The control of norm in quotient algebra 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 ${\left\| {T - A} \right\|_{L({B_1},{B_1})}} \le c$?

To start from a very simple case: If $B_1$ is a Hilbert space and $S$ is finite-dimensional such that we have
$$ \|Tv\| \le c\|v\| + \|Sv\| \quad \forall v$$
then we can find a finite-dimensional $A$ satisfying
$$\tag{1} \|Av\| \le \|Sv\|$$
and
$$\tag{2} \|(T-A)v\| \le c \|v\|$$
for every $v \in B_1$.

Proof: Write $B_1 = (\ker S)^\bot \oplus_2 \ker S$. We define $A$ separately on each summand: On $\ker S$, we can clearly choose $A = 0$. On its annihilator, we can work with an orthonormal basis: For each vector $e_n$, we set
  $$ Ae_n = \frac{\|Se_n\|}{c + \|Se_n\|} Te_n$$
  (note that we know $Se_n \ne 0$) which immediately gives us
  $$ \|Ae_n\| \le \|Se_n\|$$
  and
  $$ \|(T-A)e_n\|
= \left\|\left(1 - \frac{\|Se_n\|}{c + \|Se_n\|} \right)Te_n\right\|
\le c
$$
  Since (1) and (2) are thus satisfied both on all of $\ker S$ and whenever $v$ is member of the orthonormal basis for $(\ker S)^\bot$, i.e. $v = e_n$, they are satisfied for every $v \in B_1$.

