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Is there a Banach space $X$, $S$ an unbounded operator defined on a dense subspace $D$ of $X$ and a bounded operator $T$ on $X$ such that


is bounded? What if $T$ is assumed to be compact?

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General fact: If $G$ is a group, $H \le G$ is a subgroup, $h \in H$, and $k \notin H$, then $kh, hk \notin H$. The set of all linear operators on $D$ is an (abelian) group under addition, and the bounded operators are a subgroup. So are the compact operators. This is also the fact that an even number plus an odd number is odd. – Nate Eldredge Mar 12 '12 at 18:08
up vote 1 down vote accepted

Hint: the difference of two bounded operators is bounded.

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