Why does this prove that the range of an operator is not dense? The operator given is the right-shift operator $T$ on $l^2$. We show that $\lambda=1$ is in the residual spectrum. Therefore we show that $(I-T)$ is injective but fails to have a dense range. While injectivity is clear, I fail to understand why the following shows that the range is not dense:
Let $y=(I-T)x$. Then $y(k)+ \cdots + y(1) = x(k)$ where $x(k)$ is the $k$-th entry of the sequence. Now for every $x\in l^2$ we have $\lim_{k\rightarrow \infty} x(k) = 0$ which then forces $|y(k) + \cdots y(1)| \rightarrow 0$ as well. Now this means that $(\alpha,0,0,...)$ cannot be in the range of $(I-T)$ which implies that the range of $(I-T)$ lies in the orthogonal complement of $(1,0,0,...)$.
Now first of all: why does it imply that it lies in the complement? and second of all why does that imply that the range isn't dense?
 A: It's not correct.  In fact the range of $I-T$ is dense and $\lambda = 1$ is not in the residual spectrum, but rather in the continuous spectrum.
Indeed, suppose $y$ is in the orthogonal complement of the range of $I-T$, so that $((I-T)x, y) = 0$ for all $x$.  This implies $0 = (x, (I-T^*)y)$ so $T^* y = y$.  Now $T^*$ is the left shift operator so this can only happen if $y(k) = y(k+1)$ for all $k$, which only happens for $y \in l^2$ if $y=0$.  So $I-T$ does have dense range.  
The argument you gave does correctly show that $(1,0,0,\dots)$ is not in the range of $I-T$ and thus $\lambda = 1$ is in the spectrum of $I-T$.  It is clearly not an eigenvalue (if $Tx=x$ then $x(1) = 0$ and $x(k+1) = x(k)$ for all $k$) so it must be continuous spectrum.
A: Not a complete answer, but at least some direction: 


*

*I don't have an answer, although the fact that $T$ preserves orthogonality probably comes into it somewhere. 


But assuming this part to have been addressed, we have:


*It's not dense because, for instance, the open ball of radius $1/2$ around $(1, 0, 0, \ldots)$ is entirely outside the orthogonal complement of $(1, 0, \ldots)$ (which consists entirely of things of the form $(0, *, *, \ldots)$, of course). 

