*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](https://mathoverflow.net/questions/329811/subspaces-of-l20-1-dense-on-every-truncation-l2c-1?noredirect=1&lq=1).




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*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 $(0,1)$?*




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*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 $L^2(0,1)$. 



Are the polynomials in $\{1/t\}$ dense in $L^2(0,1)$?
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.


