Let $X,Y$ be normed vector spaces and $V \subset X$ be a non-closed subspace. Let $A : V \rightarrow Y$ be a linear map.
How can I prove that there always exists an extension of A to a linear map $\overline{V} \rightarrow Y$ and that this extension is unique iff $A$ is bounded ?
I guess one has to use Hahn-Banach extension theorem but I don't see how to proceed for the uniqueness part.
Thank you in advance for your help !
Best Answer
I think this statement is false. If there is an extension there are infinitely many whether or not $A$ is bounded. (As stated, there is no requirement that the extended linear map is bounded, even in the case $A$ is bounded).
There is a linear functional $f$ on $\overline V$ which is $0$ in $V$ but not on $\overline V$. Fix $y \neq 0$ in $Y$ and consider $B(.)+nf(.)y$ where $B$ is any extension of $A$ to $\overline V$. These are all distinct extension of $A$ to linear maps on $\overline V$.