Transpose of continuous operator bounded below is surjective

Suppose $T: X\rightarrow Y$ is a continuous linear transformation between 2 normed vector spaces such tha $\|Tx\| \geq C\|x\|$ for some $C > 0$. Why must the transpose $T^{\ast}: Y^{\ast} \rightarrow X^{\ast}$ be surjective?

• Closed range theorem. Why can you without loss of generality pretend that $X$ and $Y$ are Banach spaces? Feb 11, 2014 at 1:46

First note that $T$ is injective so there exists $T^{-1}: T(X) \to X$ such that $T^{-1}(Tx))=x$ for all $x \in X$. We note that $T^{-1}$ is continuous. For $y=Tx$, $$\|T^{-1}(y)\| = \|T^{-1}Tx\| = \|x\| \leq \frac{\|Tx\|}{c}=\frac{\|y\|}{c}$$ Then for $\lambda: X \to \mathbb{C}$ a continuous linear functional, $\lambda':T(X) \to \mathbb{C}$ defined as $y \to \lambda T^{-1}(y)$ is a continuous linear functional of a subspace of $Y$ so by Hahn-Banach, we can extend it to $\tilde{\lambda'} \in Y^*$. Then we see for all $x \in X$, $$\lambda x =\lambda T^{-1}Tx = \tilde{\lambda'}(Tx)=T^*(\tilde{\lambda'}x)$$ which means that $\lambda=T^*\tilde\lambda'$.