Show that $\operatorname{ran}(I-T)$ is dense in $\ell^2$, $T$ is a right shift operator $T$ is a right shift operator from $\ell^2 \to \ell^2$, $(\alpha_1, \alpha_2,\ldots)\mapsto (0,\alpha_1,\alpha_2,\ldots)$. I want to show that $\operatorname{ran}(I-T)$ is dense in $\ell^2$. 
Could anyone help me or give me a hint please?
 A: You can use the adjoint. Equivalently, you can prove that Ker(I-T*) is equal to {0}.
A: Let $D:=\operatorname{ran}(I-T)$, and $y\in D^{\perp}$. Let $e^{(n)}$ the sequence defined by $e^{(n)}_k=\begin{cases}1&\mbox{ if } n=k\\\  0&\mbox{ otherwise}\end{cases}$. For each $k$, $e^{(k)}-e^{(k+1)}\in D$ so $y_k-y_{k+1}=0$ and $y_k=C$ where $C$ is a constant. Since $y\in \ell^2$, $C=0$ so $y=0$.
A: Let $e_i$ be the standard $i^{\rm th}$ unit vector of $\ell_2$. Considering the images of the $e_i$ in $\ell_2$ under the map $I-T$, we see that the range of $I-T$ contains the set $\{ e_i-e_{i+1}\mid i=1,2,3,\ldots\,\}$. It will follow that the range of $I-T$ is dense in $\ell_2$ if we can show that for each $i$ and for each $\epsilon>0$, there is a vector $x\in\text{ran}(I-T)$ such that $\Vert e_i-x\Vert<\epsilon$. 
We show how this can be done for the vector $e_1$. It should then be apparent how 
prove that it can be done for the  other $e_i$.
Towards showing the statement is true for $e_1$, consider the vectors
$$
\eqalign{
& x_1=(\rlap{\phantom{-}1}\qquad,  \rlap{-1}\qquad, \rlap{\ \ \ \ 0}\qquad, \rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\ldots)\cr
& x_2=(\rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}\alpha_1}\qquad, \rlap{-\alpha_1}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\ldots)\cr
& x_3=(\rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad, \rlap{\phantom{-}\alpha_2}\qquad,\rlap{ {-}\alpha_2}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\ldots)\cr
& x_4=(\rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad, \rlap{\phantom{-}\alpha_3}\qquad,\rlap{ {-}\alpha_3}\qquad,\rlap{\ \ \ \ 0}\qquad,\rlap{\ \ \ \ 0}\qquad,\ldots)\cr
& x_5=(\rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad,  \rlap{\phantom{-}0}\qquad, \rlap{\phantom{-}\alpha_4}\qquad,\rlap{ {-}\alpha_4}\qquad,\rlap{\ \ \ \ 0}\qquad,\ldots)\cr

}
$$
$$
\vdots
$$
Note each $x_i$ is in the range of $I-T$.
Now, fix a positive integer $n$ and
set 
$ \alpha_i= 1-{i \over n}$,  for  $i=1$, $2$, $\ldots\,$, $n $.
Then the sum $x=x_1+x_2+x_3+\cdots+x_{n +1}$ is
$$ 
x=\bigl( 1 ,
\underbrace{{\textstyle-{{1\over n}}   ,-{{1\over n}},\ldots  ,-{{1\over n}}}}_{(n+1) \text{-terms}}, 0, 0, \ldots\bigr).
$$
We have
$$
\Vert e_1-x\Vert_{\ell_2} ={\sqrt{n+1}\over n}\ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ 0.
$$
A: You're trying to show that no matter how small $\varepsilon$ is, for every point $\beta=(\beta_1,\beta_2,\beta_3,\ldots)\in\ell^2$, there is some $\alpha=(\alpha_1,\alpha_2,\alpha_3,\ldots)\in\ell_2$ such that the $\ell^2$ distance between $\beta$ and $(\alpha_1,\alpha_2-\alpha_1,\alpha_3-\alpha_2,\alpha_4-\alpha_3,\ldots)$ is less than $\varepsilon$.
So I'll make a "first attempt", which is problematic, then refine it to make it right.
Let
$$\alpha_1=\beta_1,$$
$$\alpha_2 = \beta_1+\beta_2$$
$$\alpha_3=\beta_1+\beta_2+\beta_3$$
and so on.
This makes the distance $0$.
The problem is that we don't know that this $\alpha$ is in $\ell^2$.
So what we do is we truncate $\alpha$ after the first $n$ components, i.e. make all the later components $0$, and choose $n$ large enough to make the distance less than $\varepsilon$.
Later note: As noted in the comments, this answer isn't finished yet; it needs further work.
