Statement
If $X$ and $Y$ are Banach Spaces ad if $T \in \mathcal{B}(X,Y)$, then each one of the following three conditions implies the other two:
(a) $\mathcal{R}(T)$ is closed in $Y$ $\iff$ (b) $\mathcal{R}(T^*)$ is weak*-closed in $X^*$ $\iff$ (c) $\mathcal{R}(T^*)$ is norm-closed in $X^*$.
Reporting the proof and the bits I don't understand
PROOF: Suppose (a) holds. By Theorem 4.12 and (b) of Theorem 4.7, $\mathcal{N}(T)^{\perp}$ is the weak*-closure of $\mathcal{R}(T^*)$. To prove (b) it is therefore enough to show that $\mathcal{N}(T)^{\perp} \subset \mathcal{R}(T^*)$. Pick $x^* \in \mathcal{N}(T)^{\perp}$. Define the linear functional $\Lambda$ on $\mathcal{R}(T^*)$ by$$ \Lambda T x = \left\langle x, x^*\right\rangle \;\; (x \in X) $$Note that $\Lambda$ is well defined, for if $Tx = Tx'$, then $x - x' \in \mathcal{N}(T)$; hence$$ \left\langle x - x', x^* \right\rangle = 0 $$
Question : To me showing that $\Lambda$ is well defined means to show that for each $y \in \mathcal{R}(T)$ then $\Lambda y$ is uniquely defined and linear. Linearity is easy enough so I skip and I guess Rudin did as well, uniquely defined I think is what the author is going for, right? Indeed if $y \in \mathcal{R}(T)$ and $Tx = y, Tx' = y$ we have the conclusion already given, namely$$ \left\langle x - x' , x^*\right\rangle = 0 $$but more specifically this implies$$ 0 = \left\langle x - x' , x^*\right\rangle = \left\langle x , x^* \right\rangle - \left\langle x' , x^* \right\rangle \iff \left\langle x , x^* \right\rangle = \left\langle x' , x^* \right\rangle \iff \Lambda T x = \Lambda T x' $$which is I guess the wanted conclusion. Am I right?
The open mapping theorem applies to$$ T : X \to \mathbb{R}(T) $$since $\mathbb{R}(T)$ is assumed to be a closed subspace of the complete space $Y$ and is therefore complete. It follows that there exists $K < \infty$ such that to each $y \in \mathcal{R}(T)$ corresponds an $x \in X$ with $Tx = y$ and $\left\lVert x \right\rVert \leq K \left\lVert y \right\rVert$,
Question I don't get how the bound
$$ \left\lVert x \right\rVert \leq K \left\lVert y \right\rVert $$
follows from the application of the open mapping theorem, can you clarify?
and$$ \left| \Lambda y \right| = \left| \Lambda T x \right| = \left| \left\langle x, x^* \right\rangle \right| \leq K \left\lVert y \right\rVert \left\lVert x^* \right\rVert $$Thus $\Lambda$ is continuous. By the Hahn-Banach Theorem, some $y^* \in Y^*$ extends $\Lambda$. Hence$$ \left\langle Tx, y^* \right\rangle = \Lambda T x = \left\langle x, x^* \right\rangle $$This implies $x^* = T^* y^*$. Since $x^*$ was an arbitrary element of $\mathcal{N}(T)^{\perp}$, we have shown that $\mathcal{N}(T)^{\perp} \subset \mathcal{R}(T)$. Thus (b) follows from (a).
I think the remaining is straightforward, but in case I'll ask a separate question.
$\endgroup$ 4 Reset to default
