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Is there a Schwartz function $\zeta(t)$, defined on $\mathbb{R}$, satisfying the following:

  1. $\int \zeta(t)\: dt=1$,
  2. $\int t^k \zeta(t)\: dt=0$ for all $k\geq 1$,
  3. $\operatorname{supp}(\zeta)\subset (0,\infty)$.

If we drop the third property, this is easy! Just let $\zeta$ be the Fourier transform of a Schwartz function which equals $1$ on a neighborhood of $0$.

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  • $\begingroup$ $supp (\zeta) \subset (0, \infty)$ more or less means that $\zeta$ has an analytic continuation to the lower half-plane plus the convergence of the logarithmic integral. We can probably get an analytic function satisfying all these conditions, so the answer must be yes. $\endgroup$
    – Aleksei Kulikov
    Commented Jan 3 at 18:39
  • $\begingroup$ The slightly weaker assignment where we replace (3) by $\textrm{supp }\zeta\subseteq [0,\infty)$ is equivalent to finding a function $f=\widehat{\zeta}$ satisfying $f\in H^2\cap\mathcal S$, $f(0)=1$, $f^{(k)}(0)=0$. This looks potentially difficult to me, as $H^2$ functions are determined by their values on any positive measure subset of $\mathbb R$. For example, there is no such function that is $=1$ near zero. $\endgroup$
    – Christian Remling
    Commented Jan 3 at 20:23
  • $\begingroup$ One can also try to use moment problems, perhaps as follows: start out with a $g\ge 0$ with indeterminate moments $m_n=\int_0^{\infty} x^n g(x)\, dx$. Then the Stieltjes moment problem $m_0+1, m_1,m_2,\ldots$ still has solutions (obvious from the standard criteria) and should still be indeterminate (not clear to me right now how to prove it, but it feels right). In particular, there are absolutely continuous solutions $h\, dx$, and then we can take $\zeta=h-g$. This satisfies (1), (2), $\textrm{supp }\zeta\subseteq[0,\infty)$, but is perhaps not in $\mathcal S$. $\endgroup$
    – Christian Remling
    Commented Jan 4 at 3:33
  • $\begingroup$ The known conditions on Stieltjes moment problem do not ensure the existence of a measure with given moments and having a density which belongs to the Schwartz space. $\endgroup$
    – Christophe Leuridan
    Commented Jan 4 at 8:08
  • $\begingroup$ @ChristopheLeuridan: Yes, I know that this is not obvious (as I stated at the end of my long comment), but on the other hand the space of solutions is huge (Nevanlinna parametrization) and one can perhaps take advantage of this to produce a Schwartz solution. $\endgroup$
    – Christian Remling
    Commented Jan 4 at 14:40

1 Answer 1

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The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious.

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too.

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F_I=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular, $$ \int_{\mathbb R} G(t) \, dt = 1,\quad \int_{\mathbb R}G(t)t^{2k} \, dt = 0 \text{ for } k\ge 1 $$

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take $$ 2G(t)=2t H(t^2) $$ on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.

Edit: By Alexei's request, here is the proof that $PQ$ is contracting. We have $\|PQ\|^2=\|PQP\|$. The operator $PQP$ has the kernel $\chi_I(x)\chi_I(x')\int_Ie^{2\pi i(x-x')y}\,dy$, which is supported on $I\times I$ and is bounded by $I$ in absolute value. Thus, by the HS bound, or by the Schur test, we have $\|PQP\|\le |I|^2$, i.e., $\|PQ\|\le |I|$.

It is worth noting that it is actually contracting for an interval $I$ of arbitrary length or even for a set $I$ of finite Lebesgue measure, but the proofs of those facts are not so simple.

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  • $\begingroup$ Nice argument! I had something quite different and way more technical in mind (cooking a function in the unit disk and then translating it to the upper half-plane). But I think you should give some explanation/reference to the fact that $PQ$ is contractive, while it is quite old I don't think it is that well-known (I assume you have a Hilbert--Schmidt norm argument in mind?) $\endgroup$
    – Aleksei Kulikov
    Commented Jan 4 at 16:36
  • $\begingroup$ @AlekseiKulikov OK. Yeah, for short intervals it is a trivial HS-norm bound. I'll add it a bit later :-) $\endgroup$
    – fedja
    Commented Jan 4 at 17:06
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    $\begingroup$ This is too funny!! As I was writing my comment on your answer to the other question, I was thinking to myself, this question here is probably of that type too... (and I hadn't seen your answer here yet) $\endgroup$
    – Christian Remling
    Commented Jan 4 at 17:08
  • $\begingroup$ Everything is very nice. Where is a proof of the contractivity of $PQ$ for long intervals? $\endgroup$
    – Giorgio Metafune
    Commented Jan 4 at 18:23
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    $\begingroup$ @GiorgioMetafune For you I can tell it in a couple of lines. $PQP$ is a self-adjoint compact (HS) positive definite operator of norm at most $1$. The eigenvalue $1$ is impossible because the eigenvector would be a function with bounded support and spectrum simultaneously. Hence, the norm is strictly less than $1$. If you are interested in quantitative bounds, look at the book by Havin and Joricke "The Uncertainty Principle in Harmonic Analysis" for proofs and references. $\endgroup$
    – fedja
    Commented Jan 4 at 18:33

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