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Let $X$ be a Banach space and $P:X^* \to X^*$ a (bounded linear) projection, i.e., $P^2 = P$. I would like to know whether the following statement is true:

Claim. If $\operatorname{Im}(P)$ is $\sigma(X^*,X)$-closed (that is, weak$^*$-closed), then $P$ is $\sigma(X^*,X)$-continuous (weak$^*$-continuous).

My attempt / reasoning.

We have $\operatorname{Im}(P) = \ker(I-P)$. So $\ker(I-P)$ is weak$^*$-closed. My initial thought was: perhaps this implies that the linear map $I-P$ is weak$^*$-continuous (since the kernel of a continuous operator is closed), and then $P = I-(I-P)$ would be the difference of weak$^*$-continuous maps, hence weak$^*$-continuous. But I realized this direction is not valid in general: having a weak$^*$-closed kernel does not imply weak$^*$-continuity.

So my questions are:

  • Is the claim above true? If so, what would be a correct proof (or which additional hypotheses are required)?
  • If the claim is false, can someone provide a counterexample (or a reference) of a projection $P$ on $X^*$ whose range is weak$^*$-closed but which is not weak$^*$-continuous?

Any counterexample, confirmation, or reference would be greatly appreciated.

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2 Answers 2

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This is not true. For example, let $X = \ell^1$. Then $X^\ast = \ell^\infty$. Fix a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. Then $P: \ell^\infty \to \ell^\infty$, $P((a_n)_n) = (\lim_{k \to \mathcal{U}} a_k) \cdot (1)_n$ is a bounded linear projection onto the $1$-dimensional subspace $\text{span}\{(1)_n\}$ of $X^\ast$, which is clearly $\sigma(X^\ast, X)$-closed. However, $\text{ker}(P)$ contains $c_0$, which is $\sigma(X^\ast, X)$-dense, but $\text{ker}(P) \neq X^\ast$, so $\text{ker}(P)$ is not $\sigma(X^\ast, X)$-closed. Hence, $P$ is not $\sigma(X^\ast, X)$-continuous.

Let me provide a more general answer: a counterexample $P$ exists iff $X$ is non-reflexive. Clearly, if $X$ is reflexive, any bounded linear $P$ is automatically $\sigma(X^\ast, X)$-continuous, so no counterexample can exist. Otherwise, assume $X$ is non-reflexive. Then there exists $\varphi \in X^{\ast\ast} \setminus X$. Since $\varphi \neq 0$, we may pick $v \in X^\ast$ s.t. $\varphi(v) = 1$. Then $P(w) = \varphi(w) \cdot v$ is a bounded linear projection onto $\text{span}\{v\}$, which, being one-dimensional, is $\sigma(X^\ast, X)$-closed. However, since $\varphi$ is not $\sigma(X^\ast, X)$-continuous, neither is $P$.

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  • $\begingroup$ What is $\lim_{k \to \mathcal{U}} a_k$? $\endgroup$ Commented Sep 2 at 6:09
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    $\begingroup$ @DeanMiller The ultralimit of the sequence $a_k$ along the ultrafilter $\mathcal{U}$. See this Wikipedia article for a detailed exposition. In short, if $a_k$ is a sequence of elements of a topological space $S$ and $a\in S$, then $\lim_{k\to\mathcal{U}}a_k=a$ means for any open neighborhood $O$ of $a$, there exists $K\in\mathcal{U}$ s.t. whenever $k\in K$, we have $a_k\in O$. If $S$ is compact Hausdorff, then the ultralimit exists and is unique. In our case, since the sequence $a_k$ is bounded, it lies within a comapct Hausdorff space, ... $\endgroup$ Commented Sep 2 at 6:39
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    $\begingroup$ @DeanMiller ...namely, a closed disk in the complex plane. So, $\lim_{k\to\mathcal{U}}a_k$ is well-defined in our case. (If you're unfamiliar with this but are familiar with $C^\ast$-algebras, just think of $\mathcal{U}$ as a character on $\ell^\infty$ that does not correspond to evaluation at any element of $\mathbb{N}$. If you interpret it that way, $\lim_{k\to\mathcal{U}}a_k$ just means $\mathcal{U}((a_k)_k)$.) $\endgroup$ Commented Sep 2 at 6:45
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    $\begingroup$ Thanks for the explanation. I found this and this quite helpful for understanding the definition and its main properties. $\endgroup$ Commented Sep 2 at 8:10
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The answer by David Gao shows that we can find a counterexample whenever $X$ is not reflexive. In this answer we see a case where there is a positive result. We have the following.

Theorem 1. Let $X$ be a Banach space. Assume $P\colon X^{*} \to X^{*}$ is a bounded linear projection such that ${\rm ker} \, P$ and ${\rm ran} \, P$ are both closed in the weak$^{*}$ topology. Then $P\colon (X^{*}, {\rm weak}^{*}) \to (X^{*}, {\rm weak}^{*})$ is continuous.

To show this we first need a closed graph theorem.

Theorem 2. Let $X$ and $Y$ be Banach spaces. Let $T\colon X^{*} \to Y^{*}$ be a bounded linear operator such that the graph of $T$ is closed in the respective weak$^{*}$ topologies, namely closed in $(X, {\rm weak}^{*}) \times (Y^{*}, {\rm weak}^{*})$. Then $T\colon (X^{*}, {\rm weak}^{*}) \to (Y^{*}, {\rm weak}^{*})$ is continuous.

A short proof of Theorem 2 can be found in Theorem A.1 within the paper On The Fundamental Principles of Unbounded Functional Calculi by Haase (link here).

Proof of Theorem 1. The idea is to show that $P$ has a weak$^{*}$-weak$^{*}$ closed graph and then apply Theorem 2 to deduce the result. Let $(x^{*}_{i}, Px^{*}_{i})_{i\in I}$ be a net in $X^{*}\times X^{*}$ that converges to $(x^{*}, y^{*}) \in X^{*}\times X^{*}$ in $(X^{*}, {\rm weak}^{*})\times (X^{*}, {\rm weak}^{*})$. We wish to show $y = Px^{*}$ and then it will follow that $P$ has a weak$^{*}$-weak$^{*}$ closed graph.

First note that $(Px^{*}_{i})_{i\in I}$ is a net in the weak$^{*}$ closed set ${\rm ran}\, P$ that converges in the weak$^{*}$ topology to $y^{*}$. Hence we have $y^{*} \in {\rm ran} \, P$. Furthermore, since $P$ is a projection we also have that $(x^{*}_{i} - Px^{*}_{i})_{i\in I}$ is a net in the weak$^{*}$ closed set ${\rm ker} \, P$ that converges in the weak$^{*}$ topology to $x^{*} - y^{*}$. Hence we have $x^{*} - y^{*} \in {\rm ker} \, P$. We have shown so far that \begin{equation} x^{*} = (x^{*} - y^{*}) + y^{*} \end{equation} with $x^{*} - y^{*} \in {\rm ker} \, P$ and $y^{*} \in {\rm ran} \, P$. As we have $X^{*} = {\rm ker} \, P \oplus {\rm ran} \, P$ as an algebraic direct sum, we conclude $Px^{*} = y^{*}$. Hence the graph of $P$ is weak$^{*}$-weak$^{*}$ closed. We now apply Theorem 2 to conclude that $P$ is weak$^{*}$-weak$^{*}$ continuous as desired.

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