How do you quickly show whether an operator is compact? In my text on functional analysis, the author defines an operator on normed space as compact if it:


*

*continuous

*transforms bounded sets into relatively compact sets


Okay, number 1 we can work with. Proving continuity using $\epsilon-\delta$, things of that sort.
What about number 2? How do we go about picking every single bounded set possible and check if it gets mapped into relatively compact sets?
What is the general step to quickly how that an operator is compact?
 A: I'll write this since no one has yet written anything ( and, of course, I think this is helpful):
An operator between normed spaces is compact if the image of any bounded sequence has a convergent subsequence ( which, if you unravel, is equivalent to other definitions in terms of, e.g, relatively -compact sets ).
Showing the right-shift operator in $l^2$ $T(x_1,x_2,.....,x_n,...)=(0,x_1, x_2,....) $ is not compact, since there are sequences ( like $(e_j)$ , as described below ), whose image does not have a convergent subsequence:
Say , in $l^2$ , you have the sequence $e_j ; j=1, 2,3,...$ , where $ e_j$ is
the sequence that has a $1$ in the $j-$th spot and $0$ everywhere else.
Then each $e_j$ is $l^2$-bounded.  But {$T(e_j)$}= {$T(0,0,..,1,0,..)= (0,0,..,0,1,0...)$} i.e., $1$ in $j-$th place shifted to $1$ in (j+1)-st place.
The image does not have a convergent subsequence in $l^2$ , since, e.g., it is not Cauchy ( and $l^2$ is complete ). So the right-shift operator  in $l^2$ is not a compact operator.
Hope you were looking for something like this.
