Bounded linear operator with non-dense range Let $A:X\rightarrow Y$ be a bounded linear operator between Banach spaces. Suppose $\overline{R(A)}\neq Y$. How to prove that there exist $y^*\in Y^*$ s.th. for any $x\in X $ we have $\langle y^*, Ax \rangle=0$ ? (where $\langle \cdot,\cdot\rangle$ means dual pairing). I think I should use Hahn-Banach theorem, but now I don't see how to do it.
 A: I think you mean something like $y^* \circ A = 0$?
The Hahn-Banach theorem lets you extend (continuous) linear functionals defined on closed subspaces of $Y$ to (continuous) functionals defined on all of $Y$. In this case, take the zero functional on $\overline{(R(A)}$ and extend it to a functional $y^*$ on $Y$.
I think you also want to ensure that $y^* \not = 0$ identically on $Y$. You can do this using the idea above (it is easy to extend functionas from a finite codimension subspace $V$ of a Banach space $W$ to a functional on $W$ ... does this hint make sense to you?)
Edit:
More detail:
Prove the following lemma:
If $X$ is a closed subspace of finite codimension of a Banach space $Y$, then there is a functional which is zero on $X$ and nonzero on $Y$. For this, you can use the quotient map $Y / X \cong R^n$. (Actually once it is finite codimension I think it it is automatically closed.)
So take some vector not in $\overline{R(A)}$ and prove that the space spanned by $\overline{R(A)}$ and $v$ is still closed. $\overline{R(A)}$ is finite codimension here, so by the previous lemma you can find a functional which vanishes on $\overline{R(A)}$ but not on $Span (v, \overline{R(A)})$. Now extend this function to $Y$ using Hahn-Banach.
