# Application of Hahn Banach to show there exists $T \in BL(X,Y)$

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 $$\ell_{n}:u \mapsto (Su)(n)$$.

• $(Su)(n)$ denotes the $n$-th entry of $Su \in \ell^\infty$ so $Su(n)$ belongs to the $\mathbb{C}$ (or $\mathbb{R}$ if your Banach spaces are over $\mathbb{R}$). This means that $\ell_n$ is actually a linear functional defined on $U$. You should try extending that functional to all of $X$. – Rhys Steele Jul 26 at 17:34
• Try considering $T: X \to Y$ defined by $(Tx)(n) = k_n(x)$ – Rhys Steele Jul 26 at 18:51
• @RhysSteele I've updated my answer, did I get it right? – MinaThuma Jul 27 at 13:13
• Yes, this is all looks good. You should post this work as an answer to the question – Rhys Steele Jul 27 at 14:23

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 $$\vert \vert T \vert \vert = \vert \vert S \vert \vert$$.