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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 ${\left\| {T - A} \right\|_{L({B_1},{B_1})}} \le c$?

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    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?2012-12-30
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    What if $T$ is the identity operator?2013-01-09
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    @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.2013-01-15
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    @RobertIsrael If $T$ is the identity, then $c \le 1 - \varepsilon$ would imply $\|Sv\|_{B_2} \ge \varepsilon \|v\|$, rendering $S$ invertible and both $B_1$ and $B_2$ finite-dimensional. So for infinite-dimensional $B_1$ or $B_2$, we must have $c \ge 1$ and one can choose $A = 0$.2016-12-24

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