[proofplan] The four properties are verified independently. Smoothness follows from the fact that both factors of $\Phi$ -- the power $t^{-n/2}$ and the Gaussian exponential -- are $C^\infty$ on $\mathbb{R}^n \times (0,\infty)$, so their product is. The [heat equation](/page/Heat%20Equation) is established by direct computation: compute $\partial_t\Phi$ and $\Delta\Phi$ separately, then observe that the two expressions are identical and cancel. Unit mass reduces to the standard Gaussian integral $\int_{\mathbb{R}} e^{-z^2}\,d\mathcal{L}^1(z) = \sqrt{\pi}$ via a coordinate-by-coordinate substitution. The delta-[limit](/page/Limit) property -- the most substantial step -- uses the near-field/far-field decomposition: the near-field integral is controlled by continuity of the test function, and the far-field integral vanishes because the Gaussian tail shrinks to the empty set as $t \downarrow 0$. [/proofplan] [step:Verify smoothness of $\Phi$ on $\mathbb{R}^n \times (0,\infty)$] Write $\Phi$ in factored form: \begin{align*} \Phi(x,t) &= (4\pi)^{-n/2}\, t^{-n/2}\, \exp\!\left(-\frac{|x|^2}{4t}\right), \qquad (x,t) \in \mathbb{R}^n \times (0,\infty). \end{align*} The factor $(x,t) \mapsto t^{-n/2}$ is $C^\infty$ on $\mathbb{R}^n \times (0,\infty)$ since $t > 0$. The exponent $(x,t) \mapsto -|x|^2/(4t)$ is a $C^\infty$ function of $(x,t)$ for $t > 0$ (it is a rational function of $t$ with $t \neq 0$ and polynomial in $x$), and the exponential function $\exp: \mathbb{R} \to \mathbb{R}$ is entire. Therefore the composition $(x,t) \mapsto \exp(-|x|^2/(4t))$ is $C^\infty$ on $\mathbb{R}^n \times (0,\infty)$. The product of $C^\infty$ functions is $C^\infty$, so $\Phi \in C^\infty(\mathbb{R}^n \times (0,\infty))$. [/step] [step:Compute $\partial_t\Phi$ and $\Delta\Phi$ and verify cancellation] [claim:Time derivative of the heat kernel] For all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, \begin{align*} \partial_t \Phi(x,t) &= \left(-\frac{n}{2t} + \frac{|x|^2}{4t^2}\right)\Phi(x,t). \end{align*} [/claim] [proof] Write $f(t) := t^{-n/2}$ and $g(x,t) := \exp(-|x|^2/(4t))$. Then $f'(t) = -(n/2)\,t^{-n/2-1}$ and, by the chain rule applied to $\partial_t(-|x|^2/(4t)) = |x|^2/(4t^2)$: \begin{align*} \partial_t g(x,t) = g(x,t) \cdot \frac{|x|^2}{4t^2}. \end{align*} By the product rule: \begin{align*} \partial_t \Phi &= (4\pi)^{-n/2}\bigl(f'(t)\,g(x,t) + f(t)\,\partial_t g(x,t)\bigr) \\ &= (4\pi)^{-n/2}\left(-\frac{n}{2}\,t^{-n/2-1} + t^{-n/2}\cdot\frac{|x|^2}{4t^2}\right)g(x,t) \\ &= \left(-\frac{n}{2t} + \frac{|x|^2}{4t^2}\right)\Phi(x,t). \end{align*} [/proof] [claim:Laplacian of the heat kernel] For all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, \begin{align*} \Delta \Phi(x,t) &= \left(\frac{|x|^2}{4t^2} - \frac{n}{2t}\right)\Phi(x,t). \end{align*} [/claim] [proof] For each coordinate index $i \in \{1, \ldots, n\}$, compute the first spatial derivative. Since $\partial_{x_i}(-|x|^2/(4t)) = -x_i/(2t)$ and $t^{-n/2}$ is independent of $x$, the chain rule gives: \begin{align*} \partial_{x_i} \Phi &= (4\pi)^{-n/2}\,t^{-n/2}\,\exp\!