Let $A$ be a Banach algebra. Suppose that $e\in A$ such that $e^2=e$ and $eAe$ is division algebra(i.e., $eAe$ is unital and every element of $eAe$ has inverse in $eAe$). Define $T_e:A\to A$ with $T_e(a)=eae$. Also Define $S_e:Ae\to Ae$ with $S_e(ae)=eae$,(In this case, when $A$ is semi-prime, $Ae$ is a left minimal ideal of A).
Now, if $S_e$ is not a weakly compact operator, could we say that $T_e$ is not a weakly compact operator (or is not compact operator)?
Suppose that $A_1=\{a\in A: \|a\|<1\}$ and $B=\{ b\in A: \|be\|<1\}$. Obviously $Be\subset A_1$, and $S_e(Be)=T_e(Be)$. Hence $S_e(Be)\subset T_e(A_1)$, and consequently $\overline{S_e(Be)}^w\subset \overline{T_e(A_1)}^w$. But $\overline{T_e(A_1)}^w$ is weakly compact from assumption, and so $\overline{S_e(Be)}^w$ is weakly compact. But $$S_e(Be)=\{ae: a\in A, \|ae\|<1\}$$ which implies $S_e:Ae\to Ae$ is weakly compact operator.