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Suppose we have a Banach space $X$ and an idempotent operator $Q\colon X^*\to X^*$ with range isomorphic to $\ell_1$. Must $Q$ be an adjoint to some idempotent operator on $X$? In other words, is $Q$ $w^*$-$w^*$-continuous?

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No. Consider $X = c_{0} \mathbin{\oplus_{\infty}} \ell^1$ so that $X^{\ast} = \ell^{1} \mathbin{\oplus_1} \ell^{\infty}$ and take a Banach limit $L: \ell^{\infty} \to \mathbb{R}$. Then $L \in (\ell^{\infty})^\ast \smallsetminus \ell^{1}$, so it is not weak$^\ast$-continuous: the sequence $h_n = (1,\ldots,1,0,\ldots)$ with $n$ nonzero entries weak$^\ast$-converges to the constant sequence $1$ but $1 = L(1) \neq 0 =\lim_{n\to\infty}L(h_n)$.

Let us identify $\mathbb{R}$ with the subspace of constant sequences in $\ell^\infty$.

The range of the map $Q\colon \ell^{1} \mathbin{\oplus_1} \ell^{\infty} \to \ell^{1} \mathbin{\oplus_1} \ell^{\infty}$ given by $Q(g,h) = (g,Lh)$ is $\ell^{1} \mathbin{\oplus_1} \mathbb{R} \subset \ell^1 \mathbin{\oplus_1} \ell^\infty$ and $Q$ is idempotent because $L$ fixes the space of constant sequences. But $Q$ is not weak$^\ast$-continuous because $(0,1)$ is the weak$^\ast$-limit of $(0,h_n)$ while $(0,L(1)) = (0,1) \neq (0,0) = \lim_{n\to\infty}(0,Lh_n)$.

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Let $X = C[0,1]$ so $X^* = M[0,1]$, the signed Borel measures on $[0,1]$. Define $Q: X^* \to X^*$ so that $Q \mu$ is the measure supported on $\{1/n: n \in {\mathbb N}\}$ with $(Q\mu)(\{1/k\}) = \mu((1/(k+1),1/k])$.
This is not $w^*-w^*$ continuous: as $t \to 1/n+$, the point mass $\delta_t$ converges $w^*$ to $\delta_{1/n}$, but $Q \delta_t = \delta_{1/(n-1)}$ for $t \in (1/n, 1/(n-1)]$.

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  • $\begingroup$ That's a very nice example! $\endgroup$ – t.b. Jun 1 '12 at 20:48

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