Consider the operator $T:\ell^\infty({\mathbb N})\to\ell^\infty({\mathbb N})$ defined by
$$
(Tx)\_m=\sum\_{k=m+1}^\infty p\_{k,m} \ \ x\_k,
$$
where
$$
p\_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}.
$$
Then $T$ is a bounded operator of norm $\zeta(2)=\frac{\pi^2}6$ as an easy calculation shows.
I need to know the dimension of the eigenspace to the eigenvalue $1$, i.e.
$$
E=\{x\in \ell^\infty: Tx=x\}.
$$
Ideally, I would like to have $\dim(E)=1$, however, I already don't know whether this space is finite-dimensional.
So my question is, what is the dimension of the space $E$?



Eigenspace of a specific operator
Theorem.
Let $T:\ell^\infty\to\ell^\infty$ be defined by
$$
(Tx)\_m=\sum\_{k=m+1}^\infty p\_{k,m}x\_k,
$$
where $p\_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}$.
Let $L=\{x\in\ell^\infty: Tx=x\}$.
Then $\dim L= 1$.


We first show $\dim L\ge 1$.


Lemma.
The element $\alpha$ with $\alpha\_1 = 1$ and
\begin{align\*}
\alpha\_k &= (1-k)\int\_0^1 \frac{t^{k-1}}{\log(1-t)} dt\quad\text{for } k = 2,3,\dots
\end{align\*}
lies in $L$.


For $m=1$ we compute
\begin{align\*}
(T\alpha)\_1&=\sum\_{k=2}^\infty \frac 1{(k-1)^2}\alpha\_k\\
&=-\sum\_{k=2}^\infty \frac1{k-1}\int\_0^1\frac{t^{k-1}}{\log(1-t)}\,dt\\
&=-\int\_0^1\sum\_{k=2}^\infty \frac{t^{k-1}}{k-1}\frac{1}{\log(1-t)}\,dt=\int\_0^11\,dt=1=\alpha\_1.
\end{align\*}
For $N\in{\mathbb N}$, $N\ge 2$ and $x\in\ell^\infty$ we get
\begin{align\*}
\sum\_{m=1}^N(Tx)\_m&=\sum\_{m=1}^N\sum\_{k={m+1}}^\infty\frac k{(k-1)(k-m)(k+1-m)}x\_k\\
&=\sum\_{k=2}^\infty\frac{kx\_k}{k-1}\sum\_{m=1}^{\min(N,k-1)}\frac1{(k-m)(k+1-m)}\\
&=\sum\_{k=2}^\infty\frac{kx\_k}{k-1}\sum\_{m=1}^{\min(N,k-1)}\left(\frac1{k-m}-\frac1{k+1-m}\right)\\
&=\sum\_{k=2}^\infty\frac{kx\_k}{k-1}\left(\frac1{k-\min(N,k-1)}-\frac1k\right)\\
&=\sum\_{k=2}^N\frac{kx\_k}{k-1}\left(1-\frac1k\right)+\sum\_{k=N+1}^\infty\frac{kx\_k}{k-1}\left(\frac1{k-N}-\frac1k\right)\\
&=\sum\_{k=2}^Nx\_k+N\sum\_{k=N+1}^\infty\frac{x\_k}{(k-1)(k-N)}.
\end{align\*}
Now assume we have shown $(T\alpha)\_j=\alpha\_j$ for $1\le j\le N-1$.
Then we get
$$
1+(T\alpha)\_N=\alpha\_N+N\sum\_{k=N+1}^\infty\frac{\alpha\_k}{(k-1)(k-N)}.
$$
We compute
\begin{align\*}
N\sum\_{k=N+1}^\infty\frac{\alpha\_k}{(k-1)(k-N)}
&=-N\sum\_{k=N+1}^\infty\frac{1}{(k-N)}
\int\_0^1 \frac{t^{k-1}}{\log(1-t)} dt\\
&=-N\int\_0^1\sum\_{k=N+1}^\infty\frac{t^{k-N}}{(k-N)}
 \frac{t^{N-1}}{\log(1-t)} dt\\
&= N\int\_0^1t^{N-1}\,dt=1.
\end{align\*}
The lemma follows.
$\square$


In order to show $\dim L\le 1$, let now $x\in L$ with $x\_1=0$. We need to show that $x=0$. 
Let $y\_k=\frac {k+1}{k}x\_{k+1}$.
Then $y\in\ell^\infty$ and $Tx=x$ as well as $x\_1=0$ lead to
\begin{align\*}
0&=\sum\_{k=1}^\infty\frac{y\_{k}}{k(k+1)},\\
\frac m{m+1}y\_m&=\sum\_{k=m+1}^\infty\frac{y\_k}{(k-m)(k-m+1)},\quad m\ge 1.
\end{align\*}
We have to show $y=0$.
For $m\in{\mathbb N}\_0$ let $z^{(m)}\in\ell^1$ be defined by
\begin{align\*}
z^{(0)}\_k&=\frac1{k(k+1)},\\
z^{(m)}\_k&=\begin{cases}0&k<m,\\
\frac m{m+1}&k=m,\\
\frac{-1}{(k-m)(k-m+1)}&k>m.
\end{cases}
\end{align\*}
Now $\ell^\infty$ is the dual space of $\ell^1$.
Let $<.,.>$ denote the duality pairing.
The formulas above say that
$$
<z^{(k)},y>=0\qquad\text{for }k=0,1,\dots
$$
Therefore the theorem follows if we can show that the sequence $z^{(0)},z^{(1)},\dots$ spans a dense subspace of $\ell^1$.
Let $Z$ denote the span of $z^{(0)},z^{(1)},\dots$. 


