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Let $0 \leq p < \infty$ and $e_1=(1,0,0,...) \in \ell^p$. Let $W=\mathbb{F}e_1$, and define $f_W=W \to \mathbb{F}$ by $f_w(\alpha e_1)=\alpha$. It is clear that $f_W \in W'$ and that $\lVert f_W\rVert=1$ hence we know that $f_W$ can be extended to $f \in (\ell^p)'$ such that $\lVert f\rVert=1$. Describe all such extensions in term of $\ell^q$ with $q$ is conjugate exponent of $p$.

I have shown that $f_W$ is not unique, but I got stuck finding all such extensions in term of $\ell^q$.

davyjones
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I recall some basic properties of $\ell^p$ spaces. If $p<1$ then the space $\ell^p$ is not normed (it is not even locally convex), so we may not apply to it Hahn-Banach Theorem. Also in this case a number $q$ such that $1/p+1/q=1$ is negative if $p\ne 0$ and if $p=0$ then the “norm” $\|x\|_0$ of an element $x=(x_n)\in\ell^0$ should be $\max\{x_n^0\}$, which is $1$ or undefined. So further I’ll assume that $p\ge 1$. It is well-known that in this case the dual space $(\ell^p)’$ of all bounder linear functionals on $\ell^p$ is the space $\ell^q$, where $1/p+1/q=1$. Let $y=(y_n)\in\ell^q$. Then $y(x)=\sum_{n=1}^\infty x_ny_n$ for each $x=(x_n)\in\ell^p$ and $\|y\|_{\ell_p^*}=\|y\|_{\ell_q}$ (see the above Wikipedia article for $p>1$ and a comment to this question for $p=1$). From this it is easy to see that $y$ is an extension of $f_W$ iff $\|y\|_{\ell_q}$ and $y_1=1$.

Alex Ravsky
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    where is the $y_1=1$ from? @alex –  Nov 26 '17 at 18:22
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    The restriction of $y$ to $W$ should be $f_W$. So we need $y(\lambda e_1)=\lambda$. Now $y(x)=\sum_n x_n y_n$ so we get $y(\lambda e_1)=\lambda y_1=\lambda$. –  Dec 03 '17 at 15:16