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Consider $$ T\colon\ell_1\to(c_0)', (Tx)(y):=\sum\limits_{n=1}^{\infty}x_n y_n, x=(x_n)_{n\in\mathbb{N}}\in\ell_1, y=(y_n)_{n\in\mathbb{N}}\in c_0 $$ with $$ \ell_1=\left\{(t_n): t_n\in\mathbb{K}, \sum\limits_{n=1}^{\infty}\lvert t_n\rvert<\infty\right\} $$ and $$ c_0=\left\{(t_n):t_n\in\mathbb{K},\lim\limits_{n\to\infty} t_n=0\right\}. $$ Show that $T$ is surjective.

Unfortunately I have absolutely no idea how to show the subjectivity... Can you pls help me?

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Let $\lambda \in (c_0)'$. Find a sequence $x \in \ell^1$ that represents $\lambda$ by looking at what $\lambda$ does on the subspace of $c_0$ consisting of the sequences with only finitely many non-zero terms. – Daniel Fischer Aug 27 '13 at 11:08
I do not know exactly what you mean. Let $\lambda\in (c_0)'$, i.e. $\lambda\colon c_0\to\mathbb{K}$. The aim is to find a $x\in\ell_1: Tx=\lambda$ i.e. $(Tx)(y)=\sum\limits_{n=1}^{\infty}x_n y_n=\lambda(y)~\forall~y\in c_0$. – math12 Aug 27 '13 at 11:18
Yes. Supposing you have found such an $x$, how can you characterise $x_n$? – Daniel Fischer Aug 27 '13 at 11:20
Do you mean $(Tx)(e_n)=\sum\limits_{i=1}^{\infty}x_i\cdot e_n^{i}=x_n=\lambda(e_n)$ with $e_n=(0,0,...,1,0,0,...)$ with the 1 on the n-th position? – math12 Aug 27 '13 at 11:25
Yes. So, given $\lambda \in (c_0)'$, your candidate sequence is $(x_n)$ with $x_n = \lambda(e_n)$. Now you need to show a) $x \in \ell^1$, and b) $Tx = \lambda$. – Daniel Fischer Aug 27 '13 at 11:27
up vote 1 down vote accepted

To show the surjectivity of $T$, we need to find, for every $\lambda \in (c_0)'$, an $x \in \ell^1$ with $Tx = \lambda$.

That consists of two parts,

  1. finding a candidate sequence $x = (x_n)_{n\in\mathbb{Z}^+}$,
  2. showing that the $x$ found in 1. is an element of $\ell^1$ with $Tx = \lambda$.

We use the "standard basis" $\{e^{(n)} : n\in\mathbb{Z}^+\}$, with $e^{(n)}_k = \delta_{nk}$ (with the Kronecker-$\delta$) for that.

An easy computation shows $Tx(e^{(n)}) = x_n$ for every $x\in\ell^1$, so the candidate sequence for the given (arbitrary) $\lambda$ is $x = \bigl(\lambda(e^{(n)})\bigr)_{n \in\mathbb{Z}^+}$.

To show that this sequence belongs to $\ell^1$, note that we have

$$\left\lvert \lambda\left(\sum_{n=1}^N y_n\cdot e^{(n)} \right)\right\rvert \leqslant \lVert \lambda\rVert\cdot \left\lVert \sum_{n=1}^N y_n\cdot e^{(n)}\right\rVert = \lVert\lambda\rVert\cdot\max_{1\leqslant n\leqslant N} \lvert y_n\rvert\tag{1}$$

for all choices of $y_n$. Letting

$$\rho(z) = \begin{cases}0 &, z = 0\\ \frac{\lvert z\rvert}{z} &, z \neq 0, \end{cases}$$

we note that $\lvert \rho(z)\rvert \leqslant 1$ and $\rho(z)\cdot z = \lvert z\rvert$ for all $z\in\mathbb{K}$. Hence, choosing $y_n = \rho\bigl(\lambda(e^{(n)})\bigr)$ in $(1)$, we find

