Let $u(x,t)=(f*\mathcal{H}_t)(x)$ for $t>0$ where $f$ is a function in the Schwartz space and $\mathcal{H}_t$ is the heat kernel. Then we have the following estimate (from Stein and Shakarchi's Fourier Analysis):
$$|u(x,t)|\le \int_{|y|\le|x|/2}|f(x-y)|\mathcal{H}_t(y)dy+\int_{|y|\ge|x|/2}|f(x-y)|\mathcal{H}_t(y)dy \\ \le \frac{C_N}{(1+|x|)^N}+\frac{C}{\sqrt{t}}e^{-cx^2/t}.$$
To get the first inequality, the text says "Indeed,since $f$ is rapidly decreasing, we have $|f(x-y)|\le C_N/(1+|x|)^N$ when $|y|\le |x|/2$."
However, this is what I don't understand. I don't know how to get this specific form of inequality given $|y|\le |x|/2$.
A related inequality given in the text is that if $g$ is rapidly decreasing then by considering the two cases $|x|\le 2|y|$ and $|x|\ge 2|y|$, we have $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l.$ I think this one can be shown by similar reasoning as the above one, but I really have no idea how to show this by considering the two cases.
A function $f$ is rapidly decreasing if it is indefinitely differentiable and $$\sup_{x\in R} |x|^k |f^{(l)}(x)|<\infty \; \text{for every} \; k,l\ge 0.$$
I would greatly appreciate it if anyone could show this inequality.
Use the binomial theorem and the rapidly decreasing property of $f$ to prove that for all $u\in \Bbb R^N$, $(1/2 + \lvert u\rvert)^N\lvert f(u)\rvert \le c_N$ for some constant $c_N$ depending only on $N$. Then $\lvert f(x - y)\rvert \le c_N (1/2 + \lvert x - y\rvert)^{-N}$. Since $\lvert y\rvert \le \lvert x\rvert/2$, the reverse triangle inequality gives $\lvert x - y\rvert \ge \lvert x\rvert/2$. Thus $(1/2 + \lvert x - y\rvert)^{-N} \le 2^N (1 + \lvert x\rvert)^{-N}$ and so $\lvert f(x - y)\rvert \le C_N (1 + \lvert x\rvert)^{-N}$, where $C_N = 2^Nc_N$.