Here is what I know/proved so far:
Let $c_0\subset\ell^\infty$ be the collection of all sequences that converge to zero. Prove that the dual space $c_0^*=\ell^1$.
$Proof$: Let $x\in c_0$ and let $y\in\ell^1$. We claim that $f_y(x)=\sum_{k=1}^\infty x_ky_k$ is a bounded linear functional. Clearl...
@Koro I might have missed something, but it seems that from the sequence $c_n=(-1)^n$ (which belongs to $\ell_\infty$ we get a linear functional which is unbounded.
@Koro But $c$ belongs to $\ell_\infty$. You're trying to show that for every $c\in\ell_\infty$ you get a linear continuous functional (which has the same norm as $c$.) In any case, I would suggest to continue in another chatroom so that it is not mixed with the other conversations here. (But not today - it is already late in my timezone.)
Let $\varphi$ be the map which assigns to a sequence $x=(x_n)$ the sequence of its partial sums, i.e., $\phi(x)=s$ where $s_n=\sum_{k=1}^n x_k$. (We can consider this map, for
example, on the set of all real sequences.)
Isn't then the restriction $\varphi|_X:X\mapsto c$ an isometric isomorphism between $X$ and $c$. (Here $c$ denotes the space of all convergent sequences endowed with the sup-norm.)
If $X\cong c$, then we also have $X^*\cong c^*$. And we can get dual of $X$ by looking at the dual of $c$ and transferring it via the map $\varphi^{-1}$.