$$\sum_{n=1}^N \lvert x_n\rvert = \sum_{n=1}^N \rho\bigl(\lambda(e^{(n)})\bigr)\lambda(e^{(n)}) = \lambda\left(\sum_{n=1}^N \rho\bigl(\lambda(e^{(n)})\bigr)e^{(n)}\right) \leqslant \lVert\lambda\rVert\max_{1\leqslant n \leqslant N} \bigl\lvert \rho\bigl(\lambda(e^{(n)})\bigr)\bigr\rvert \leqslant \lVert\lambda\rVert,$$

and that holds for all $N$, so $x \in \ell^1$ with

$$\lVert x\rVert = \sum_{n=1}^\infty \lvert x_n\rvert \leqslant \lVert\lambda\rVert.$$

It remains to show that indeed $Tx = \lambda$. That follows since $Tx$ and $\lambda$ are both continuous linear forms that coincide on the dense (show that!) subspace $\operatorname{span} \{ e^{(n)} : n \in\mathbb{Z}^+\}$ by construction.

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THANKS, now I understood why $x\in\ell_1$! Nevertheless I have one more question: Why is it enough to show that $Tx=\lambda$ on $\mbox{span}\left\{e_n:n\in\mathbb{N}\right\}$? Okay, it is dense, because: Let $x=(x_n)\in c_0$, then for every $\varepsilon > 0$ $\exists N: \lvert x_n\rvert\leq\varepsilon~\forall n\geq N$. Now consider $\tilde{x}=(\tilde{x}_n)$ with $\tilde{x}_n:=\sum\limits_{i=1}^{n}x_ie_i$. Then it is $\lVert\tilde{x}_n-x\rVert=\sup\limits_{n\geq N}\lvert x_n\rvert\leq\varepsilon$ - - But why is it enough to show the identity of $Tx$ and $\lambda$ on a dense subset? – math12 Aug 27 '13 at 12:58
That's a consequence of continuity. $\mathbb{K}$ is (with the usual topology) a Hausdorff space, so $\mu^{-1}(\{0\})$ is a closed subspace of $c_0$ for all $\mu \in (c_0)'$ (as the inverse image of a closed subspace of $\mathbb{K}$). Take $\mu = Tx - \lambda$ to see that $\ker \mu$ is a closed subspace containing the dense $\operatorname{span} \{e^{(n)}\}$, hence is the entire space. – Daniel Fischer Aug 27 '13 at 13:03
Ok, I think I do not understand that in all depth; nevertheless I want to show that $Tx=\lambda$ on $\mbox{span}\left\{e_n:n\in\mathbb{N}\right\}$. So let $y=(y_n)\in c_0$, then $\lambda(y)=\lambda(\lim\limits_{N\to\infty}\sum_{i=1}^{N}y_ie_i)=\lim\limits_{N‌​\to\infty}\sum_{i=1}^{N}y_i\lambda(e_i)$ because of the continuty/ linearity of $\lambda$. But now? – math12 Aug 27 '13 at 13:32
$= \lim\limits_{N\to\infty} Tx\left(\sum_{i=1}^N y_ie_i\right) = Tx(y)$, the first equality because $Tx\lvert_{\operatorname{span}\{e_i\}} = \lambda\lvert_{\operatorname{span}\{e_i\}}$, the second is the continuity of $Tx$. – Daniel Fischer Aug 27 '13 at 13:35
Ah, sorry, misunderstood. We have $Tx(e_i) = x_i = \lambda(e_i)$ by construction of $x$. So by linearity, we have $Tx(y) = \lambda(y)$ for each $y$ of the form $\sum_{i=1}^N y_i\cdot e_i$. But those are precisely the elements of $\operatorname{span}\{e_i\}$, the finite linear combinations of the $e_i$. The so-called "standard basis" is not a basis in the algebraic sense (a Hamel basis), it's a "basis" in a topological sense (a Schauder basis, if you want to look it up). – Daniel Fischer Aug 27 '13 at 13:52

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