# Proving that an operator is bounded from below

Hi again Stackexchange,

This is a question I had on an exam today that I could not answer, and it deeply disturbs me.

Let $X$ be a Banach space and $T:X\to X$ a surjective linear map. Show that the adjoint $T^*:X^*\to X^*$ is bounded from below, i.e. there is a $c$ such that $c\Vert g\Vert\leq\Vert T^*\Vert\Vert l\Vert$.

It feels like it should be easy but my mind is frozen.

Apply the open mapping theorem to $T$, then restate its conclusion as a bound on $T^*$. –  user53153 Dec 20 '12 at 20:24
Do you mean that $T$ is a bounded linear map and $c|l|\le |T^*l|$? –  23rd Dec 20 '12 at 20:28
@richard; bounded linear and surjective. @PavelM; Ok thanks, that was quite easy. But I have to say it feels like it should be even simpler. Shouldn't it be possible to exploit the fact that $T^*$ is injective whenever $T$ is surjective? –  user25470 Dec 20 '12 at 20:33
It's a bit more than just that $T^*$ is injective: you also need the fact that $T^*$ has closed range. Then $T^*$ is an isomorphism onto its image, and the result follows. –  Robert Israel Dec 20 '12 at 20:40
Since $T$ is surjective operator on Banach space, by open mapping theorem $T$ it is open. Hence there exist $c>0$ such that $T(\operatorname{Ball}_X)\supset c\operatorname{Ball}_X$. In this case \begin{align} \Vert T^*(g)\Vert &=\sup\{|T^*(g)(x)|:x\in\operatorname{Ball}_X\}\\ &=\sup\{|g(T(x))|:x\in\operatorname{Ball}_X\}\\ &\geq\sup\{|g(y)|:y \in c\operatorname{Ball}_X\}\\ &=c\sup\{|g(y)|:y \in \operatorname{Ball}_X\}\\ &=c\Vert g\Vert \end{align}