Let $\displaystyle\mathcal{H}_{t}(x)=\frac{1}{(4\pi t)^{1/2}}e^{-x^{2}/4t}$ be the Heat Kernel.
The imposed initial condition for the heat equation on a real line is $u(x,0)=f(x)$ a function belongs to Schwartz Space.
Then, $u(x,t)=(f*\mathcal{H}_{t})(x)$ for $t>0$
In my textbook, the author uses the following estimation to bound $u(x,t)$:
$$\displaystyle|u(x,t)|\leq \int_{|y|\leq |x|/2}|f(x-y)|\mathcal{H}_{t}(y)\,dy+\int_{|y|\geq |x|/2}|f(x-y)|\mathcal{H}_{t}(y)\,dy$$ $$\displaystyle\leq\frac{C_{N}}{(1+|x|)^{N}}+\frac{B}{\sqrt{t}}e^{-Dx^{2}/t}$$ where $C_{N},B,D>0$ are constants.
My question is why using the cases $|y|\leq|x|/2$ and $|y|\geq|x|/2$? How come it bounds the second integral like this?
Notation: I'll probably be a little messy, and absorb a lot of constants as is standard with this sort of thing.
If $f$ is in the Schwartz space, then $f$ is infinitely differentiable (in particular continuous), and $f$ (as well as its derivatives) decay faster than any polynomial. Now, we wish to split the integral $$ u(x,t) = \int_{\mathbb{R}^n} f(x-y)H_t(y)~dy = A + B $$ into two integrals $A$ and $B$ so that we have good control over both $A$ and $B$. If we choose to split the integral into a compact domain and its complement, then since we have good control on $g$ on compact sets and global control on $|f(x-y)|$ (since $f$ is a Schwartz function) it's easy to control $A$ by $|x|$ on the compact domain; the problem is to deal with $f(x-y)H_t(y)$ near infinity.
With that said, how should we choose our compact domain? $H_t(y)$ decays quickly as $|y|\to\infty$, so the problem is how to deal with $|f(x-y)|$, and $|f(x-y)|$ is bounded: $\sup|f(z)|<\infty$ since $f$ is a Schwartz function. Since we need control over the integral in terms of $|x|$ this doesn't help us, but we can pass the dependence on $x$ over to $H_t$ via $$ \int_\Omega f(x-y)H_t(y)~dy = \int_\Omega f(y)H_t(x-y)~dy. $$ So this integral is bounded by $$ \left|\int f(y)H_t(x-y)~dy\right| \leq C\left|\int H_t(x-y)~dy\right| \leq \frac{C}{\sqrt{t}}\left|\int e^{\frac{-|x-y|^2}{4t}}~dy\right|. $$ This last term is $$ \left|\int_\Omega e^{(-|x|^2 + 2x\cdot y - |y|^2)/4t}~dy\right| \leq e^{-\frac{-|x|^2}{4t}}\left|\int_\Omega e^{\frac{2|x||y|}{4t}}e^{\frac{-|y|^2}{4t}}~dy\right| = e^{-\frac{-|x|^2}{4t}}\int_\Omega e^{|y|(2|x|-|y|)/4t}~dy. $$ Now we need only control $e^{|y|(2|x|-|y|)/4t}$, and this is where we can choose our $\Omega$. When $|y|$ is large compared to $|x|$, $2|x|-|y| \sim -|y|$ is negative and the integrand behaves like $$ e^{|y|(2|x|-|y|)/4t} \sim e^{-|y|^2/4t}, $$ which we know is integrable. Via this heuristic we let $\Omega = \{y:|y|\geq |x|/m\}$. Then on $\Omega$, $$ e^{|y|(2|x|-|y|)/4t} \leq e^{|y|(2m|y|-|y|)/4t} = e^{(\frac{1}{2}m-\frac{1}{4})|y|^2/t} $$ and to make $\frac{1}{2}m - \frac{1}{4}$ negative we choose $m < \frac{1}{2}$, say $m = \frac{1}{2+\epsilon}$. So $\Omega = \{y:|y| > (2+\epsilon)|x|\}$ will work. So we split up the integral into $$ u(x,t) = \int_{|y| \leq (2+\epsilon)|x|} f(x-y)H_t(y)~dy + \int_{|y| \geq (2+\epsilon)|x|} f(x-y)H_t(y)~dy = A + B. $$ $A$ is bounded by $C/(1+|x|)^N$ because the behavior of this integrand is dominated by $|f(x-y)|$ and $f$ is a Schwartz function, while $B$ is bounded by $$ \frac{C}{\sqrt{t}}e^{-|x|^2/4t}\int_{|y| \geq (2+\epsilon)|x|} e^{-D|y|^2/t}~dy \leq \frac{C}{\sqrt{t}}e^{-|x|^2/4t} $$ where $D>0$. This gets us the desired bounds $$ |u(x,y)| \leq \frac{C}{(1+|x|)^N} + \frac{B}{\sqrt{t}}e^{-D|x|^2/t}. $$
Caution, $D$ depends on the choice of $\epsilon$. If one is more careful one might be able to get rid of the $\epsilon$ in the definition of $\Omega$, but if you just want these bounds then the precise value of $m$ isn't important so long as it's small enough.