Let $X,Y$ be real normed vector spaces, and suppose that $T:X \to Y$ is a bounded linear operator with closed image.

Let $\tilde X,\tilde Y$ be the completions of $X,Y$, and let $\tilde T:\tilde X \to \tilde Y$ be the natural continuous extension of $T$ (which is unique).

Is $\text{Image } \tilde T$ closed in $\tilde Y$?

I proved that $\, \, \overline{\text{Image } \tilde T}^{\tilde Y}= \overline{\text{Image } T}^{\tilde Y} $, however, this doesn't seem to resolve the question. (We only know $\text{Image } T$ is closed in $Y$; it does not need to be closed in $\tilde Y$).

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    $\begingroup$ Indeed it is not true. Take $X$ to be Banach and $T':X\to X$ any map that does not have a closed image. Let $Y=\mathrm{im}(T')$ and $T$ the restriction of codomain of $T'$ to $Y$. Now $T$ has a closed image but if you change the codomain to the completion of $Y$ this is not true anymore. $\endgroup$ – s.harp Oct 7 '18 at 7:31
  • $\begingroup$ Thanks! that seems to answer the question. $\endgroup$ – Asaf Shachar Oct 7 '18 at 9:13

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