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Added. My question in the title was solved (in the negative) by Nik Weaver (in the answer below) and Mateusz Kwaśnicki (in the comments). In both solutions, the reason is that the $L^2$ density fails even on the smaller interval $[1/3,1]$, where the function $\{1/t\}$ is piecewise glued by $1/t - 2$ (from $1/3$ to $1/2$) and $1/t - 1$ (from $1/2$ to $1$). In view of this feature that I had neglected (existence of an obstruction already on some compact subsegment $[c,1] \subset (0,1]$), and also in view of my own example below the line, I still wanted to raise the follow-up in this linked question.

The original question. (Which turned out easy.)

The question of the title:

Is $\{1/t\}^k$, $k = 0, 1, \ldots$, an $L^2$ spanning set on the interval $(0,1)$?

Background. For amusement, let me include some background on what leads me to raise this kind of question (besides idle curiosity). The following has no strictly mathematical bearing on the actual question above, which could for all I know turn out to be easy and unrelated to RH.

I had an observation some time ago that the Riemann Hypothesis implies (and is reversely implied by; which is the trivial part from the viewpoint of the Mellin transform) the $L^2$ density of the linear span of the countable sequence of functions $$ \mathcal{B}: \quad t^{d}\big(B_{d+1}(\{1/t\}) - B_{d+1}(1/t)\big) - t^{d-1}\big(B_d(\{1/t\}) - B_d(1/t)\big), \quad d = 1, 2, \ldots. $$ Here $B_m(x)$ are the Bernoulli polynomials; so this (RH conditional) $L^2$ spanning set consists of certain polynomials in $\{1/t\}$ and $t$. Observe that the Mellin transform of this linear space is equal to $$ \{ p(s) \zeta(s) \big/ s(s+1) \cdots (s+N) \quad \mid \quad p(s) \in \mathbb{R}[s], \, \deg{p} \leq N; \quad N = 0, 1, \ldots \}. $$

I was somehow led to think that this should probably be straightforward to prove for anyone who cared enough about such an implication. (I could be wrong to think that. Anyhow it may hopefully be appropriate to leave this statement here as an exercise "for the interested reader.") I did however find this to be quite different in at least one regard than the classical Nyman-Beurling result, which expressed RH as the question of the $L^2$ density of linear span of the continuum of functions $$ \mathcal{N}: \quad \Big\{ \alpha \{1/t\} - \{ \alpha / t\} \mid \alpha \in [0,1] \Big\}. $$ In either question, the full problem reduces to expressing the constant function $1$ as an $L^2$-limit of linear combinations from the designated set of functions. But while in Nyman-Beurling one knows that $1$ is in the $L^2$ closure of span of $\mathcal{N}$ if and only if it is already in the $L^2$ closure of the linear subspace spanned by the countable sequence $\alpha = 1, 1/2, 1/3, \ldots$, the latter countable sequence is known to not be a topological spanning set for the full $L^2(0,1)$; indeed it has a rather small closure. In contrast, under RH, the Bernoulli sequence $\mathcal{B}$ is a countable spanning set for all of $L^2(0,1)$.

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    $\begingroup$ It seems like what we need here is the multiplicative system lemma, which is sort of the "measurable" version of Stone-Weierstrass. I think the only nontrivial part is showing that the family of functions $\{1/t\}^k$ generates the Borel $\sigma$-algebra, which corresponds roughly to "separates points". $\endgroup$
    – Nate Eldredge
    Commented Apr 23, 2019 at 22:38
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    $\begingroup$ I may be missing something obvious here: how can one expect $g(t)=f(\{1/t\})$ to be $L^2$-close to, say, $h(t)=\mathbb{1}_{(1/2,1)}(t)$? If $\|g-h\|_2<\delta$, then, by looking at $t\in (1/2,1)$, we have $\|f-1\|_2<C_1\delta$, while by looking at $t\in (1/3,1/2)$ we have $\|f-0\|_2<C_2\delta$, which leads to a contradiction if $\delta>0$ is small enough. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 23, 2019 at 22:58
  • $\begingroup$ @MateuszKwaśnicki: It looks like you are right. Thanks! The characteristic function of [0.5,1] shouldn't be a limit, for the reason you point out, that the set of values on $[1/2,1]$ or $[1/3,1/2]$ are the same. We should have a more interesting sequence of polynomials in $\{1/t\}$ and $t$, like in the conditional observation below the line. (How about, say, $t^k \{1/t\}^k$, or $t^k + \{1/t\}^k$?) $\endgroup$
    – Vesselin Dimitrov
    Commented Apr 23, 2019 at 23:09
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    $\begingroup$ Now following up on my previous comment in view of Nik's answer, this family of functions evidently does not separate points, neither literally nor in an a.e.-sense, for the basic reason that $\{1/t\}$ is (in an essential way) not one-to-one. $\endgroup$
    – Nate Eldredge
    Commented Apr 24, 2019 at 0:03

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They are not. The function $g(t) = \begin{cases}\frac{-1}{(1-t)^2}& \frac{1}{3} < t < \frac{1}{2}\cr 1& \frac{1}{2} < t < 1\end{cases}$ is orthogonal to all of them. That is because $\{\frac{1}{t}\} = \frac{1}{t} - 2$ on the first interval and $\frac{1}{t} - 1$ on the second. So integrating $g$ against $\{\frac{1}{t}\}^k$ on $[\frac{1}{3}, \frac{1}{2}]$ yields $-\int_{1/3}^{1/2} (\frac{1}{t} - 2)^k \frac{dt}{(1-t)^2}$, and integrating the product on $[\frac{1}{2}, 1]$ yields $\int_{1/2}^1 (\frac{1}{t} - 1)^k dt$. A simple change of variable with $s = \frac{t}{1-t}$ (so $\frac{1}{s} = \frac{1}{t} - 1$) turns the first integral into minus the second, so their sum is zero.

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  • $\begingroup$ Thank you for this clean answer. $\endgroup$
    – Vesselin Dimitrov
    Commented Apr 23, 2019 at 23:11
  • $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Commented Apr 23, 2019 at 23:14
  • $\begingroup$ By the way, I wonder if this kind of argument would also answer my follow-up question about $t^k \{1/t\}^k$ or $t^k + \{1/t\}^k$ (or other simple families of polynomials in $t$ and $\{1/t\}$). $\endgroup$
    – Vesselin Dimitrov
    Commented Apr 23, 2019 at 23:18
  • $\begingroup$ It does apply to $t^k\{1/t\}^k$, I'll have to think about the second one. $\endgroup$
    – Nik Weaver
    Commented Apr 23, 2019 at 23:26

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