\left(-\frac{|x|^2}{4t}\right)\cdot\left(-\frac{x_i}{2t}\right) = -\frac{x_i}{2t}\,\Phi. \end{align*} Differentiating again in $x_i$ by the product rule: \begin{align*} \partial_{x_i x_i} \Phi &= -\frac{1}{2t}\,\Phi + \left(-\frac{x_i}{2t}\right)\left(-\frac{x_i}{2t}\right)\Phi = \left(\frac{x_i^2}{4t^2} - \frac{1}{2t}\right)\Phi. \end{align*} Summing over $i = 1, \ldots, n$: \begin{align*} \Delta \Phi &= \sum_{i=1}^{n} \partial_{x_i x_i} \Phi = \left(\frac{\sum_{i=1}^n x_i^2}{4t^2} - \frac{n}{2t}\right)\Phi = \left(\frac{|x|^2}{4t^2} - \frac{n}{2t}\right)\Phi. \end{align*} [/proof] Subtracting the Laplacian from the time derivative: \begin{align*} \partial_t \Phi - \Delta \Phi &= \left(-\frac{n}{2t} + \frac{|x|^2}{4t^2}\right)\Phi - \left(\frac{|x|^2}{4t^2} - \frac{n}{2t}\right)\Phi = 0. \end{align*} [/step] [step:Reduce the unit mass integral to the standard Gaussian via coordinate substitution] We show $\int_{\mathbb{R}^n} \Phi(x,t)\,d\mathcal{L}^n(x) = 1$ for every $t > 0$. Since the exponent separates as $|x|^2 = x_1^2 + \cdots + x_n^2$, the integral factors by Fubini's theorem (the integrand is non-negative, so Tonelli's theorem applies without integrability verification): \begin{align*} \int_{\mathbb{R}^n} \Phi(x,t)\,d\mathcal{L}^n(x) &= (4\pi t)^{-n/2}\prod_{j=1}^{n}\int_{-\infty}^{\infty}\exp\!\left(-\frac{x_j^2}{4t}\right)d\mathcal{L}^1(x_j). \end{align*} For each factor, substitute $z_j := x_j/(2\sqrt{t})$. Then $x_j = 2\sqrt{t}\,z_j$ and $d\mathcal{L}^1(x_j) = 2\sqrt{t}\,d\mathcal{L}^1(z_j)$, and the domain $x_j \in (-\infty, \infty)$ maps to $z_j \in (-\infty, \infty)$: \begin{align*} \int_{-\infty}^{\infty}\exp\!\left(-\frac{x_j^2}{4t}\right)d\mathcal{L}^1(x_j) &= 2\sqrt{t}\int_{-\infty}^{\infty}e^{-z_j^2}\,d\mathcal{L}^1(z_j) = 2\sqrt{t}\cdot\sqrt{\pi} = 2\sqrt{\pi t}, \end{align*} where $\int_{-\infty}^{\infty} e^{-z^2}\,d\mathcal{L}^1(z) = \sqrt{\pi}$ is the standard Gaussian integral. Taking the product of $n$ identical factors: \begin{align*} (4\pi t)^{-n/2}\cdot\bigl(2\sqrt{\pi t}\bigr)^n &= (4\pi t)^{-n/2}\cdot 2^n\cdot\pi^{n/2}\cdot t^{n/2} = \frac{2^n \cdot \pi^{n/2}\cdot t^{n/2}}{2^n\cdot\pi^{n/2}\cdot t^{n/2}} = 1. \end{align*} [guided] The unit mass property says that $\Phi(\cdot, t)$ is a probability density on $\mathbb{R}^n$ for each $t > 0$. The computation exploits the product structure of the Gaussian: the $n$-dimensional integral factors into $n$ identical one-dimensional integrals, each of which is a rescaled version of the standard Gaussian integral $\int e^{-z^2}\,d\mathcal{L}^1(z) = \sqrt{\pi}$. The substitution $z_j = x_j/(2\sqrt{t})$ normalises each factor to the standard form. The measure transforms as $d\mathcal{L}^1(x_j) = 2\sqrt{t}\,d\mathcal{L}^1(z_j)$ (the Jacobian of the affine map $x_j \mapsto x_j/(2\sqrt{t})$ is $1/(2\sqrt{t})$, and the inverse function theorem for Lebesgue measure gives the factor $2\sqrt{t}$). Each one-dimensional integral contributes $2\sqrt{\pi t}$, and the product $(2\sqrt{\pi t})^n$ exactly cancels the prefactor $(4\pi t)^{-n/2}$: \begin{align*} (4\pi t)^{-n/2} \cdot (2\sqrt{\pi t})^n = \frac{2^n \pi^{n/2} t^{n/2}}{(4\pi t)^{n/2}} = \frac{2^n \pi^{n/2} t^{n/2}}{2^n \pi^{n/2} t^{n/2}} = 1. \end{align*} Fubini's theorem is invoked to factor the $n$-dimensional integral into a product of one-dimensional integrals. Since the integrand $\Phi(x,t) \geq 0$ for all $x \in \mathbb{R}^n$ and $t > 0$, Tonelli's theorem applies directly without needing to verify integrability first. [/guided] [/step] [step:Prove