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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.

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1 Answer 1

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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 $C_N = 2^Nc_N$.

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  • $\begingroup$ This perfectly answers the first part. I've tried to modify it to prove the second one, that is $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l$ but your answer doesn't work here because we're not multiplying it by a fractional power. Could you please help me with this part as well? I've really been struggling to understand this inequality for a long time. $\endgroup$ Oct 2, 2016 at 8:06
  • $\begingroup$ Actually, by the same line of reasoning, in both cases I can get $|x|^l |g(x-y)|\le A_l (1+|y|)^{-l}$, but then clearly $(1+|y|)^{-l}\le (1+|y|)^l$. I'm not sure if this is what is intended though, as why didn't the authors write $-l$th power in the first place? $\endgroup$ Oct 2, 2016 at 8:19
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    $\begingroup$ You don't need to use the binomial theorem, just note that $|x|\leq (|x-y|+|y|)\leq 2|x-y|$ when $|x|\geq 2|y|$ and $1\leq (1+|y|)^{l}$ since $l\geq 0$. $\endgroup$ Jun 13, 2019 at 19:47

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