# Does the completion of an operator with closed image also have a closed image?

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 $$\tilde Y$$).

• 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. – s.harp Oct 7 '18 at 7:31
• Thanks! that seems to answer the question. – Asaf Shachar Oct 7 '18 at 9:13