# Density property fractional heat kernel

Let us consider $$p_t^{(n+2)}(\tilde x) ,$$ where $$x = (x_1, \ldots, x_n) \in \mathbb R^n$$, $$\tilde x = (x_1, \ldots, x_n, 0, 0) \in \mathbb R^{n+2}$$, and $$p_t^{(n)}(x)$$ is the heat kernel for $$(-\Delta)^s$$ in $$n$$ dimensions.

It is well known that we can write $$p_t^{(n+2)}(\tilde x)= t^{-(n+2)/2s}k(|\tilde x|/t^{1/(2s)})$$. Is it true that the subalgebra generated by $$\Big(f_t(r)\Big)_{t>0} \text{ is dense in C_0(0,\infty)},$$ where $$f_t(r) =r^{n}k(r/t^{1/(2s)})$$?

I wanted to use the locally compact version of Stone-Weierstrass. But, since the expression of $$f_t$$ is not explicit, I don't know how to prove that the subalgebra separates points and vanishes nowhere. Also, I'm not sure if the fact that $$\tilde x=(x_1,…,x_n,0,0)$$ (i.e. the last two coordinates are fixed) is a problem or not.

• Doesn't the usual Stone–Weierstrass argument work here? Make your space compact by gluing $0$ and $\infty$, adjoin a constant function and you are good to go. Jan 31, 2021 at 14:18
• @MateuszKwaśnicki Yes, I wanted to use the locally compact version of Stone-Weierstrass en.wikipedia.org/wiki/…. But, since the expression of $f_t$ is not explicit, I don't know how to prove that the subalgebra separates points and vanishes nowhere. Also, I'm not sure if the fact that $\tilde{x} = (x_1, \dots, x_n,0,0)$ (i.e. the last two coordinates are fixed) is a problem or not
– Jay
Jan 31, 2021 at 18:28
• @MateuszKwaśnicki By the way, why do you mention the gluing of $0$ and $\infty$?
– Jay
Feb 2, 2021 at 10:41

The answer is affirmative, and it does not depend on any properties of $$k(x)$$.
Take any non-zero continuous function $$f$$ on $$(0, \infty)$$ such that $$f$$ goes to zero at $$0^+$$ and at $$\infty$$. Define $$f_k(x) = f(k x)$$. Then the family of functions $$\{f_k : k > 0\}$$ separates points of $$(0, \infty)$$: indeed, if $$f_k(x_1) = f_k(x_2)$$ for all $$k > 0$$, then $$f(x)$$ is a periodic function of $$\log x$$ (with period $$\log (x_1 / x_2)$$), which is of course not possible. Furthermore, for every $$x_0 > 0$$ there is $$k > 0$$ such that $$f_k(x_0) \ne 0$$.
Let $$K$$ be the one-point compactification of $$(0, \infty)$$ (obtained by glueing $$0$$ and $$\infty$$), and let $$\mathcal A$$ be the algebra generated by a constant $$1$$ and the family of functions $$f_k$$. Then $$\mathcal A$$ separates the points of $$K$$ and contains constants, and hence it is dense in $$C(K)$$.
Now every function in $$C(K)$$ can be written as $$g(x) + c$$ for $$g$$ continuous on $$(0, \infty)$$ and vanishing at $$0$$ and $$\infty$$, and it is easy to see that $$g$$ belongs to the algebra $$\mathcal A'$$ generated by the family of functions $$f_k$$. Thus, $$\mathcal A'$$ is dense in $$C_0((0, \infty))$$, as desired.