on $B(0, \varepsilon)^c$, the divergence theorem on the sphere $\partial B(0, \varepsilon)$, and [the fact that $\Phi$ solves Laplace's equation away from the origin](/theorems/3). The integration by parts requires only that $\varphi$ has two continuous derivatives, so the identity extends to $\varphi \in C_c^{2}(\mathbb{R}^n)$. For fixed $x$, define $\varphi_x(y) := f(x - y)$. We verify: - $\varphi_x \in C^{2}(\mathbb{R}^n)$ because $f \in C^{2}(\mathbb{R}^n)$ and $y \mapsto x - y$ is smooth. - $\varphi_x$ has compact support: $\varphi_x(y) \neq 0$ only when $x - y \in \operatorname{supp} f \subset B(0, R)$, i.e., $y \in B(x, R)$. - The Laplacian with respect to $y$ agrees with the Laplacian with respect to $x$: $\partial_{y_i} f(x - y) = -\partial_{x_i} f(x - y)$, so $\partial_{y_i}^{2} f(x - y) = \partial_{x_i}^{2} f(x - y)$. Summing over $i$: $\Delta_y \varphi_x(y) = \Delta_x f(x - y)$. Applying the distributional identity to $\varphi_x$: \begin{align*} \Delta u(x) &= \int_{\mathbb{R}^n} \Phi(y) \, \Delta_x f(x - y) \, d\mathcal{L}^n(y) \\ &= \int_{\mathbb{R}^n} \Phi(y) \, \Delta_y \varphi_x(y) \, d\mathcal{L}^n(y) \\ &= -\varphi_x(0) \\ &= -f(x). \end{align*} Rearranging: $-\Delta u(x) = f(x)$ for every $x \in \mathbb{R}^n$. [/guided] [/step]