Showing that $ker(A)^\perp$ is the weak-* closure of $im(A*)$

Question

I am trying to show that $ker(A)^\perp$ equals the weak-* closure of $im(A^*)$ (Denoted by $cl(im(A^*))$), where $A: X \to Y$ is a continuous linear map between Banach spaces, where $A^*$ denotes the adjoint operator of $A$.

I have shown that $cl(im(A^*)) \subset ker(A)^\perp$. I am also pretty confident about this inclusion.

However, for the other inclusion I would like some verification of my proof.

Let $x' \in ker(A)^\perp$, then define $\tilde T: im(A) \to \mathbb{K}$, where $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$, by $\tilde T(y) = x'(x)$ if $Ax=y$. It can then be shown that $\tilde T$ is continuous and can thus be extended to a continuous linear functional $T: Y \to \mathbb{K}$. Then $(A^* T)(x) = T(Ax) = \tilde T (Ax) = x'(x)$ and thus $ (A^* T) = x'$. Thus, $x' \in im(A^*)$, thus in particular in $cl(im(A^*))$. Which concludes the proof.

I am unsure if this is correct, since I have not used the closure in the second part. Am I missing something. I would appreciate some feedback.

Answer 1

Suppose for a contradiction that the weak$^*$ closure $cl(im(A^*))$ is not equal to $\ker(A)^\perp$. Then, there exists $f\in\ker(A)^\perp$, $f\notin cl(im(A^*))$. By Hahn-Banach theorem, there exists a weak$^*$ continuous linear functional $\mu:X^*\to\mathbb{K}$ such that $$\mathbf{\mu(f) = 1}, \hspace{4mm} cl(im(A^*))\subseteq \ker\mu. $$ Since $\mu$ is weak$^*$ continuous, $\exists x_0\in X \hspace{2mm} \mu(f)=f(x_0)\hspace{2mm}\forall f\in X^*$.

$cl(im(A^*))\subseteq ker\mu \Rightarrow \mu(A^*g) = 0 \hspace {2mm} \forall g\in Y^* \Rightarrow g(Ax_0) = 0 \hspace {2mm} \forall g\in Y^* \Rightarrow Ax_0=0 \Rightarrow \mathbf{f(x_0)=0}$. Contradiction.

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Note: For the weak$^*$ continuous linear functionals on $X^*$, please see what follows:

Let $Z$ be a locally convex topological vector space with the set of seminorms $\mathcal{F}$ generating its topology. Let $Y$ be a Banach space, $T:Z\to Y$ be a linear operator.

1. $T$ is continuous $\Leftrightarrow\exists E\subseteq\mathcal{F}$ finite $\exists M>0\hspace{2mm}\forall x\in X \hspace{3mm} \|Tx\|\leq M\sup_{p\in E}p(x)$.

Proof: $(\Rightarrow)$ Since $T$ is continuous at $0$, there is an open neighborhood $V$ of $0$ such that $\forall x\in V$, $\|Tx\|\leq 1$. WLOG $$V=\bigcap_{p\in E} p^{-1}(B_r(0))=r\bigcap_{p\in E} p^{-1}(B_1(0))$$ for some $r>0$ and finite $E\subseteq\mathcal{F}$. Then, $\|Tx\|\leq 1/r$ whenever $p(x)<1$ $\forall p\in E$.

For each $\delta>0$, let $y_{\delta}=x/\left(\delta+\sup_{p\in E}p(x)\right)$. Clearly, $p(y_{\delta})<1$ so $\|Ty_{\delta}\|\leq 1/r$. It follows that, $$\|Tx\|\leq \frac{1}{r} \left(\delta+\sup_{p\in E}p(x)\right).$$ Letting $\delta \to 0$, we obtain the result with $M=1/r$.

$(\Leftarrow)$ Let $(x_{\alpha})$ be a net such that $x_{\alpha}\to x$. By definition, $p(x_{\alpha})\to p(x)$ for all $p\in\mathcal{F}$. Hence $\|Tx_{\alpha} - Tx\|\to 0$.

2. Suppose $\mathcal{F}$ is a set of linear functionals on $Z$, i.e., every $p\in\mathcal{F}$ is of the form $p(x)=|f(x)|$ for some linear $f:Z\to\mathbb{C}$. The following are equivalent.

• i. $Tx=\sum_{k=1}^n f_k(x)y_k$ for some $f_1,\dots,f_n\in\mathcal{F}$ and $y_1,\dots,y_n\in Y$.
• ii. $Tx = W(f_1(x),\dots,f_n(x))$ for some linear $W:\mathbb{C}^n\to Y$.
• iii. $\exists M>0\hspace{2mm}\forall x\in X\hspace{3mm}\|Tx\|\leq M\sup_k|f_k(x)|$. (that is, $T$ is continuous)
• iv. $\ker{T}\supseteq\bigcap_{k=1}^n\ker{f_k}$.

Proof: $(i\Rightarrow ii\Rightarrow iii\Rightarrow iv)$ is straightforward. The reader only needs to prove $(iv\Rightarrow i)$.

3. Thus, when $Y=\mathbb{C}$ in (2.), then $T\in\mathcal{F}$. Additionally, when $Z=X^*$ and $\mathcal{F}=X$, then $T:X^*\to\mathbb{C}$ is weak$^*$ continuous iff $T\in X$.