Suppose that $T \in L(V,W)$. Prove that if Img$(T)$ is dense in $W$ then $T^*$ is one-to-one.
Answer
Suppose $T^*$ is not one-to-one. Then there exists $f_1,f_2 \in W^*$ with $f_1\neq f_2$ such that \[ T^*(f_1)=T^*(f_2). \] Thus \[ f_1(Tv)=f_2(Tv), \forall v\in V. \] and hence $(f_1-f_2)(Tv)=0$, $\forall v \in V$, i.e. \[f_1-f_2=0, \forall w\in Img (T).\] Since $f_1-f_2$ is continuous, it also vanishes on $\overline{Img(T)}=W$, and therefore $f_1-f_2=0$ on $W$, which contradicts the assumption $f_1 \neq f_2$. The proof is complete.
Savionf
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The answer is accepted.
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