Corollarie (b) after theorem 4.12 of Rudin's Functional Analysis states

$\mathscr{R}(T)$ is dense in $Y$ if and only if $T^*$ is one-one.

Where $T:X \rightarrow Y$ is a bounded linear operator between topological vector spaces.

The proof given is: $\mathscr{R}(T)$ is dense if and only if $\mathscr{R}(T)^{\perp}=\{0\}$; in that case, $\mathscr{N}(T^*)=\{0\}$. Note that the Hahn-Banach theorem 3.5 was tacitly used in the proof.

Theorem 3.5 states

Suppose $M$ is a subspace of a locally convex space $X$, and $x_0 \in X$. If $x_0$ is not in the closure of $M$, then there exists $\Lambda \in X^*$ such that $\Lambda x_0=1$ but $\Lambda x=0$ for every $x\in M$.

I don't see how this is used to prove the corollarie.


It is used in the statement: "$\mathscr{R}(T)$ is dense if and only if $\mathscr{R}(T)^{\perp}=\{0\}$." Namely in the proof of the implication: $\mathscr{R}(T)$ is dense, if $\mathscr{R}(T)^{\perp}=\{0\}$. Here is the elaboration:

Suppose $\mathscr{R}(T)$ is not dense in $Y$. Then by the definition of dense there is some $y_0 \in Y$ that is not in the closure of $\mathscr{R}(T)$. Hence by 3.5 there is a functional $\Lambda$ that annihilates $\mathscr{R}(T)$, but sends $y_0$ to $1$, which implies $\Lambda \neq 0$ and $\Lambda \in \mathscr{R}(T)^{\perp}$, hence $\mathscr{R}(T)^{\perp} \neq \{0\}$

  • $\begingroup$ And the converse is trivial? $\endgroup$ – simon Jan 3 '14 at 1:44
  • $\begingroup$ Yes, by the definitions of continuity and closure. $\endgroup$ – thomas Jan 3 '14 at 1:49

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