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$T$ 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?

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This is a pretty simple question, but what does $ran$ stand for? I saw it referenced in a geometry paper as well... – Patch Mar 17 '12 at 20:20
@Patch The range of the operator. – David Mitra Mar 17 '12 at 20:38
up vote 2 down vote accepted

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 $$ \Vert e_1-x\Vert_{\ell_2} ={\sqrt{n+1}\over n}\ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ 0. $$

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Yes,I think it's correct. Thank you very much. I haven't thought of proving a subspace is dense by showing its closure contains all e1,e2,e3,... – Long Mai Mar 17 '12 at 22:54

You can use the adjoint. Equivalently, you can prove that Ker(I-T*) is equal to {0}.

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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 $y=0$.

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This is similar to the Bostons answer – Norbert Mar 17 '12 at 18:55
Very nice. A good reminder that when trying to prove a subspace of a Hilbert space is dense, one should consider proving that the orthogonal complement is zero. – Nate Eldredge Mar 17 '12 at 23:21

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.

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I don't understand why the distance goes to 0, as n goes to infinity. Given ($\alpha_1, \alpha_2,\alpha_3,..)=(\beta_1, \beta_1+ \beta_2,...,\beta_1+\beta_2+...+\beta_n,0,0,0,0...)$, the distance = ($\beta_1+\beta_2+...+\beta_n)^2$ + $\beta_{n+1}^2+\beta_{n+2}^2+...$. Why does the first part ($\beta_1+\beta_2+...+\beta_n)^2$ goes to 0 as n goes to 0? – Long Mai Mar 17 '12 at 17:10
The distance between $(\beta_1,\beta_2,\beta_3,\ldots,\beta_n,\beta_{n+1},\beta_{n+2},\ldots)$ and $(\alpha_1,\alpha_2-\alpha_1,\alpha_3-\alpha_2,\ldots,\alpha_n-\alpha_{n-1},0,0,‌​0,\ldots)$ is the distance between $(\beta_1,\beta_2,\beta_3,\ldots,\beta_n,\beta_{n+1},\beta_{n+2},\ldots)$ and $(\beta_1,\beta_2,\beta_3,\ldots,\beta_n,0,0,0,0,\ldots)$. That distance is $|\beta_{n+1}|^2+|\beta_{n+2}|^2+|\beta_{n+3}|^2+\cdots$. That distance approaches $0$ as $n\to\infty$ because $\beta\in\ell^2$. – Michael Hardy Mar 17 '12 at 22:12
OK, I think I see what you're saying: We want $(\alpha_1,\alpha_2-\alpha_1,\alpha_3-\alpha_2,\ldots,\alpha_n-\alpha_{n-1},0-\a‌​lpha_n,0,0,0,\ldots)$. – Michael Hardy Mar 17 '12 at 22:17

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