Surjectivity of $T: l_2 \to l_2$ defined as $(x_n)_{n=1}^\infty \mapsto (x_n+x_{n+1})_{n=1}^\infty$

Question

\begin{aligned} T: l_{2}(\mathbb{C}) &\longrightarrow l_{2}(\mathbb{C}) \\ \left(x_{n}\right)_{n=1}^{\infty} &\longmapsto\left(x_{n}+x_{n+1}\right)_{n=1}^{\infty} \end{aligned}

Any element of $l_2$ can be uniquely expressed in the basis $\{e_n\}_{n=1}^{\infty}$:

$$ \left(x_{n}\right)_{n=1}^{\infty}=\lim _{k \rightarrow \infty} \sum_{n=1}^{k} x_{n} e_{n}, \quad e_{n}=\left(\delta_{n_{j}}\right)_{j=1}^{\infty} \\ $$

I found these $e_n$ in Im$(T)$:

\begin{aligned} v_{1} & =(1,0,0,0, \ldots) & \longmapsto & &(1,0,0,0, \ldots) \\ v_{2} & =(-1,1,0,0, \ldots) & \longmapsto & &(0,1,0,0, \cdots) \\ v_{3} & =(1,-1,1,0, \ldots) & \longmapsto & &(0,0,1,0, \cdots) \\ \vdots\\ v_{n} & =\left((-1)^{n+1},(-1)^{n+2}, \ldots,(-1)^{n+n}, 0, \ldots\right) & \longmapsto & &(0, \ldots, 0,1,0, \ldots) \\ \vdots \end{aligned}

Now, for any $\left(y_{n}\right)_{n=1}^{\infty} \in l_2$,

$$ \sum_{n=1}^{\infty} y_{n} e_{n}=\sum_{n=1}^{\infty} y_{n} T\left(v_{n}\right)=T\left(\sum_{n=1}^{\infty} y_{n} v_{n}\right). $$ However, the limit \begin{equation} \lim _{k \rightarrow \infty} \sum_{n=1}^{k} y_{n} v_{n}=\left(y_{n}^{\prime}\right)_{n=1}^{\infty} \end{equation}

would have to exist, every element $y_{n}^{\prime}$ of the sequence,

\begin{equation} \begin{aligned} y_{1}^{\prime} & =y_{1}-y_{2}+y_{3}-y_{4} \cdots \\ y_{2}^{\prime} & =y_{2}-y_{3}+y_{4}-y_{5} \cdots \\ \vdots\\ \end{aligned} \end{equation}

would have to converge and, finally, that \begin{equation} ||\left(y_{n}^{\prime}\right)_{n=1}^{\infty}||=\sum_{n=1}^{\infty}\left|y_{n}^\prime\right|^{2}<\infty \end{equation}

I have arrived to the conclusion that $T$ is not surjective, but I am afraid I could be wrong. Also, as $T$ is inyective,there is an inverse that is again defined, for every element of the transformed sequence, with another alternating series of elements of the input sequence...

Thank you; I hope you find it enjoyable!

Answer 1

Throughout this posting I consider any element $\mathbf{x}\ell_2$ as a function $\mathbf{x}:\mathbb{N}\rightarrow\mathbb{R}$ such that $\sum_n|\mathbf{x}(n)|^2<\infty$. For each $n\in\mathbb{N}$, se denote by $\mathbf{e}_n=\mathbb{1}_{\{n\}}$, that is, the sequence such that $\mathbf{e}_n(m)=0$ if $m\neq n$ and $\mathbb{e}_n(n)=1$.

The map $T$ on $\ell_2$ is injective (as the OP mentioned) for if $T(\mathbf{y})=\mathbf{0}$, then $$\mathbf{y}(n)=-\mathbf{y}(n+1), \qquad n\in\mathbb{N}$$ Then $\mathbf{y}(n)=(-1)^n\mathbf{y}(1)$ which means that $\mathbf{y}(1)=0$ and so, $\mathbf{y}=\mathbf{0}$.

As in the OP, define $$\mathbf{v}_n(m)=(-1)^{n+m}\mathbb{1}_{\{1,\ldots,n\}}(m)$$ and notice that $T(\mathbf{v}_n)=\mathbf{e}_n$.

Suppose $T$ is surjective. Then $T$ admits a bounded inverse (consequence of the open map theorem). Now, for any $\mathbf{y}\in\ell_2$, we have that $$\mathbf{y}_n:=\sum^n_{k=1}\mathbf{y}(k)\mathbf{e}_k\xrightarrow{n\rightarrow\infty}\mathbf{y}\qquad\text{in}\quad \ell_2$$

Consequently, the sequence $$\sum^n_{k=1}\mathbf{y}(k)\mathbf{v}_k=T^{-1}\mathbf{y}_n$$ converges in $\ell_2$. In particular, $\big(T^{-1}\mathbf{y}_n(1):n\in\mathbb{N}\big)$ is a Cauchy sequence in $\mathbb{R}$.

Consider the sequence $\mathbf{x}(m)=\frac{(-1)^m}{m}$. Clearly $\mathbf{x}\in\ell_2$, but $$\Big(\sum^n_{k=1}\mathbf{x}(k)\mathbf{v}_k\Big)(1)=\sum^n_{k=1}\frac{1}{k}.$$ The sequence $n\mapsto\sum^n_{k=1}\frac{1}{k}$ however is not Cauchy in $\mathbb{R}$. Therefore, $T$ is not surjective!

Answer 2

One way is to indicate an element in $\ell^2,$ which does not belong to the range of $T.$ Another way is to observe that $0\in\sigma(T).$ Thus injectivity of $T$ implies that $T$ cannot be surjective. In order to prove that $0\in \sigma(T)$ it suffices to show that $0\in\sigma(T^*).$ For $v_z=\{z^n\}_{n=1}^\infty,$ $|z|<1,$ we have $T^*v_z=(1+z)v_z.$ Hence $1+z\in\sigma(T^*)$ for any $|z|<1.$ Since the set $\sigma(T^*)$ is closed then $1+z\in \sigma(T^*)$ for $|z|\le 1.$ In particular $0=1+(-1)\in \sigma(T^*).$