# The transpose in Banach spaces is bounded below

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?

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The important part seems to be missing: Is it clear to you why $R_{T'}$ is closed? It is not assumed that $T$ is injective which you would need in order to conclude that $T'$ is surjective. So, you need to know that the range is closed. If you know that then $T'$ is a continuous bijection from one Banach space onto another, so bounded below follows from the open mapping theorem. – Martin Dec 25 '12 at 19:20
It is not clear to me why $T$ injective implies $T'$ surjective. Or why it is enough to show that the range is closed, can u please expand? – Johan Dec 25 '12 at 20:09

Since $X$ is a Banach space and $T$ is surjective then by open mapping theorem $T$ is open, i.e. $$\{x\in X: \Vert x\Vert\leq c\}\subset \{T(x):\Vert x\Vert\leq 1\}\tag{1}$$ for some $c>0$. Hence \begin{align} \Vert T'(x')\Vert &=\sup\limits_{\Vert x\Vert\leq 1}|T'(x')(x)|\\&=\sup\limits_{\Vert x\Vert\leq 1}|x'(T(x))|\\ &\geq\sup\limits_{ \Vert x\Vert\leq c}|x'(x)|\tag{1}\\ &=\sup\limits_{ \Vert x\Vert\leq 1}|x'(cx)|\\ &=c\sup\limits_{ \Vert x\Vert\leq 1}|x'(x)|\\ &=c\Vert x'\Vert \end{align}
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...
should it be $\|y\| \leq c\|x\|$ and $\|T'y'\|$ ? – Johan Dec 26 '12 at 13:37
It should be $\|T'y'\|$ (I'll edit). But I did mean $\|x\| \le c \|y\|$. – Robert Israel Dec 26 '12 at 20:50