Let $X$ be a Banach space and let $T:X \rightarrow X$ be a bounded linear map. Show that: If $T$ is surrjective then its transpose $T':X' \rightarrow X'$ is bounded below.
My try: We know that $R^\perp_M = N_{M'}$ and since X is surrjective
$R_M = X$ hence $R_M^\perp = N_{M'} = 0$ so $M'$ is invertible and bounded below.
Am I missing some details? Is invertible and bounded below true?
Best Answer
If $T$ is surjective, by the Open Mapping Theorem there is some $c > 0$ such that for every $y \in X$ there is $x \in X$ with $Tx = y$ and $\|x\| \le c \|y\|$. Use this to get a lower bound on $\|T'y'\|$ for $y' \in X'$, taking $y$ such that $|y'(y)|$ is not too small...