[proofplan] The proof uses only the explicit heat kernel representation and the separation between the observation point $x$ and the support of $f$. Since $f$ is supported in $\overline{B}(0,R)$ and $|x| \ge 2R$, every point $y$ contributing to the convolution satisfies $|x-y| \ge |x|/2$. This converts the Gaussian factor into a uniform tail factor, after which the remaining integral is exactly bounded by the $L^1$ norm of $f$. [/proofplan] [step:Use the support condition to separate $x$ from all contributing points] Fix $t > 0$ and $x \in \mathbb{R}^n$ with $|x| \ge 2R$. Let $S := \operatorname{supp} f \subset \overline{B}(0,R)$ denote the closed support of $f$. For every $y \in S$, the triangle inequality gives \begin{align*} |x-y| \ge |x| - |y|. \end{align*} Since $y \in \overline{B}(0,R)$, we have $|y| \le R$. Since $|x| \ge 2R$, we also have $R \le |x|/2$. Therefore \begin{align*} |x-y| \ge |x| - R \ge \frac{|x|}{2}. \end{align*} Squaring this non-negative inequality gives \begin{align*} |x-y|^2 \ge \frac{|x|^2}{4}. \end{align*} [guided] We fix $t > 0$ and $x \in \mathbb{R}^n$ satisfying $|x| \ge 2R$. The only points $y$ that can contribute to the convolution integral are points in the support of $f$, because $f(y)=0$ outside $\operatorname{supp} f$. Define \begin{align*} S := \operatorname{supp} f. \end{align*} By hypothesis, $S \subset \overline{B}(0,R)$, so every $y \in S$ satisfies $|y| \le R$. The geometric estimate we need is a lower bound on the distance from $x$ to each such $y$. By the triangle inequality in $\mathbb{R}^n$, \begin{align*} |x| \le |x-y| + |y|. \end{align*} Rearranging gives \begin{align*} |x-y| \ge |x| - |y|. \end{align*} Since $|y| \le R$, this implies \begin{align*} |x-y| \ge |x| - R. \end{align*} The assumption $|x| \ge 2R$ is used exactly here: it gives $R \le |x|/2$, and hence \begin{align*} |x-y| \ge |x| - R \ge \frac{|x|}{2}. \end{align*} Both sides are non-negative, so squaring preserves the inequality: \begin{align*} |x-y|^2 \ge \frac{|x|^2}{4}. \end{align*} This is the point where compact support becomes a Gaussian tail estimate: the support condition converts spatial separation into a uniform lower bound inside the exponential. [/guided] [/step] [step:Convert the distance lower bound into a Gaussian upper bound] For every $y \in S$, the estimate from the previous step gives \begin{align*} -\frac{|x-y|^2}{4t} \le -\frac{|x|^2}{16t}. \end{align*} Since the exponential function is increasing on $\mathbb{R}$, it follows that \begin{align*} \exp\left(-\frac{|x-y|^2}{4t}\right) \le \exp\left(-\frac{|x|^2}{16t}\right). \end{align*} Because $f=0$ $\mathcal{L}^n$-a.e. on $\mathbb{R}^n \setminus S$, the same bound is sufficient for the full integral after multiplication by $|f(y)|$. [/step] [step:Estimate the heat convolution by the $L^1$ norm] Using the defining convolution formula for $e^{t\Delta}f$, the triangle inequality for the [Lebesgue integral](/page/Lebesgue%20Integral), and the non-negativity of the heat kernel, \begin{align*} |e^{t\Delta}f(x)| \le (4\pi t)^{-n/2}\int_{\mathbb{R}^n} \exp\left(-\frac{|x-y|^2}{4t}\right)|f(y)|\,d\mathcal{L}^n(y). \end{align*} Since $f=0$ $\mathcal{L}^n$-a.e. outside $S$, we may restrict the integral to $S$: \begin{align*} |e^{t\Delta}f(x)| \le (4\pi t)^{-n/2}\int_S \exp\left(-\frac{|x-y|^2}{4t}\right)|f(y)|\,d\mathcal{L}^n(y). \end{align*} Applying the Gaussian bound from the previous step pointwise on $S$ gives \begin{align*} |e^{t\Delta}f(x)| \le (4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{16t}\right)\int_S |f(y)|\,d\mathcal{L}^n(y). \end{align*} Finally, since $S \subset \mathbb{R}^n$, \begin{align*} \int_S |f(y)|\,d\mathcal{L}^n(y) \le \int_{\mathbb{R}^n} |f(y)|\,d\mathcal{L}^n(y) = \|f\|_{L^1(\mathbb{R}^n)}. \end{align*} Combining the last two estimates yields \begin{align*} |e^{t\Delta}f(x)| \le (4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{16t}\right)\|f\|_{L^1(\mathbb{R}^n)}. \end{align*} This is the desired bound for the fixed $t > 0$ and $x \in \mathbb{R}^n$ with $|x| \ge 2R$, and hence the theorem follows. [/step]