Extension of a bounded linear operator '' Let T be a bounded linear map of a subspace F into B(S), where B(S) denote the space of bounded functions on a set S, and F is a subspace of a normed space E. Prove that there is an extension S from E to B(S) of T with ||S|| = ||T||.
Prove also that a subspace of a normed space which is isomorphic to B(S) is complemented.'' 
I have this question. For the first part I know that I must use Hahn Banach Theorem but here T is not a functional. So, I could not use Hahn Banach directly. How can I use it? Can you help me? For the second part, we can write E = F + span(E\F) in a direct sum. Then how can we use the extension S of T to show that a subspace of a normed space which is isomorphic to B(S) is complemented?
 A: For each $s \in S$ define $T_s(x)=T(x) (s)$. This is a continuous linear functional on $F$. By Hahn - Banach Theorem there exists a continuous linear functional $\overset {-} {T_s}$  on $E$ such that $\overset {-} {T_s}=T_s$ on $F$ and $\sup \{|\overset {-} {T_s}(x):\|x|| \leq 1\} =\|T_s||$. Define $\overset {-} {T}$ by $\overset {-} {T}(x) (s)=\overset {-} {T_s}(x)$. For any $s \in S$ and any $x$ with $\|x\|\leq 1$ we have $|\overset {-} {T}(x) (s)| \leq \|\overset {-} {T_s}\|=\|T\|$. It follows that $\|\overset {-} {T}\| \leq \|T\|$. The reverse inequlaity is automatic because $\overset {-} {T}$ extends $T$. 
Second part: let $T:M \to B(S)$ be linear isomorphism where $M$ is a subspace of $X$. Let $U$ be an extension of $T$ to $X$ with $\|U\|=\|T\|$(whose existence has just been proved). Claim: $X$ is the direct sum of $M$ and $ker(U)$. For this pick any $x \in X$ and let $y=Ux$. Since $T$ is an isomorphism there exists $z \in M$ such that $Tz=y$. Since $Ux=y=Tz=Uz$  it follows that $x-z \in ker (U)$. Since $x =(x-z)+z$ we have proved that $X =ker (U)+M$. Now we prove that $ker (U)\cap  M=\{0\}$. Let $x \in  ker (U)\cap  M$. Then $Tx=Ux=0$. Since $T$ is injective we get $x=0$. This completes the proof. 
