How do we prove the "infinite propagation speed" property of the solution of the heat equation? Let $d\in\mathbb N$, $$p_t(x):=\left.\begin{cases}\frac{e^{-{\|x\|^2}{4|t|}}}{(4\pi|t|)^{\frac d2}}&\text{, if }t>0\\0&\text{, if }t<0\end{cases}\right\}\;\;\;\text{for }t\in\mathbb R\setminus\{0\}\text{ and }x\in\mathbb R^d$$ and $$p_t(x,y):=p_t(x-y)\;\;\;\text{for }t\in\mathbb R\setminus\{0\}\text{ and }x,y\in\mathbb R^d.$$
Moreover, let $\lambda$ denote the Lebesgue measure on $\mathbb R$ and $$\kappa_t(x,B):=\int_Bp_t(x,y)\:\lambda^{\otimes d}({\rm d}y)\;\;\;\text{for }(x,B)\in\mathbb R^d\times\mathcal B(\mathbb R)^{\otimes d}$$ for $t\in\mathbb R\setminus\{0\}$.
If $u_0\in C(\mathbb R^d)\cap\mathcal L^\infty(\mathbb R^d)$, we can easily see that $$u(t,x):=(\kappa_tu_0)(x)=\int\kappa_t(x,{\rm d}y)u_0(y)\;\;\;\text{for }(t,x)\in(0,\infty)\times\mathbb R^d$$ is a solution of the heat equation in $\mathbb R^d$, i.e. $$\frac{\partial u}{\partial t}=\Delta_xu\tag1.$$ Moreover, for all $x_0\in\mathbb R^d$, $$u(t,x)\xrightarrow{\begin{align}t&\to0+\\x&\to x_0\end{align}}u_0(x_0)\tag2.$$

Question 1: I've read that "$u(t,x)$ depends on the values of $u_0(y)$ at all points $y\in\mathbb R^d$ no matter how far $y$ and $x$ are away" and "even when $u_0$ is compactly supported, tthe domain of influence on the solution $u$ is still all of $x\in\mathbb R^d$ as long as $t>0". What precisely does that mean and how do we prove this rigorously?


Question 2: In light of $(2)$, are we able to extend $u$ to $[0,\infty)\times\Omega$ and show that $$u(0,x)=u_0(x)\tag3$$ for all $x\in\mathbb R^d$?


Question 3: If we don't consider all of $\mathbb R^d$, but a bounded open subset $\Omega\subseteq\mathbb R^d$ instead, is the solution of $$\frac{\partial v}{\partial t}=\Delta_xv\;\;\;\text{in }(0,\infty)\times\Omega\tag4$$ still of the previous form? And what would need to assume on $u_0:\Omega\to\mathbb R$ such that this solution satisfies $(3)$ for all $x\in\Omega$ as well?

 A: This is too long for a comment, but maybe not as thorough as an answer ought to be. I'll put it here regardless, hopefully it's useful :)
Regarding Q1, the best I can think of formalizing the statement is as follows: Suppose $u_0 \in C_c^\infty(\mathbb{R}^d)$ (smooth functions which vanish identically outside of some ball) and suppose $u_0\geq 0$ throughout. Then you can easily check that $u$ as you construct it is always positive (this is because the kernel $p_t$ is strictly positive for positive times). The way to think about this is that no matter how localized your initial temperature $u_0$ is (remember it's zero outside of some ball), at any given time $t>0$, the solution $u(t,\cdot)>0$ throughout $\mathbb{R}^d$, so that even points very far from where $u_0$ was concentrated feel it's effect no matter how close $t$ is to $0$.
For Q2 the best I can say is yes, since you have convergence you can define $u(0,x)=u_0$ and the resulting function is continuous on $[0,\infty)\times \Omega$.
For Q3 the answer is no, not in general and the issue is a little delicate. It turns out, even in $\Omega=\mathbb{R}^d$, the way your question is written, the answer is no: There exist solutions to the heat equation that are not identically zero, but with zero initial data (these functions are wild, and they grow pretty fast as $|x|\to \infty$ for $t>0$). So the issue is that you needs to restrict what kind of solutions you're working with. One way of doing it is by looking at bounded solutions.
Of course, given the nature of the solutions I mentioned above, you can't expect the representation formula to be true in the bounded domain case either. But, if you impose some boundary conditions on $(0,\infty)\times\partial\Omega$, then you're closer to something. It turns out $p_t$ alone still won't give you the solution as before (your boundary conditions may not be satisfied), but it's not too bad. In this regard I'd suggest looking at Green's functions.
