Let $X,Y$ be Banach spaces and $U$ a closed subspace of $X$, and further $S\in BL(U,Y)$.
Show in the case $Y=\ell^{\infty}$ that there exists a $T\in BL(X,Y)$ so that $T\vert_{U}=S$ and $\vert\vert T \vert \vert = \vert \vert S \vert \vert$
As a hint, I am supposed to use Hahn Banach on the functionals $\ell_{n}:u \mapsto (Su)(n)$.
Consider any $u \in U$, we have that $\ell_{n} $ defines a functional on $U$ in the $\ell_{n}(Su)=(Su)(n)$ so the projection onto $n-$th coordinate. since $Su\in \ell^{\infty}$ for all $u \in U$, we have that $\vert \ell_{n}(u)\vert=\vert (Su)(n)\vert\leq \vert\vert S\vert \vert\times \vert \vert u\vert\vert_{\infty}$ so $\vert\vert \ell_{n} \vert \vert_{*}\leq\vert\vert S\vert \vert$ so for any $n \in \mathbb N$ we have $\ell_{n}$ is a bounded linear functional. All of these $\ell_{n}$ can be extended to $X$ by Hahn-Banach while maintaining the same norm, i.e. extend $\ell_{n}$ to $k_{n}:X \to \mathbb K$ where $\vert\vert k_{n}\vert\vert_{*}= \vert\vert \ell_{n}\vert \vert_{*}$.
Now consider the operator $T: X \to \ell^{\infty}$ where $x \mapsto (k_{n}(x))_{n}$
Note that for any $u \in U$ we have that $Tu=(k_{n}(u))_{n}=(\ell_{n}(u))_{n}=(Su(n))_{n}=Su$ so $T\vert_{U}=S$
Now that show that $\vert \vert T \vert \vert =\vert\vert S \vert \vert$.
$T$ is an extension of $S$ so $\vert \vert T \vert \vert \geq\vert\vert S \vert \vert$ is trivial. And for $\leq$:
Note that for any $\vert \vert x\vert \vert=1$, we have that $\vert\vert Tx\vert \vert_{\infty}= \vert\vert (k_{n}(x))_{n}\vert \vert_{\infty}\leq\sup\limits_{n \in \mathbb N} \vert \vert k_{n}\vert\vert_{*}=\sup\limits_{n \in \mathbb N} \vert \vert \ell_{n}\vert\vert_{*}\leq \vert \vert S\vert \vert_{*}$
Hence $\vert \vert T \vert \vert = \vert \vert S \vert \vert$.