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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.

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

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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 $\|S\|=\|T\|$.

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  • $\begingroup$ Thanks for your help, but I actually made an error in the original post so your answer was based off of that error, apologies $\endgroup$
    – turnip_man
    Feb 15, 2022 at 23:31
  • $\begingroup$ 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 $\endgroup$
    – Kavi Rama Murthy
    Feb 15, 2022 at 23:48
  • $\begingroup$ 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 $T(E)$? $\endgroup$
    – turnip_man
    Feb 16, 2022 at 10:36
  • $\begingroup$ @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 $\ell^{\infty}$. $\endgroup$
    – Kavi Rama Murthy
    Feb 16, 2022 at 11:31
  • $\begingroup$ Thanks for your explanation, just one last point, how do know that $Sx$ is a bounded sequence? $\endgroup$
    – turnip_man
    Feb 16, 2022 at 12:18

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