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Lee and Ni show in their work Link Here that the heat semigroup $e^{t \Delta}u_0$ has decay as $t^{-\min \{a, n\} /2}$, $t \to \infty$ if $u_0 = C(1+|x|^2)^{a/2}$ if $a \neq n$. I'm trying to understand if this minimum in power always happens for other types of initial value $u_0$, that is, the fastest it can decay is $t^{- n /2}$. I'm trying to understand the point he made, I've put an image below. He says to estimate the integral in two parts

$$ t^{-n/2} \int_0^{\infty} e^{-r/4t} (1+r)^{-a/2}r^{n/2-1}dr = t^{-n/2} \int_0^{1} e^{-r/4t} (1+r)^{-a/2}r^{n/2-1}dr+ t^{-n/2} \int_1^{\infty} e^{-r/4t} (1+r)^{-a/2}r^{n/2-1}dr $$

I'm not completely sure about how he bound the second integral, but it seems to me that in the first is bounded by $$ t^{-a/2} \int_0^{1/4t} e^{-r} r^{n/2-a/2-1}dr$$ So it is necessary that $n/2-a/2-1>-1$, or $n>a$. So I believe that for this type of radial data the growth really is this. But there is no other type of data in which the decay is faster? Perhaps even considering this same type of radial data, but with a value of $0$ on the unit ball, to remove the integrability problem.

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    $\begingroup$ If the initial data is positive, then indeed $t^{-n/2}$ is the fastest possible decay. This is the decay obtained for example if the initial data is positive and supported in the unit ball. On the other hand if the initial data is allowed to oscillate then the decay can be faster. If for example the initial data decays sufficiently rapidly and satisfies $\int u p = 0$ for all polynomials $p$ of degree less than $d$, then the decay goes as $t^{-(n+d)/2}$. $\endgroup$
    – felipeh
    Commented Sep 7 at 2:26
  • $\begingroup$ @felipeh thanks for responding. Would you know where I can find this theorem you mentioned? $\endgroup$
    – Ilovemath
    Commented Sep 7 at 11:40
  • $\begingroup$ I don't know a standard reference but I wrote out the proof in an answer below. Hope this is helpful. $\endgroup$
    – felipeh
    Commented Sep 9 at 0:51

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I'm expanding my comment as an answer in case it is helpful. Suppose that $u$ satisfies $$ \int u(x) p(x) \,dx = 0 $$ for all polynomials of degree less than $d$ (so in particular, if $d=1$ then this means that $\int u = 0$) and that $$ \int |x|^d |u(x)| \,dx = M_d < \infty. $$ Then I will show that there exists an absolute constant $C$ such that $$ |e^{t\Delta}u(0)| \leq C t^{-(n+d)/2} M_d. $$ To see this, we write $$ e^{t\Delta}u(0) = (4\pi t)^{-n/2} \int e^{-|y|^2/4t} u(y)\,dy. $$ Now we expand $e^{-|y|^2/4t}$ using the Taylor series to obtain $$ e^{-|y|^2/4t} = p_t(y) + O(t^{-d/2} |y|^d) $$ where $p_t$ is the Taylor polynomial of degree $d-1$, and $t^{-d/2}$ appears as it bounds the $C^d$ norm of $e^{-|y|^2/4t}$. In particular, $\int p_t(y) u(y)\,dy = 0$. Therefore $$ |e^{t\Delta}u(0)| \leq C t^{-(n+d)/2} \int |y|^d |u(y)|\,dy \leq C t^{-(n+d)/2} M_d, $$ as desired.

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