If a linear operator commutes with right shift in $\ell^p$, then it is continuous I'm studying for a PHD qualifying exam and I got stuck with this problem:

Let $p\ge 1$ and $l^p:= L^p(\mathbb{N},\mu)$, where $\mu$ is the counting measure. Let $S:\ell^p\rightarrow \ell^p$ the right shift operator defined by $$S(x_1,x_2,\cdots)=(0,x_1,x_2,\cdots)$$ and suppose that the linear operator $T:\ell^p\rightarrow\ell^p$ satisfies $TS=ST$. Prove that $T$ is continuous.

I tried the following approach:
First I defined $\{e_j\}_{j=1}^\infty$ as the subset of $l^p$ given by $e_j(n)=\delta_{jn}$, where $\delta_{jn}=1$ if $j=n$ and $\delta_{jn}=0$ otherwise. Noticing that $S^m e_j=e_{j+m}$ I managed to prove $||T(e_{j+1})||_p\le ||T(e_1)||_p$ using the fact that $ST=TS$. I chose to name $C:=||T(e_1)||_p$. Then I let $||x||=1$ and tried to prove $||Tx||_p \le K$ for some constant $K$ by doing the following:
$$||Tx||_p=||T(\sum_{n=1}^\infty x_n e_n)||_p = ||\sum_{n=1}^N T(x_ne_n)+T\sum_{n=N+1}^\infty x_n e_n||_p\le ||\sum_{n=1}^N x_n T(e_n)||_p+||T\sum_{n=N+1}^\infty x_ne_n||_p\le \sum_{n=1}^N  ||x_n T(e_n)||_p+||T\sum_{n=N+1}^\infty x_n e_n||_p \le \sum_{n=1}^n |x_n| ||T(e_n)||_p+||T\sum_{n=N+1}^\infty x_n e_n||_p\le \sum_{n=1}^N C|x_n|+ ||T \sum_{n=N+1}^\infty x_n e_n ||_p =C\sum_{n=1}^N |x_n|+ ||T\sum_{n=N+1}^\infty x_n e_n ||_p =C\sum_{n=1}^N |x_n|+||T\sum_{n=N+1}^\infty x_n e_n||_p. $$
Then, if I let $N\rightarrow \infty$ I get $||Tx||_p\le C||x||_1$. This works if $p=1$, but fails for greater values of $p$ since I have no guarantee that $x\in \ell^1$ if $p>1$. 
I looked for books and articles about linear operators that commute with the shift operator but I only found papers with several lemmas that I couldn't figure out while taking the test with a time limit. I Also tried to generalize the previous discussion by placing $p$ in different parts of the expresions, but I didn't get anything useful. The last approach I tried was noticing that if $L$ is the left shift operator we get $T=LTS$, also $||T||$ and $||S||=1$. Sadly, I didn't get anything nice with that.
I'm looking for a hint that allows me to generalize my proof for $p\ge 1$ or a totally different solution or hint to solve it. The problem isn't clear about if I need to prove it for $\ell^\infty$, but I'll be happy if I understand the finite case. Any help is appreciated.
 A: Here is a proof, taken from ''Continuity of linear operators commuting with shifts'' by Richard J Loy, Journal of Functional Analysis
Volume 17, Issue 1, September 1974, Pages 48-60, http://dx.doi.org/10.1016/0022-1236(74)90003-2
For $x\in l^p$, denote by $x|_n:=(x_1,\dots,x_n,0,0,\dots)$ the restriction to the first $n$ elements of $x$. Define 
$$
T_nx:= (Tx)|_n \quad x\in l^p, \ n\in \mathbb N.
$$
Then $T_nx\to Tx$ for all $x$, and if all the $T_n$ are continuous, it follows that $T$ is continuous as well. Assume that for some $k$ the operator $T_k$ is not continuous.
Then we can choose a sequence $(x_n)$ inductively such that $\|x_n\|\le 2^{-n}$ and
$$
\|T_kx_n\|\ge n + \sum_{i=1}^{n-1}\|Tx_i\|.
$$
Define $y:=\sum_{n=1}^\infty S^{2kn} x_n$, and $Q_m:=T_{(2m+1)k}$. Then it holds
$$
Q_m (S^{2km} x_m) = (TS^{2km} x_m)|_{(2m+1)k} = (S^{2km} Tx_m)|_{(2m+1)k} = 
 S^{2km} T_kx_m
$$
since $(S^iTx)|_{i+j}=S^iT_jx$.
Moreover it holds with $z=\sum_{n=m+1}^\infty S^{2k(n-m-1)}x_n$
$$
Q_m\left(\sum_{n=m+1}^\infty S^{2kn}x_n \right)
=Q_m S^{2k(m+1)}z 
=T_{(2m+1)k} S^{2k(m+1)}z = \left(S^{2k(m+1)}T z\right)_{(2m+1)k}=0
$$
since $2k(m+1)=2km+2k>(2m+1)k$.
Then it follows
$$
\|Ty\| \ge \|Q_my\| = \left\| Q_m \left(\sum_{n=1}^m S^{2kn}x_n
\right)\right\|\\
\ge \|Q_m S^{km}x_m\| - \sum_{n=1}^{m-1}\|Q_mS^{2kn}x_n\|
\ge \|T_kx_m\| - \sum_{n=1}^{m-1}\|Tx_n\| \ge m
$$
for all $m$, which is impossible.
[Hope this makes sense.]
As you can see from this proof, the principal idea is to consider truncated sequences in the image space (not in the domain of $T$).
