As we know, continuous spectrum and residual spectrum are two cases in the spectrum of an operator, which only appear in infinite dimension. 


If $T$ is a operator from Banach space $X$ to $X$, $aI-T$ is injective, and $R(aI-T)$ is not $X$. If $R(aI-T)$ is dense in $X$, then $a$ belongs to the continuous specturm, it not, $a$ belongs to the residual spectrum. 


1) I want to know why do we care about whether $R(aI-T)$ is dense.


2) It seems as we just emphasize this difference for the operator, but not for the Banach algebra, why? 



Why do we distinguish the continuous spectrum and the residual spectrum?
I'm afraid that this is more or less a reformulation of what you asked, but: a bounded operator $T$ on a Banach space $X$ is invertible if and only if it is bounded below (i.e. there is some constant $C>0$ such that $||Tx|| \geq C||x||$ for all $x\in X$) and has dense range. Bounded below implies injective, and it also implies that the range is closed; the range is closed and dense, hence is everything.


So these two different types of spectrum distinguish different ways that the operator can fail to be invertible.


Why is this distinction not made for Banach algebras in general? Well, it doesn't make sense to ask this unless the Banach algebra is presented as a subalgebra of operators on some Banach space. I suppose that you could always consider the left-regular representation of the algebra on itself and then interpret those two conditions in that setting. If $A$ is a Banach algebra, let $a \mapsto L\_a$ denote the left-regular representation $L\_ab = ab$.


If $L\_a$ does not have dense range then $\overline{aA}$ is not all of $A$, which condition I don't think can be reduced to anything simpler.


If $L\_a$ is not bounded below, that means that there is a sequence $(x\_n)$ in $A$ with $||x\_n||=1$ such that $ax\_n \to 0$, i.e. $a$ is what is called a (left?) topological zero divisor.


