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If $X$ is a normed linear space and $X^*$ be its completion, consider a linear functional $f$ belonging to $X'$ which is a closed map. By Hahn-Banach extension there exists a linear functional $f_1 \colon X^*\to\mathbb R$, s.t. $f_1$ restricted to $X$ is $f$ and $||f_1||=||f||$. Question: is $f_1$ also closed?

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What nls stand for? –  Davide Giraudo Feb 13 '12 at 18:38
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This is not 4chan, please write your questions without any abbreviations. –  Norbert Feb 13 '12 at 18:39
    
@DavideGiraudo normed linear space –  Norbert Feb 13 '12 at 18:39
    
Maybe I misunderstand something, but $X$ is isometric to a dense subset of $X^*$ so we don't need Hahn-Banach to extend $f$ to $f_1$ and $f_1$ is unique. We can check that for a closed set $F$ of $X^*$ $f_1(F)$ is sequentially closed using the density of $X$. –  Davide Giraudo Feb 13 '12 at 18:51
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If you do not assume $f$ is bounded, then what sense does it make to write $||f_1||=||f||$ ?? Of course, if your extension $f_1$ is closed, then it is bounded by the Closed Graph Theorem. –  GEdgar Feb 13 '12 at 22:38
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