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For $X, Y$ Banach spaces, and $T$ a bounded linear operator $T:X \to Y$, if $T$ isn't bounded below then $T^*:Y^* \to X^*$ isn't an open map.

What methods can be used to prove something like this? The open mapping theorem is the only theorem I know that deals with open maps, and it feels all but useless here.

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  • $\begingroup$ Argue by contradiction. If $T^*$ is open, then one can find some constant $C$ such that $T^*(CB_{Y^*})$ contains $B_{X^*}$. Use this and Hahn-Banach to show that $\Vert x\Vert\leq C\, \Vert Tx\Vert$ for every $x\in X$. $\endgroup$ – Etienne Oct 5 '15 at 10:41
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You don't need open mapping theorem here. Assume $T^*$ is open: this means there is $C>0$ such that every unit norm functional $f\in X^*$ can be written as $g\circ T$ for some $g\in Y^*$ with $\|g\|\le C$. Following Etienne's hint, apply the above to the norming functional of an element $x\in X$; i.e., a unit functional $f$ such that $f(x)=\|x\|$. Conclude that $\|x\|=f(x)=g(T(x))\le C\|Tx\|$, as desired.

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