Let $X$ be a Banach space. A projection $P$ is a continuous map $:P:X\to X$ such that $P^2=P$ The existence of a projection it's equivalent to the decomposition of $X= M \oplus N$ where $M,N$ are "closed" subspaces of $X$ in fact if $P$ exist then $M= Ker (P-I)$ and $N= Ker P$ (are clearly closed). And if there exist such a decomposition we can define $P(m+n)=m$ and prove that it's a projection.
I'm trying to prove that the identity operator on $c_0$ (i.e the space of sequence that converges to 0) does not have a continuous linear extension to $l^{\infty}$ except from the identity itself.
I may use that there exist no projection (remember that it's continuous) from $l^{\infty}$ to $c_0$
When we say that $ P:X\to X$ is a projection from $X$ to $Y\le X$ we mean that $\mathrm{Image}(P)=Y$
Clearly I want to show that the existence of such extension implies the existence of a continuous projection, but I'm a little stuck. Let me explain why: Let $P: l^{\infty} \to l^{\infty}$ such that $P$ be such continuous extension, such that $P\ne I$ i.e $Ker (P-I)$ it's a proper and closed subset of $l^{\infty}$. Clearly $ c_0 \subset Ker(P-I)$ but if it's a proper subset, then the function $P$ only defines a projection from $l^{\infty}$ to $Ker(P-I)$, and that's not a contradiction. Please help me to construct a projection to $c_0$ :(
This is not an exact duplicate of the other question, I have to use that result to prove this new one..... but needs a trick, I explained the reason....
Such extensions exist. Let $\phi:l^\infty\to \mathbb R$ be a nonzero bounded linear functional such that $c_0\subset \ker \phi$. Then $x\mapsto x+\phi(x)e_1$ is a continuous linear operator on $l^\infty$ whose restriction to $c_0$ is the identity map. The vector $e_1=(1,0,\dots,0)$ here could be replaced with any other nonzero vector in $l^\infty$.
The construction of $\phi$ is standard: first define $\phi(x)=\lim_{n\to\infty} x_n$ for the convergent sequences $x$, then extend by Hahn-Banach.