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Possible Duplicate:
The identity cannot be a commutator in a Banach algebra?

If $X$ is non empty normed space and let $S,T : X\to X $ is a linear map with $ST-TS=id_X$ , then how do i show that either $S$ or $T$ is discontinuous ? What i have been thinking is that if the operator has to be discontinuous then one of them is not bounded . so if i just substitute i get the form $$ST^{n+1}-T^{n+1}S=(n+1)T^n \forall n\in \mathbb N$$ Does this mean anything to what i am looking for ? I am looking for some help.

And is it true that set of operators from banch space to itself form a open set in norm topology ?

I would be glad to get ideas invovled . Thanks

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    See: [The identity cannot be a commutator in a Banach algebra?](http://math.stackexchange.com/q/54397)2012-11-11
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    @MattN. : i didn't get you Matt N. I think i am registered . Please let me know if there is a problem :)2012-11-11
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    @Norbert : I have actually one more question along with it . I got the first one . Do you suggest me to delete the first question ?2012-11-11
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    The first question is already answered, the second is triviail - the set $\mathcal{B}(X,X)$ is open in $\mathcal{B}(X,X)$ by definition. of topology.2012-11-11

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