Extending operator from subspace of Banach space to whole space

Question

Let $Y$ be subspace of Banach space $X$. $T:Y \rightarrow \ell_{\infty}$ is continuous linear operator. Show that there is linear function $S:~X \rightarrow \ell_{\infty}$ such that $S\vert _{Y}=T$ and $\| S \|=\| T \|$

Answer 1

You have to use the Hahn Banach theorem.

Comsider the linear function $\phi_n$ on $l_{\infty}$ given by $\phi_n(a_1,..a_n,..)=a_n$. Then $\phi_noT$ is a linear functional on $Y$ and by the Hahn Banach theorem, it has an unique extension to a linear functional $\psi_n$(say). Define $S(x)=(\psi_1(x),..\psi_n(x),...)$. Check that $S$ extends $T$ and satisfies that $||S||=||T||$.

Space with such extension property have different names: injective, isometrically injective or $\mathcal{P}_1$-space. There is a deep result due to Nachbin, Goodner, Kelly and others stating that every such space is isometrically isomorphic to the space $C(K)$ of continuous functions on extremelly disconnected compact $K$. For example $\ell_\infty\cong_1 C(\beta\mathbb{N})$, where $\beta\mathbb{N}$ is a Stone-Chech compactification of $\mathbb{N}$.