Example for $\phi$

Question

Can some one give an example for $\phi$ in the
The answaer of following question ?Existence of smooth extension of a function defined on a closed interval

Answer 1

I tried for quite some time to produce a reasonably explicit and simple construction, but I think it is actually very hard. Indeed, the problem is very closely related to the Stieltjes Moment Problem, which seeks for a smooth, rapidly decaying functions vanishing on $(-\infty, 0]$ that has a given sequence of moments. A general existence result can be found here: https://www.ams.org/journals/proc/1989-107-03/S0002-9939-1989-0984787-0/S0002-9939-1989-0984787-0.pdf

A quite elaborate construction of a function satisfying your requirements can be found in Section 4.2 here: http://www.math.uconn.edu/~rogers/Preprints/thesis.pdf

The closest I came to a simple and explicit construction is the following classical example. It does not really satisfy all your requested properties: The following will be a construction of a rapidly decaying function $\phi$ such that $\phi \in C^\infty(\mathbb R)$, $\phi = 0$ on $(\infty, 0]$, $\int\phi(t)dt \neq 0$ and $$ \int t^k \phi(t) dt = 0 $$ for all $k \in \mathbb N$.

We set $\phi(t) = t^{-1}\sin(4 \pi \log(t)) e^{-\log^2(t)}$ if $t >0$ and $\phi(t) = 0$ otherwise.

Let us check the properties:

1. $\phi \in C^\infty$:

It is clear that $\phi\in C^\infty((-\infty, 0) \cup (0, \infty))$. Hence we simply need to check if all derivatives vanish for $t \to 0$. This is not hard to see since $e^{-\log^2(t)} = t^{-\log(t)}$.

2. $\phi$ is rapidly decreasing:

We have that $|t^k \phi(t)| \leq |t|^{k-\log(|t|)-1} \to 0$ for $|t| \to \infty$.

3. Orthogonality on monomials:

We have for $k \in \mathbb N$ that

\begin{eqnarray} \int t^k \phi(t) dt &=& \int_0^\infty t^k \sin(4 \pi \log(t)) e^{-\log^2(t)} dt/t\\ &=& \int_0^\infty e^{k \log(t)} \sin(4 \pi \log(t)) e^{-\log^2(t)} dt / t\\ &=& \int_0^\infty e^{-\left(\log(t) - \frac{k}{2}\right)^2} e^{\frac{k^2}{4}}\sin(4 \pi \log(t)) dt / t. \end{eqnarray} where we used the binomial formula in the last step. By substitution this yields that \begin{eqnarray} \int t^k \phi(t) dt &=& e^{\frac{k^2}{4}} \int e^{-u^2} \sin\left(4 \pi \left(u + \frac{k}{2}\right)\right) dt\\ &=& e^{\frac{k^2}{4}} \int e^{-u^2} \sin(4 \pi u) dt, \end{eqnarray} where the last step is due to the $2\pi$ periodicity of $\sin$. Since $e^{-u^2} \sin(4 \pi u)$ is an odd function, we can conclude that the integral is zero.