For $m\in{\mathbb N}$ let
$$
h^{(m)}\_k=\begin{cases}0&1\le k\le m-1,\\
\frac m{(k-m+1)(k+1)}&k\ge m. \end{cases}
$$
Then $h^{(1)}=z^{(0)}$ and a quick computation shows 
$h^{(m)}=h^{(m-1)}-z^{(m-1)}$ for $m\ge 2$.
Therefore, by induction we have $h^{(m)}\in Z$ for every $m$.
Next, let
$$
v^{(m)}=\frac1m h^{(m)}+z^{(m)}.
$$
then $v^{(m)}\in Z$ for every $m\in{\mathbb N}$ and
$$
v^{(m)}\_k=\begin{cases}0&1\le k\le m-1,\\
1& k=m,\\
\frac{-(m+1)}{(k+1)(k-m)(k-m+1)}&k\ge m+1.\end{cases}
$$
Let $n\in{\mathbb N}$ and let $e\_n$ be the $n$-th standard basis element of $\ell^1$, i.e., $e\_n=(0,\dots,0,1,0,\dots)$ with the $1$ at the $n$-th position.
We show that $e\_n$ lies in the closure of the span of $v^{(n)},v^{(n+1)},\dots$.
For this purpose let $a^{(n)}=v^{(n)}$ and for $m>n$ let $a^{(m)}=a^{(m-1)}-a^{(m-1)}\_mv^{(m)}$.
Then $a^{(m)}\in Z$ and 
$$
a^{(m)}\_k=\begin{cases}0&1\le k\le m,\ k\ne n,\\
1 &k=n,\\
a^{(m-1)}\_k-a\_m^{(m-1)}v^{(m)}\_k&k\ge m+1.\end{cases}
$$
Iterating the recursive relation yields for $k>m$,
\begin{align\*}
a^{(m)}\_k&=a^{(n)}\_k-\sum\_{j=n}^{m-1}a^{(j)}\_{j+1}v\_k^{(j+1)}\\
&=\frac{-(n+1)}{(k+1)(k-n)(k-n+1)}
+\sum\_{j=n}^{m-1}a\_{j+1}^{(j)}\frac{j+2}{(k+1)(k-j-1)(k-j)}.
\end{align\*}
For $m\ge n$ set $b\_m=a^{(m)}\_{m+1}$, then in particular,
$$
b\_m=\frac{-(n+1)}{(m+2)(m-n+1)(m-n+2)}+\sum\_{j=n}^{m-1}\frac{b\_j(j+2)}{(m+2)(m-j)(m-j+1)}.
$$
 Lemma.
For every $m\ge n$ we have $|b\_m|\le \frac{n+1}{m+2}$.


Proof of the lemma.
For $m=n$ we have
\begin{align\*}
|b\_m|=\frac{n+1}{n+2}\frac12\le \frac{n+1}{n+2}=\frac{n+1}{m+2}.
\end{align\*}
For the induction step we assume $m>n$ and the claim proven for smaller indices. Then
\begin{align\*}
|b\_m|&\le \frac{n+1}{(m+2)(m-n+1)(m-n+2)}+\sum\_{j=n}^{m-1}\frac{|b\_j|(j+2)}{(m+2)(m-j)(m-j+1)}\\
&\le \frac{n+1}{m+2}\left(\frac1{m-n+1}+\sum\_{j=n}^{m-1}\frac1{m-j)(m-j+1)}\right)\\
&= \frac{n+1}{m+2}\left(\frac1{m-n+1}+1-\frac1{m-n+1}\right)=\frac{n+1}{m+2}.
\end{align\*}
$\square$


One has
\begin{align\*}
\parallel{e\_n-a^{(m)}}\parallel\_1&=\sum\_{k=m+1}^\infty |a\_k^{(m)}|\\
&\le \sum\_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)}\\
&\ \ \ +\sum\_{k=m+1}^\infty \sum\_{j=n}^{m-1}|b\_j| \frac{j+2}{(k+1)(k-j-1)(k-j)}.
\end{align\*}
We estimate the first sum
\begin{align\*}
\sum\_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)}
&\le \frac{n+1}{m+2}\sum\_{k=m+1}^\infty \frac1{(k-n)(k+1-n)}\\
&= \frac{n+1}{m+2}\frac1{m+1-n}=A(m,n)\to 0,\quad m\to\infty.
\end{align\*}
For the second sum we have
\begin{align\*}
\sum\_{k=m+1}^\infty \sum\_{j=n}^{m-1}|b\_j| \frac{j+2}{(k+1)(k-j-1)(k-j)}
&\le \sum\_{k=m+1}^\infty \sum\_{j=n}^{m-1} \frac{n+1}{(k+1)(k-j-1)(k-j)}\\
&=\sum\_{k=m+1}^\infty \frac{n+1}{k+1}\left(\frac1{k-m}-\frac1{k-n}\right),
\end{align\*}
which tends to zero as $m\to\infty$.
We have shown that
$$ \parallel{e\_n-a^{(m)}}\parallel\_1\to 0
$$ as $m\to\infty$. The theorem follows.
$\square$


