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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 31, 2021 at 14:18
  • $\begingroup$ @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 $\endgroup$
    – Jay
    Jan 31, 2021 at 18:28
  • $\begingroup$ @MateuszKwaśnicki By the way, why do you mention the gluing of $0$ and $\infty$? $\endgroup$
    – Jay
    Feb 2, 2021 at 10:41

1 Answer 1

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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.

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