I want to investigate the decay of $L(x)$:
$$L(x) := \int_{\mathbb{R}^3_+} \nabla_x \Phi(x-y,1/2)(\eta(y)g(y))dy,$$
where $g:\mathbb{R}^3_+ \rightarrow \mathbb{R}^3$ is infinitely smooth away from the origin with $|g(y)|\leq\frac{1}{|y|}$ for all $y$. Here $\eta(y) \in C_c^\infty(\mathbb{R}^3)$ is a cut-off function such that $\eta(y)=1$ for $|y|<1$.
Now, the fundamental solution $\Phi$ of the heat equations is as usual, $$ \Phi(x,t)=\frac{1}{(4\pi t)^{3/2}}e^{-\frac{|x|^2}{4t}}$$ for any $(x,t)\in\mathbb{R}^3 \times (0,\infty)$.
Claim: $L(x)$ is bounded roughly by $C(1+|x|)^{-2},$ or probably by $C e^{-x^2/4}$ for some constant $C$ not depending on $|x|$. Here $x\in\mathbb{R}^3_+$.
What I have done is the following:
One can divide the half-space into two components $|x-y|\geq |x|/2$ and $|x-y|<|x|/2$, say $L = L_1 + L_2$. Then, concerned with $L_1$, $$ L_1 \lesssim e^{-\frac{|x|^2}{4}} \int_{M} \frac{|y-x|}{|y|} dy \lesssim (1+|x|)e^{-\frac{|x|^2}{4}},$$ where $M$ is the compact support of $\eta$. My problem is that the decay may not be optimal!
Geometrically, the other case implies $|y|>|x|/2$, so $$ L_2 \lesssim |x|(1/|x|)\int_M e^{-|x-y|^2/2}dy \lesssim \int_M e^{-|x|^2/4}dy \lesssim e^{-|x|^2/4},$$ which seem to be optimal.
Let me know how to obtain the faster decay when $|x-y|>|x|/2$ or suggest a theme that I can work on!
Observe if $|x|\gg |y|$ then we see that \begin{align} \left|\int_{B_1(0)} \frac{\nabla_x e^{-|x-y|^2}}{|y|}\ dy\right| \sim& \left|\int_{B_1(0)} \frac{e^{-|x-y|^2}}{|y|}(x-y)\ dy\right| \geq \int_{B_1(0)}\frac{e^{-|x-y|^2}}{|y|}(|x|-|y|)\ dy\\ \geq&\ \frac{|x|}{2}\int_{B_1(0)}\frac{e^{-|x-y|^2}}{|y|}\ dy \sim |x| e^{-|x|^2}. \end{align}