the delta-limit $T_{\Phi(\cdot,t)} \to \delta_0$ in $\mathcal{D}'(\mathbb{R}^n)$ via near-field/far-field decomposition] We must show that for every [test function](/page/Test%20Function) $\varphi \in C_c^\infty(\mathbb{R}^n)$: \begin{align*} \lim_{t \downarrow 0}\int_{\mathbb{R}^n}\Phi(x,t)\,\varphi(x)\,d\mathcal{L}^n(x) = \varphi(0). \end{align*} By the unit mass property (previous step), $\varphi(0) = \varphi(0)\int_{\mathbb{R}^n}\Phi(x,t)\,d\mathcal{L}^n(x)$ for every $t > 0$. Define the error \begin{align*} I(t) &:= \int_{\mathbb{R}^n}\Phi(x,t)\bigl(\varphi(x) - \varphi(0)\bigr)\,d\mathcal{L}^n(x). \end{align*} Fix $\varepsilon > 0$. Since $\varphi$ is [continuous](/page/Continuity) at the origin, there exists $\rho > 0$ such that $|x| \leq \rho$ implies $|\varphi(x) - \varphi(0)| < \varepsilon$. Decompose $I(t) = I_1(t) + I_2(t)$ where \begin{align*} I_1(t) &:= \int_{B(0,\rho)}\Phi(x,t)\bigl(\varphi(x) - \varphi(0)\bigr)\,d\mathcal{L}^n(x), \\ I_2(t) &:= \int_{\mathbb{R}^n \setminus B(0,\rho)}\Phi(x,t)\bigl(\varphi(x) - \varphi(0)\bigr)\,d\mathcal{L}^n(x). \end{align*} [claim:Near-field estimate] For all $t > 0$, $|I_1(t)| \leq \varepsilon$. [/claim] [proof] Since $\Phi(x,t) > 0$ for all $(x,t) \in \mathbb{R}^n \times (0,\infty)$, and $|\varphi(x) - \varphi(0)| < \varepsilon$ for $x \in B(0,\rho)$: \begin{align*} |I_1(t)| &\leq \int_{B(0,\rho)}\Phi(x,t)\,|\varphi(x) - \varphi(0)|\,d\mathcal{L}^n(x) < \varepsilon\int_{B(0,\rho)}\Phi(x,t)\,d\mathcal{L}^n(x). \end{align*} Since $\Phi \geq 0$, we enlarge the domain of integration from $B(0,\rho)$ to $\mathbb{R}^n$: \begin{align*} \varepsilon\int_{B(0,\rho)}\Phi(x,t)\,d\mathcal{L}^n(x) \leq \varepsilon\int_{\mathbb{R}^n}\Phi(x,t)\,d\mathcal{L}^n(x) = \varepsilon, \end{align*} where the equality is the unit mass property. [/proof] [claim:Far-field estimate] $I_2(t) \to 0$ as $t \downarrow 0$. [/claim] [proof] Set $M := \|\varphi\|_{L^\infty(\mathbb{R}^n)} + |\varphi(0)|$, which is finite since $\varphi \in C_c^\infty(\mathbb{R}^n)$. Then $|\varphi(x) - \varphi(0)| \leq M$ for all $x \in \mathbb{R}^n$, so \begin{align*} |I_2(t)| &\leq M\int_{\mathbb{R}^n \setminus B(0,\rho)}\Phi(x,t)\,d\mathcal{L}^n(x). \end{align*} Substitute $y := x/(2\sqrt{t})$. Then $d\mathcal{L}^n(x) = (2\sqrt{t})^n\,d\mathcal{L}^n(y)$, the condition $|x| \geq \rho$ becomes $|y| \geq \rho/(2\sqrt{t})$, and \begin{align*} \int_{\mathbb{R}^n \setminus B(0,\rho)}\Phi(x,t)\,d\mathcal{L}^n(x) &= (4\pi t)^{-n/2}\cdot(2\sqrt{t})^n\int_{|y| \geq \rho/(2\sqrt{t})}e^{-|y|^2}\,d\mathcal{L}^n(y) \\ &= \pi^{-n/2}\int_{|y| \geq \rho/(2\sqrt{t})}e^{-|y|^2}\,d\mathcal{L}^n(y). \end{align*} The function $y \mapsto e^{-|y|^2}$ is integrable on $\mathbb{R}^n$ (its integral is $\pi^{n/2}$). As $t \downarrow 0$, the lower bound $\rho/(2\sqrt{t}) \to \infty$, so the domain of integration $\{|y| \geq \rho/(2\sqrt{t})\}$ shrinks to the empty set. Since $e^{-|y|^2} \geq 0$, the Gaussian tail integral decreases monotonically to $0$. Therefore $|I_2(t)| \to 0$ as $t \downarrow 0$. [/proof] Combining the two estimates: \begin{align*} \limsup_{t \downarrow 0} |I(t)| \leq \limsup_{t \downarrow 0}\bigl(|I_1(t)| + |I_2(t)|\bigr) \leq \varepsilon + 0 = \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\lim_{t \downarrow 0} I(t) = 0$, which gives $T_{\Phi(\cdot,t)} \to \delta_0$ in $\mathcal{D}'(\mathbb{R}^n)$. [guided] The delta-limit property says that $\Phi(\cdot, t)$ concentrates at the origin as $t \downarrow 0$: it acts like a [Dirac delta](/page/Distribution) in the distributional sense. Concretely, for every test function $\varphi \in C_c^\infty(\mathbb{R}^n)$, we must show \begin{align*} \lim_{t \downarrow 0}\int_{\mathbb{R}^n}\Phi(x,t)\,\varphi(x)\,d\mathcal{L}^n(x) = \varphi(0). \end{align*} The proof uses the near-field/far-field decomposition. Using the unit mass property $\int_{\mathbb{R}^n}\Phi(x,t)\,d\mathcal{L}^n(x) = 1$, rewrite the difference as \begin{align*} I(t) := \int_{\mathbb{R}^n}\Phi(x,t)\bigl(\varphi(x) - \varphi(0)\bigr)\,d\mathcal{L}^n(x). \end{align*} Fix $\varepsilon > 0$. Continuity of $\varphi$ at the origin gives $\rho > 0$ such that $|x| \leq \rho$ implies $|\varphi(x) - \varphi(0)| < \varepsilon$. Split $I(t) = I_1(t) + I_2(t)$ where $I_1$ integrates over $B(0,\rho)$ and $I_2$ over $\mathbb{R}^n \setminus B(0,\rho)$. For the near-field piece $I_1(t)$: on $B(0,\rho)$ the continuity bound gives $|\varphi(x) - \varphi(0)| < \varepsilon$, and since $\Phi(x,t) > 0$ we can enlarge the domain of integration from $B(0,\rho)$ to $\mathbb{R}^n$: \begin{align*} |I_1(t)| \leq \int_{B(0,\rho)} \Phi(x,t)\,|\varphi(x) - \varphi(0)|\,d\mathcal{L}^n(x) < \varepsilon \int_{B(0,\rho)} \Phi(x,t)\,d\mathcal{L}^n(x) \leq \varepsilon \int_{\mathbb{R}^n} \Phi(x,t)\,d\mathcal{L}^n(x) = \varepsilon. \end{align*} For the far-field piece $I_2(t)$: set $M := \|\varphi\|_{L^\infty} + |\varphi(0)|$ so that $|\varphi(x) - \varphi(0)| \leq M$ for all $x$. Substitute $y = x/(2\sqrt{t})$, giving $d\mathcal{L}^n(x) = (2\sqrt{t})^n\,d\mathcal{L}^n(y)$ and $|x| \geq \rho$ becomes $|y| \geq \rho/(2\sqrt{t})$: \begin{align*} |I_2(t)| &\leq M \int_{\mathbb{R}^n \setminus B(0,\rho)} \Phi(x,t)\,d\mathcal{L}^n(x) = M\,(4\pi t)^{-n/2}(2\sqrt{t})^n \int_{|y| \geq \rho/(2\sqrt{t})} e^{-|y|^2}\,d\mathcal{L}^n(y) \\ &= M\,\pi^{-n/2}\int_{|y| \geq \rho/(2\sqrt{t})} e^{-|y|^2}\,d\mathcal{L}^n(y). \end{align*} As $t \downarrow 0$, the cutoff $\rho/(2\sqrt{t}) \to \infty$, and since $y \mapsto e^{-|y|^2}$ is integrable on $\mathbb{R}^n$ (with total integral $\pi^{n/2}$), the tail integral vanishes: \begin{align*} \int_{|y| \geq R} e^{-|y|^2}\,d\mathcal{L}^n(y) \to 0 \quad \text{as } R \to \infty. \end{align*} This is where the Gaussian decay of $\Phi$ is essential: a kernel with slower decay (e.g., polynomial tails) would not produce a vanishing tail. More precisely, since $e^{-|y|^2} \geq 0$ and the sets $\{|y| \geq R\}$ decrease to $\varnothing$ as $R \to \infty$, the monotone convergence theorem (applied to the complements) guarantees the tail integral decreases monotonically to $0$. The far-field estimate therefore gives $|I_2(t)| \to 0$ as $t \downarrow 0$. Combining both estimates via the triangle inequality: \begin{align*} \limsup_{t \downarrow 0} |I(t)| \leq \limsup_{t \downarrow 0}\bigl(|I_1(t)| + |I_2(t)|\bigr) \leq \varepsilon + 0 = \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\lim_{t \downarrow 0} I(t) = 0$, which gives $\lim_{t \downarrow 0}\int_{\mathbb{R}^n}\Phi(x,t)\,\varphi(x)\,d\mathcal{L}^n(x) = \varphi(0)$, i.e., $T_{\Phi(\cdot,t)} \to \delta_0$ in $\mathcal{D}'(\mathbb{R}^n)$. [/guided] [/step]