Subspaces of $L^2(0,1)$ dense on every truncation $L^2(c,1)$

Question

It may be better to move this to a separate question.

Let me call a linear subspace $V \subset L^2(0,1)$ to be tame if, for every linear subspace $W \subset V$, either $W$ is dense in $L^2(0,1)$, or else there exists a $c > 0$ such that already the restriction $W|_{[c,1)}$ is not dense in $L^2(c,1)$. (We think of $0$ as a ''point at infinity,'' cf. the example in (ii*) below; it is a matter of convention whether to formulate it this way or for $L^2(0,\infty)$ and its finite truncations $L^2(0,T)$.) It is not a priori clear to me whether there is any tame and dense subspace of $L^2(0,1)$, nor indeed that this is a reasonable notion.

In relation to this question and its solutions, I wonder:

Question. (i) Is the space of polynomials tame in $L^2(0,1)$? Or is there another explicit example of a tame and dense subspace of $L^2(0,1)$?

(ii*) Is the space $$ V := \big\{ P(t,\{ 1/t \}) \mid P \in \mathbb{R}[x,y] \big\} \subset L^2(0,1) $$ tame? (Here, $\{\cdot\}$ denotes the fractional part function.)

A motivation for asking this is that, as described in the linked question, the Riemann Hypothesis can be expressed as an $L^2(0,1)$ density of the particular linear subspace $$ \mathcal{B} := \big\{ \sum_d c_d \cdot t^d (B_{d+1}(\{1/t\}) - B_{d+1}(1/t)) \quad \sum_d c_d = 0 \big\} $$ of the space $V$ in (ii*). (To give an indication of why the $L^2$ density of this latter space is equivalent to the RH, observe that the Mellin transform of the space is equal to the multiples of the Riemann zeta function by the rational functions of the form $p(s) / s(s+1) \cdots (s+N)$ with $\deg{p} \leq N$; and that $t^{\rho-1} \in L^2(0,1)$ if and only if $\mathrm{Re}(\rho) > 1/2$ strictly.)

Regarding (i), observe at least that by the Müntz–Szász theorem, the condition to be checked is fulfilled for the particular case that $W$ is the span of a subset of the monomials.

Regarding (ii*), it can be shown that $\mathcal{B}|_{[c,1)}$ is $L^2(c,1)$-dense for every $c > 0$ and hence a positive answer would imply RH. That made me to wonder about constructing any subspace $W \subset V$ of the space $V$ in (ii*) such that $W|_{(c,1)}$ is $L^2$-dense on every truncation $(c,1)$ but $W$ is not $L^2$-dense on the full segment $(0,1)$.

Lastly, the example of $\mathcal{B}$ (resp. its restriction to $d \geq 3$) shows easily that the answer in (ii*) certainly becomes negative if the Hilbert space $L^2(0,1)$ gets replaced by any of $p > 2$ Banach spaces $L^p(0,1)$, or by $C[0,1]$. The same questions may be asked also for $L^1(0,1)$, a variant which is perhaps more plausible but without interesting implications.

Answer 1

It is not a priori clear to me whether there is any tame and dense subspace of $L^2(0,1)$.

Indeed, no such space exists. To see it, choose any sequence of functions $f_k\in L^2([0,1])$ such that $f_k|_{[c,1]}$ are dense in $L^2([c,1])$ for all $c>0$ but $\int_0^1 f_k=0$ for all $k$. Now if $V$ is dense in $L^2([0,1])$, then we can find $g_k\in V$ such that $\|g_k-f_k\|_{L^2([0,1])}<\frac 1k$ and $\int_0^1 g_k=0$ (just approximate $f_k\pm \frac 1{3k}$ with precision $\frac 1{6k}$ and take an appropriate convex combination of these 2 approximations). Then the linear span of $g_k$ is dense in any $L^2([c,1])$ but not in $L^2([0,1])$.

Unfortunately, I cannot offer an equally trivial proof of RH :-)