Let $F$ be a Banach space and $E$ be a subspace of $F$. Let $e_{0}\in E$ be an element of norm $ 1$ and suppose that

span $\{f\in F^{*}:\|f\|=f(e_{0})=1\}=F^{*}$, where $F^{*}$ is the dual space of $F$ (such an element $e_{0}$ of $F$ is called a unitary element of $F$). Then it is to be shown that

span $\{f\in E^{*}:\|f\|=f(e_{0})=1\}=E^{*}$, i.e. $e_{0}$ is a unitary element of $E$.

For this, let $f\in E^{*}$. Then $f$ can be extended to an element $g$ of $F^{*}$ with $\|f\|=\|g\|$ by the Hahn- Banach theorem. By the assumption,

$g=\sum_{i=1}^{n}\alpha_{i}g_{i}$, with $g_{i}\in F^{*}, \|g_{i}\|=g_{i}(e_{0})=1.$

Now each $g_{i}\in F^{*}\subseteq E^{*}$, hence $f$ can be written as a linear combination as above. However, we may not have $\|g_{i}\|_{E}=1$, only $\|g_{i}\|_{E}\leq 1$.

Is there a different way to prove that $f$ can be written as a linear combination of elements $f_{i}$ of $E^{*}$ with $f_{i}(e_{0})=\|f_{i}\|=1$?

  • $\begingroup$ is $e_0$ supposed to have norm 1? $\endgroup$ – user178826 Aug 28 '15 at 10:38
  • $\begingroup$ Please, Arundhathi, review my answer $\endgroup$ – user178826 Aug 28 '15 at 10:52

Is $e_0$ supposed to have norm 1? Notice that if we suppose that, we obtain $1=|g_i(e_0)|\leq \Vert g_i\Vert_E\Vert e_0\Vert=\Vert g_i\Vert$, hence $\Vert g_i\Vert_E =1$.

  • $\begingroup$ Yes, $e_{0}$ is of norm one! Sorry I left that out of the question. Thank you very much! $\endgroup$ – Arundhathi Aug 28 '15 at 11:25

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