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I'm practicing some old exams for my functional analysis exam tomorrow, and i'm having trouble with the following:

Let $X$ be a reflexive Banach space and let $Y$ be a normed space. Assume there exists a linear map $T: Y\to X$ which is an isometry.

A completion $(Z, i)$ of a normed space $Y$ consists of a Banach space $Z$ and an isometry $i : Y \to Z$ such that $\overline{i(Y)} = Z$.

a) Construct a completion $Z$ of $Y$ using $T$.

b) Prove that the completion $Z$ in a) is a reflexive space.

ronalddb89
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  • Are you sure about the formulation? $X=\lbrace 0\rbrace$ is a reflexive Banach space and the zero map $T:X\to Y$ is an isometry which certainly does not help to construct a completion of $Y$. – Jochen Jun 29 '16 at 11:44
  • I checked again, i literally copied the question... But $T$ is not going from a reflexive space to a normed space with an isometry, but the other direction. – ronalddb89 Jun 29 '16 at 11:57
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    What is $J_X$??? It looks to me like there is something wrong. You are probably supposed to find a completion of $X$, and it is $Y$ which is a reflexive Banach space, not $X$. – tomasz Jun 29 '16 at 12:07
  • Sorry, i should have specified $J_X : X \to X'' : x \mapsto (f \mapsto f(x))$ – ronalddb89 Jun 29 '16 at 12:11
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    But you wrote $T:X\to Y$. For an isometry $T:Y\to X$ you can just take $Z=\overline{T(Y)}$ and $i=T$ (if you are very scrupulous, define $i:Y\to Z$ by $i(y)=T(y)$ -- it is a different map because of the codomain). $Z$ is reflexive because closed subspaces of reflexive Banach spaces are reflexive. This can be proved, e.g., using the characterization of reflexivity by weak compactness of the closed unit ball. – Jochen Jun 29 '16 at 12:14
  • Thanks @Jochen. Since $X$ is Banach and $\overline{T(Y)}$ is closed, it is also Banach. Way easier than my solution! Thanks – ronalddb89 Jun 29 '16 at 12:19

1 Answers1

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a) Let $Z = \overline{T(Y)}$. Then $Z$ is a closed subspace of a Banach space and hence a Banach space. Define $i = T$. Then $i$ is an isometry and $\overline{i(Y)} = \overline{T(Y)} = Z$. So $(Z,i)$ is a completion of $Y$.

b) Z is a closed subspace of a reflexive space, so it is reflexive.

ronalddb89
  • 141