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I have a question regarding dual spaces.

We know that the dual space of $\ell^1$ is isomorphic to $\ell^\infty$, and the dual space of $c_0$ is isomorphic to $\ell^1$. Here $c_0$ refers to the normed space of sequences of (real or complex) numbers that converge to $0$, and the norm is induced by the sup norm of $\ell^\infty$.

How do I show explicitly that the canonical mapping for $c_0$ is not surjective?

Kindly provide the details as I am new to functional analysis.

Thank you very much for the response.

Zev Chonoles
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    You've said it: take some $l^\infty$ sequence which isn't $c_0$ (e.g., a constant sequence); it defines a continuous functional on $l^1$ which isn't derived from an element of $c_0$. – Jonathan Y. Feb 16 '14 at 16:25
  • Thank you. Is it something like: Take a constant non-zero sequence $(b){n\in \mathbb{N}}$. A linear functional on $l^1$ will be $g(x)=\sum{k=1}^{\infty}b\alpha_k$, where $(\alpha_k){k\in\mathbb{N}}\in l^1$. A linear functional on $c_0$ will be $f(x)=\sum{k=1}^{\infty}\gamma_k\alpha_k$, where $(\gamma_k){k\in\mathbb{N}}\in c_0$ and $(\alpha_k){k\in\mathbb{N}}\in l^1$. Since $(\gamma_k){k\in\mathbb{N}}\in c_0$, $(\gamma_k){k\in\mathbb{N}}$ is bounded and hence $(\gamma_k){k\in\mathbb{N}}\in l^\infty$. $(b){n\in \mathbb{N}}$ does not belong to $c_0$. Hence mapping not surjective. – Novice Feb 16 '14 at 16:59

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