Let $X,Y$ be banach spaces and $T: X \to Y$ a bounded linear map. I'm trying to show that if $T^*$ (adjoint of $T$) is surjective, then $T$ is bounded below, i.e. ${\exists} c>0$ s.t $||T(x)|| \ge c||x||$ for all $x \in X$.
I'm completely lost on this one, tried to use the open mapping theorem but getting nowhere. Any ideas?
It suffices to show that there is a constant $M$ such that $\|Tx\| \le 1$ implies $\|x\| \le M$.
You were on the right track thinking about the open mapping theorem. Instead we'll use Banach-Steinhaus.
Let $A = \{x \in X : \|Tx\| \le 1\}$. For each $x \in A$ define the functional $Jx \in T^{**}$ by $\langle Jx,x^*\rangle = \langle x^*,x\rangle$ for any $x^* \in X^*$. The usual norm formulas tell you $\|x\| = \|Jx\|$.
Fix $x^* \in X^*$. Since $T^*$ is onto there exists $y^* \in Y^*$ with $x^* = T^*y^*$. For each $x \in A$ you obtain $$\langle Jx,x^* \rangle = \langle x^*,x \rangle = \langle T^*y^*,x\rangle = \langle y^*,Tx\rangle \le \|y^*\|.$$ Thus $$\sup_{x \in A} \langle Jx,x^*\rangle \le \|y^*\|$$ so that the family of functionals $\{Jx : x \in A\}$ is pointwise bounded. The Banach-Steinhaus theorem implies $$\sup_{x \in A}\|Jx\| < \infty.$$ If you denote that supremum by $M$ you find that $$\|Tx\| \le 1 \implies x \in A \implies \|Jx\| \le M \implies \|x\| \le M$$ which is the statement we needed.