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Let $X$, $Y$, and $Z$ be Banach spaces with $Z \subset X$. Suppose $T$ is a bounded linear operator with domain $X$ and range $Y$. Must $T(Z)$ be a Banach space?

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Take X to be $Z\oplus Y$, and suppose there is an injective map $S: Z \to Y$ which has non-closed range. Let $T(z,y) = S(z)-y$, then $T: X \to Y$ is surjective, but $T(Z) = S(Z)$ is not closed in $Y$.

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    $\begingroup$ To be a bit more explicit, let $Z = \ell^1$, $Y = \ell^2$ and let $S$ be the inclusion. $\endgroup$
    – t.b.
    Mar 15, 2012 at 4:18

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