I need to find what it wrong with my logic and Ii will be glad if someone can told me what I do wrong.
Define $C$ be the subspace of $ l^{\infty} $ that consists of convergent sequences and let $C_0$ be the subspace of $C$ that consists of sequences that converge to 0.
I need to show that $C_0^*$ is isometric to $l^1$ .
I already show that $C_0$ is isometric to $l^1$ and I want to use the fact that if $Z = X \oplus Y $ then $Z^* = X^* \oplus Y^* $.
So I claim that $ C \cong C_0 \oplus \mathbb C$ because every converge sequence I can write like this:
let $\{x_n\}_{n=1} ^ \infty$ that converge to $d$ then we can look at the following sequence $y_n =(x_n- d) + d $ by this observation I conclude that $C \cong C_0 \oplus \mathbb C$ because every constant sequence we can id by it first element.
Then by what I write above I get $C^* \cong C_0^* \oplus \mathbb C ^* $.
So I get $C^* \cong l^1 \oplus \mathbb C ^*$ and to get that is isometric to $l^1$ i think that it wrong because I think that there not such a function that will be 1-1 in this case.
So my questions are
1) what I doing wrong, because I need to prove this and not to say that its wrong?
2) If all that I write is correct can I conclude from this that $C_0$ not isometric to $C$ (because $C \cong C_0 \oplus \mathbb C$)?
thank
When you use sums of Banach spaces it is better emphasize the type of norm you are using on this sum. With slight modifications your proof reads as follows $$ c^*\cong(c_0\oplus_\infty\mathbb{C})^*\cong_1 c_0^*\oplus_1\mathbb{C}^*\cong_1 \ell_1\oplus_1\mathbb{C} $$ It is remains to show that $\ell_1\oplus_1\mathbb{C}\cong_1 \ell_1$, but it is really easy - just consider isometric bijective linear operator $$ I:\ell_1\oplus_1\mathbb{C}\to\ell_1:(x_1,x_2,\ldots)\oplus_1 z \mapsto(z,x_1,x_2,\ldots) $$