Problem $5.9$, Rudin's Real and Complex Analysis.
Let $c_0$, $\ell^1$, and $\ell^\infty$ be the Banach spaces consisting of all complex sequences $x = \{\xi_i\}$, $i = 1,2,3,\ldots$, defined as follows:
$$x\in \ell^1 \text{ if and only if } \|x\|_1 = \sum |\xi_i| < \infty.$$
$$x\in \ell^\infty \text{ if and only if } \|x\|_\infty = \sup|\xi_i| < \infty.$$
$c_0$ is the subspace of $\ell^\infty$ consisting of all $x\in \ell^\infty$ for which $\xi_i \to 0$ as $i\to\infty$. Prove the following four statements:
- If $y = \{\eta_i\} \in \ell^1$, and $\Lambda x = \sum \xi_i \eta_i$ for every $x\in c_0$, then $\Lambda$ is a bounded linear functional on $c_0$, and $\|\Lambda\| = \|y\|_1$. Moreover, every $\Lambda\in (c_0)^*$ is obtained in this way. In brief, $(c_0)^* = \ell^1$. (More precisely, these two spaces are not equal; the preceding statement exhibits an isometric vector space isomorphism between them.)
- In the same sense, $(\ell^1)^* = \ell^\infty$.
- Every $y\in \ell^1$ induces a bounded linear functional on $\ell^\infty$, as in $(a)$. However, this does not give all of $(\ell^\infty)^*$, since $(\ell^\infty)^*$ contains nontrivial functionals that vanish on all of $c_0$.
- $c_0$ and $\ell^1$ are separable, but $\ell^\infty$ is not.
My work:
- I have shown that $\|\Lambda\| = \|y\|_1$. It remains to show that there is an isomorphism between $(c_0)^*$ and $\ell^1$. $\|\Lambda\| = \|y\|_1$ tells us that the map $T: y\mapsto \Lambda$ from $\ell^1$ to $(c_0)^*$ is an isometry, hence it is injective. How do I show that it is also surjective? $\color{blue}{\text{Update: Done.}}$
- I start with similar construction as in the previous part. Consider $T: y\mapsto \Lambda$ such that if $y = \{\eta_i\} \in \ell^\infty$, then $\Lambda x = \sum \xi_i \eta_i$ for every $x\in \ell^1$. Linearity of $\Lambda$ is clear, and for boundedness I get $\|\Lambda\| \le \sup_i |\eta_i|$. I haven't yet been able to show that $\|\Lambda\| = \sup_i |\eta_i|$. Once this is done, injectivity is once again established as earlier, and the surjectivity bit remains. $\color{blue}{\text{Update: Done.}}$
- $\color{blue}{\text{Update: Done.}}$ (Thanks to the accepted answer.)
- $\color{blue}{\text{Update: Done.}}$
Thanks a lot!
Best Answer
The first part of (3) is just an application of Holder's inequality:
If $f\in\ell_1$, define $\Lambda_f:\ell_\infty\rightarrow\mathbb{C}$ as $\Lambda_f x=\sum_n f(n)x(n)$. Clearly $f\mapsto\Lambda_f$ is linear, and $\Lambda_f$ is linear on $\ell_\infty$. By Hölders's inequality $$|\Lambda_f x|\leq\|f\|_1\|x\|_\infty$$ and so, $\Lambda_f\in\ell_1$. Furthermore, by taking $x\in\ell_\infty$ such that $x(n)f(n)=|f(n)|$ and $|x(n)|=1$, we have that $\|\Lambda_f\|=\|f\|_1$.
For the last part, consider the subspace $\mathcal{c}\subset\ell_\infty$ of sequences in $\mathbb{C}$ that are convergent, and define $\Lambda:\mathcal{c}\rightarrow\mathbb{C}$ as $\Lambda x=\lim_{n\rightarrow\infty}x(n)$. This is a linear functional and clearly $|\Lambda x|\leq\|x\|$. In fact, by taking $x\equiv1$, we get that $\|\Lambda\|_{\mathbf{c}}=1$. An application of the Hahn-Banach theorem (see Theorem 5.16 in Rudin's Real and Complex Analysis, 3rd edition) shows that $\Lambda$ can be extended to all of $\ell_\infty$ as a bounded linear functional so that $\|\Lambda\|=1$. Thus $\Lambda\in(\ell_\infty)^*$. It is not difficult to see that $\Lambda$ has no representation as an element of $\ell_1$, that is, for no $f\in\ell_1$ is $\Lambda=\Lambda_f$.
Comment: The use of the Hahn Banach theorem seems inevitable. See this posting and links