Sequence with bounded sums

Question

We define the sum operator $T : \mathbf{a} \in \mathbb{R}^{\mathbb{N}} \longmapsto \big( \sum \limits_{k=0}^n a_k \big)_{n \ge 0}$ and for $\mathbf{a}$ bounded, $||\mathbf{a}||_{\infty} = \sup_k |a_k|$.

1. Does there exist a non null sequence $\mathbf{a} \in \mathbb{R}^{\mathbb{N}}$ such that for all $p \ge 0$, $T^p (\mathbf{a})$ is bounded ?

2. Given a sequence $(M_n)_{n \ge 0} \in \mathbb{R}_+^{\mathbb{N}}$, does there exist $\mathbf{a} \in \mathbb{R}^{\mathbb{N}}$ such that for all $p \in \mathbb{N}$, $M_p \le ||T^p(\mathbf{a})||_{\infty} < \infty$ ?

3. Does the set of sequences satisfying the previous condition contain an uncountable linearly independant family ?

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I wrote below a possible construction ; do not hesit to post other solutions which could be more abstract or simple.

Answer 1

There's a somewhat different construction for 2 based on the Prouhet–Thue-Morse sequence.

(For 3, given a sequence $(\mathbf a^i)_{i\geq 0}$ such that $T^p(\mathbf a^i)$ is bounded for each $p,i\geq 0,$ we can plug $$M_n=\max_{\lambda_0,\dots,\lambda_{n-1}\in[-n,n] }\left\|T^n\left(\sum_{i=0}^{n-1}\lambda_i\mathbf a^i\right)\right\|_{\infty}+1$$ into condition 2 to get a new sequence that is not a finite linear combination of the $\mathbf a^i.$)

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The $M_n$ can be taken to be positive integers. Define $A_0$ to be sequence of length $1$ consisting of the single number $M_0.$ Given $A_{n-1},$ define $A_n$ to be $A_{n-1}^{M_n}(-A_{n-1})^{M_n},$ that is, the concatenation of $M_n$ copies of $A_{n-1}$, and $M_n$ copies of the sequence obtained by negating each element of $A_{n-1}.$

We will take $\mathbf a$ to be the unique infinite sequence such that $A_n$ is a prefix of $\mathbf a$ for each $n.$ For example, if $M_n=1$ for all $n$ then we get $\mathbf a=1,-1,-1,1,-1,1,1,-1,\cdots,$ so $T(\mathbf a)=1,0,-1,0,-1,0,1,0,\cdots$ and $T^2(\mathbf a)=1,1,0,0,-1,-1,0,0,\cdots.$

We can define $T$ on finite sequences. By induction $T^p(A_n)$ ends with a zero for $1\leq p\leq n$ - the $A_{n-1}$ and $(-A_{n-1})$ parts cancel out. This means that for each $p\geq 1$ the sequence $T^p(\mathbf a)$ is an infinite concatenation of copies of the fixed sequence $T^p(A_p)$ and its negation $(-T^p(A_p)).$ (The choice of which of these is used each is a pattern similar to the Thue-Morse sequence but with repeats.) In particular, it's bounded. And $T^n(A_n)$ is non-negative, again by induction, so its sum is at least Its first element $M_0\geq 1.$ So the sum of $A_{n-1}^{M_n}$ is at least $M_n,$ giving $M_n\leq \|T^n(\mathbf a)\|_\infty<\infty.$

Answer 2

I provide here a positive answer for 2. (which implies 1. btw). We only need the following lemma :

Lemma : given $m \le n$ integers and $\mathbf{a} \in \mathbb{R}^{\mathbb{N}}$ such that the only non zero terms of $\mathbf{a}$ are among $a_m,...,a_{n}$, for all $p \in \mathbb{N}$ there exists a unique $\mathbf{b} \in \mathbb{R}^{\mathbb{N}}$ s.t. $T^p (\mathbf{b})=\mathbf{a}$, and it satisfies
a) the only non zero terms of $\mathbf{b}$ are $b_m,...,b_{n+p}$, $\quad \quad \quad$ b) $||\mathbf{b}||_{\infty} \le 2^p || \mathbf{a} ||_{\infty}$.

Proof : induction is done easily once we notice that $T^{-1}(\mathbf{a}) = (a_0,\ a_1-a_0,\ ..., a_{n+1}-a_n, ...)$

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We will define a solution $\mathbf{a}$ to 2. as a sum. Let $\mathbf{a}^0 = (M_0+M_1+1,0,0,0,...)$. With $p \ge 1$, assume we have defined $\mathbf{a}^0$, ..., $\mathbf{a}^{p-1}$ such that with $v = \sum \limits_{i=0}^{p-1} \mathbf{a}^i$, $T^{p-1} v$ is positive everywhere and zero after some index $n_0$, and $||T^{p-1} v||_{\infty} \ge M_{p-1}$, $\lim (T^p v) \ge M_p$ (this sequence is increasing, and constant after $n_0$). Note that for $p=0$, this is the case.

We can define $\mathbf{a}^p$ such that $T^{p-1} a_p$ is zero everywhere except for indices $n_0+p+1,...,n_0+p+k$ where its value is $-\frac{\lim (T^p v)}{k}$, with $k$ large enough. Let $w := v + \mathbf{a}^p$. It is crucial to note that for $0 \le i < p$, the non zero terms of $T^i v$ are left unchanged in $T^i w$ - there are only more finitely many terms after the previously non zero ones (see part a) of the lemma).

Then, if $k$ is big enough (see part b) of the lemma), for $0 \le i < p$, $||T^i w||_{\infty} = ||T^i v||_{\infty} \ge M_i$. Then for $j \le n_0+p$, $(T^p w)_j = (T^p v)_j$ ; for $1 \le j \le k$, $(T^p w)_{n_0+p+j} = \lim (T^p v) - i \cdot \frac{\lim T^p v)}{k}$ and for $j > n_0+p+k$, $(T^p w)_j = 0$. Hence $T^p w$ is positive and zero after some index and $||T^p w||_{\infty} \ge \lim(T^p v) \ge M_p$. Finally, for $j \ge 1$, $|(T^{p+1} w)_{n_0+p+k+j}| \ge \sum \limits_{i=1}^k \lim (T^p v) - i \cdot \frac{\lim (T^p v)}{k}$, so if $k$ is large enough, $||T^{p+1} w||_{\infty} \ge M_{p+1}$.

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Finally, note that $T^p (\sum \limits_{i \ge 0} \mathbf{a}^i ) = \sum \limits_{i \ge 0} T^p (\mathbf{a}^i)$, simply because $T^p (\mathbf{a}^i)$ for $i \ge 0$ do not overlap by construction. This enables us to conclude that with $\mathbf{a} := \sum \limits_{i \ge 0} \mathbf{a}^i$, for all $p$, $$||T^p \mathbf{a}||_{\infty} = ||T^p \mathbf{a}^p||_{\infty} \in [M_p,+\infty[.$$

Thus we can give a positive answer to 2.

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The redaction is approximate, but the construction is not really simple and I can't make it shorter while still justifying almost everything...