Set of $(x,t)$ such that $e^{- \frac{|x|^2}{4t}} \geq c$ is compact

Question

Let $\phi(x,t) = (4 \pi t)^{-\frac{n}{2}} e^{- \frac{|x|^2}{4t}}$ be the fundamental solution of the heat equation in $\mathbb R^n$. What is the exact reasoning that the set $\{(x,t) \in \mathbb R^n \times[0,\infty) | \phi(x,t) \geq t^{-\frac{n}{2}}\}$ is compact? Well, $\phi(x,t) \geq t^{-\frac{n}{2}}$ iff $e^{- \frac{|x|^2}{4t}} \geq (4 \pi)^{\frac{n}{2}}$, a constant. Now why is the set of $(x,t)$ for which this holds compact? What is the analytic property of the function $e^{- \frac{|x|^2}{4t}}$ to assure this? I mean I can picture it that this function is concentrated near $0$ but how to argue mathematical here?

Answer 1

Let $$S:=\left\{(x,t) \in \mathbb R^n \times[0,\infty) | \phi(x,t) \geq t^{-\frac{n}{2}}\right\}.$$ Observe that if $(x,t)\in C$, then $\left(\lambda x,\lambda^2t\right)$ belongs to $S$ for any positive $\lambda$. Therefore, $S$ is not bounded and cannot be compact.