# Extension of linear operator into $l^\infty$

Stuck on this homework question, would appreciate any help thanks.

Suppose $$E$$ is a normed space and $$E_0$$ is a subspace of $$E$$, all over a field $$\mathbb{K}$$. Let $$T_0 : E_0 \rightarrow l^\infty$$ be a bounded linear operator, where as usual $$l^\infty$$ is the banach space of all bounded sequences in $$\mathbb{K}$$ with norm $$||(x_n)_{n \in \mathbb{N} } || = sup_{n \in \mathbb{N}} |x_n |$$. Then use the Hahn–Banach theorem to show that there is an extension of $$T_0$$ to $$E$$, call it $$T$$, where $$|| T || = || T_0 ||$$.

I think this would be really easy if instead of $$l^\infty$$ it was just $$\mathbb{K}$$ ( precisely the statement of the hahn-banach theorem) but not sure how this still works when we change to a different normed space.

• $E_0$ is a subspace of $E$ and $T_0 : E_0 \rightarrow l^\infty$ is just a bounded linear operator from $E_0$ into $l^\infty$. Commented Feb 15, 2022 at 23:23

Let $$T_nx$$ be the $$n-$$th coordinate of $$Tx$$. There exists an extension $$S_n$$ of $$T_n$$ to $$E$$ with $$\|T_n\|=\|S_n\|$$. Let $$Sx=(S_1x,S_2x,...)$$. It is fairly easy to check that $$S$$ an extension with $$\|S\|=\|T\|$$.
• My answer is correct. I have changed notations slightly: my $T$ is your $T_0$ and my $S$ is your $T$. That should not cause any confusion. @turnip_man Commented Feb 15, 2022 at 23:48
• I'm a bit confused, just wondering if you could clear up the following; $Tx$ is a bounded sequence in $l^\infty$, so when you take the $n^{th}$ coordinate, are you mapping from $l^\infty$ into $\mathbb{K}$? In that case, when using Hahn-Banach is the domain starting from $T(E_0)$ and being extended to $T(E)$? Commented Feb 16, 2022 at 10:36
• @turnip_man You are extending the $n-th$ coordinate from $E_0$ to $E$. Do this for each $n$ and put the extended maps from $E$ into $F$ together to get a new map from $E$ into $\ell^{\infty}$. Commented Feb 16, 2022 at 11:31
• Thanks for your explanation, just one last point, how do know that $Sx$ is a bounded sequence? Commented Feb 16, 2022 at 12:18