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Let $M$ be a closed subspace of a Banach space $X$. Let $q:X\to X/M$ be the quotient map, let $q^*: (X/M)^*\to X^*$ be the conjugate operator of $q$, and let $(z^*_n)$ be a sequence in $(X/M)^*$.

Is it true that $(z^*_n)$ is $w^*$-convergent in $(X/M)^*$ $\Leftrightarrow(q^* z^*_n)$ is $w^*$-convergent in $X^*$?

Are the following arguments correct?

The direct implication follows from the $w^*$-$w^*$ continuity of $q^*$.

For the converse, if $(q^* z^*_n)$ is $w^*$-null, then $z^*_n(q\, x) =(q^* z^*_n)(x)$ converges to $0$.

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Let $(z_n^*)$ be a sequence in $(X/M)^*$. By definition of $q^*$, $$ q^*(z_n^*)(x) = z_n^*(q(x)) \quad \forall x \in X,$$ If $(z_n^*)$ is $w^*$-convergent, then the right-hand side converges, then both sides converge, proving $\Rightarrow$.

Now assume that the mapped sequence $q^*(z_n^*)$ is $w^*$-convergent. By surjectivity, $q$ has a right inverse, which we denote by $q_r$. For $y \in (X/M)$, we choose $x = q_r(y)$ in the equation above, which gives $$ q^*(z_n^*)(q_r(y)) = z_n^*(y)$$ By assumption the left-hand side converges, so the right-hand side converges as well, and since $y$ was arbitrary this proves $\Leftarrow$.

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  • $\begingroup$ Is the right inverse $q_r$ linear and continuous? $\endgroup$ – M.González Jan 20 '17 at 19:46
  • $\begingroup$ Why does it matter? $\endgroup$ – Roberto Rastapopoulos Jan 20 '17 at 21:06

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