Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $B_1,B_2$ be two Banach spaces and $L(B_i,B_j),K(B_i,B_j)(i,j=1,2)$ spaces of bounded and compact linear operator between them respectively. If $T \in L(B_1,B_1)$, we have a $S \in K(B_1,B_2)$ and a constant $c>0$ such that for any $v \in B_1$,$${\left\| {Tv} \right\|_{{B_1}}} \le c{\left\| v \right\|_{{B_1}}} + {\left\| {Sv} \right\|_{{B_2}}}.$$

My question is, can we find a $A \in K(B_1,B_1)$, such that ${\left\| {T - A} \right\|_{L({B_1},{B_1})}} \le c$?

share|cite|improve this question
The case when finite-rank operators are dense in $K(B_1,B_2)$ should be easier; do you have a proof in this case? – user53153 Dec 30 '12 at 20:06
What if $T$ is the identity operator? – Robert Israel Jan 9 '13 at 20:47
@RobertIsrael If $T$ is the identity, then the problem reduces to showing that the displayed inequality forces $c\ge 1$. The latter is easy to see when the pair $(B_1,B_2)$ has the approximation property (hence my previous comment), but perhaps not as easy in general. – user53153 Jan 15 '13 at 18:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.