Invertibility of an operator with an invertible dual (adjoint) on a Banach space

Question

I have the following problem:

Let $T:X \rightarrow X$ be a linear operator on a Banach space X and let T be its dual. Show that $\forall x \in X$ the following holds:

$$ ||Tx||≥||(T')^{−1}||^{−1}||x||$$ and prove that $T$ is invertible

What I've shown is that there is the following relationship:

$$(T')^{-1} = (T^{-1})'$$

I'd assume we have to use the Hahn-Banach Theorem but can't really see how.

Answer 1

Suppose $x\in X$. Pick an element $g\in X'$ such that $\|g\|=1$ and $g(x) = \|x\|$ (a norming functional for $x$). Let $f=(T')^{-1}g$, and note $\|f\|\le \|(T')^{-1}\|$. Then $$ \|x\| = g(x) = (T'(f))(x) = f(Tx) \le \|f\|\|Tx\| \le \|(T')^{-1}\|\|Tx\| $$ proving the first claim.

You also want to show that $T$ is surjective. By the above, the range of $T$ is closed (the existence of a lower bound $\|Tx\|\ge c\|x\|$ implies that). So if $T$ is not onto, there exists a nonzero functional $f\in X'$ that vanishes on the range of $T$. Thus, $f(Tx)=0$ for all $x\in X$, which means $T'f=0$, contradicting the invertibility of $T'$.