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?
Extension of closed linear functionals...
1
$\begingroup$
functional-analysis
-
0What nls stand for? – 2012-02-13
-
2This is not 4chan, please write your questions without any abbreviations. – 2012-02-13
-
0@DavideGiraudo normed linear space – 2012-02-13
-
0Maybe 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 $X$. – 2012-02-13
-
0Most closed functionals are not bounded, do you assume $f$ is? Every bounded linear functional on a *complete* normed space is continuous, and therefore closed. – 2012-02-13
-
0@GEdgar No, I have not made any assumptions on the boundedness of the linear functional. My motivation is the following : If the linear functional f were continuous, then we would have a unique Hahn-Banach extension of f which is also continuous. So, given that f is closed, is there a similar unique extension of f , which is closed? – 2012-02-13
-
1If 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. – 2012-02-13
-
0And so $f$ would be bounded too. So this cannot happen, unless $f$ was already bounded. – 2012-02-14