Under what conditions a bounded linear map can be extended ? - MathOverflow

Under what conditions a bounded linear map can be extended ?

2011-06-25T22:40:02Z

I have two questions after reading the Hahn-Banach theorem from Conway's book ( I have googled to know the answer but I have not found any result yet. Also I am not sure that whether my questions have been asked here somewhere on this forum - so please feel free to delete them if they are not appropriate )

Here are my questions:

We know that if $M$ is a linear subspace of $X$ and $f :M\to\mathbb{F}$ and $f$ is linear,bounded by a seminorm $p$ then $f$ can be extended onto $X$ by some functional $F$. Can $F$ be unique ? Under what condition $F$ will be an unique extension? It would be appreciate if you could give me one example that $F$ could not be unique.
If the above $\mathbb{F}$ is replaced a Banach space $Y$, i.e, let $M$ be a closed subspace of a Banach space $X$, and $f :M\to Y$ be a bounded linear operator, can we extend $f$ by a bounded operator $F :X\to Y$ ? if not, what condition should be put on $Y$ to have a such extension?

thanks so much

Answer by Gerald Edgar for Under what conditions a bounded linear map can be extended ?

2011-06-26T00:50:50Z

$Y$ is called an injective Banach space if the extension exists for all $X$, $M$, and $f$. An example is $Y = l^\infty$. (Should be in Banach space text books. Here's a paper: http://www.jstor.org/pss/1998210 )

Answer by godelian for Under what conditions a bounded linear map can be extended ?

2011-06-26T01:03:36Z

Note quite what you asked, but related:

Continuous extensions of (continuous) functionals from $M$ are unique if and only if $M$ is a dense subspace of $X$. Otherwise its closure is a proper closed subspace and therefore there exists a nonzero bounded functional $\phi$ vanishing on the closure, which implies that $F+\phi$ is bounded and also extends $f$.

For the second question, it is easier to put conditions on $M$ so that for every $Y$, every map from $M$ to $Y$ can be extended. As mentioned in the comments, a necessary and sufficient condition is that $M$ is a complemented subspace of $X$.