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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:

  1. 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.)
  2. In the same sense, $(\ell^1)^* = \ell^\infty$.
  3. 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 $c_0$.
  4. $c_0$ and $\ell^1$ are separable, but $\ell^\infty$ is not.

My work:

  1. 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? $\color{blue}{\text{Update: Done.}}$
  2. 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. $\color{blue}{\text{Update: Done.}}$
  3. $\color{blue}{\text{Update: Done.}}$ (Thanks to the accepted answer.)
  4. $\color{blue}{\text{Update: Done.}}$

Thanks a lot!

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  • $\begingroup$ (4) The subsets if $\mathbb{N}$ is uncountable. If $A,B\subset\mathbb{N}$ and $A\neq B$, then $\|\mathbb{1}_A-\mathbb{1}_B\|_\infty=1$. $\endgroup$ Jul 11, 2021 at 13:59
  • $\begingroup$ @OliverDiaz Thanks for the comment. That appears later as Theorem $6.16$, which proves that $L^q(\mu)$ is isometrically isomorphic to the dual space of $L^p(\mu)$, for $1 \le p < \infty$. I agree, that immediately gives $(\ell^1)^* = \ell^\infty$ after taking $\mu$ as the counting measure on $\mathbb N$. Very nice observation! $\endgroup$ Jul 11, 2021 at 13:59
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    $\begingroup$ (1) Given $L\in c^*_0$, let $\gamma_n=L(\mathbf{e}_n)$. For $f\in c_0$ define $f_n=\sum^n_{k=1} f(k)\mathbf{e}_k$. Show that $\|f_n-f\|_\infty\rightarrow0$. Then, notice that $L(f)=\sum_n\gamma_nf(n)$ $\endgroup$ Jul 11, 2021 at 14:24
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    $\begingroup$ @epsilon-emperor: The collection of rational finite linear combinations of the $\mathbf{e}_n$ defined by OliverDiaz is a countable dense set in both $\mathcal{c}_0$ and $\ell_1$ (or any $\ell_p$ with $0<p<\infty$ for that matter) $\endgroup$ Jul 11, 2021 at 20:24
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    $\begingroup$ @epsilon-emperor: it does prove subjectivity. Notice that it gives an $ll_1$ representative for any functional in $(\mathcal{c}_0)^*$. $\endgroup$ Jul 11, 2021 at 20:38

2 Answers 2

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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

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  • $\begingroup$ The answer looks great. I'm still trying to prove that there is no $f\in \ell^1$ for which $\Lambda = \Lambda_f$. Suppose such an $f\in \ell^1$ exists, to get a contradiction. Since $\Lambda_f$ is a functional on $\ell^\infty$, and $c \subset \ell^\infty$, I think when you say $\Lambda = \Lambda_f$, we are really considering the restriction of $\Lambda_f$ to $c \subset\ell^\infty$. We would get $$\Lambda_f x = \sum_n f(n) x(n) = \lim_{n\to\infty} x(n) = \Lambda x \quad (x\in c)$$ How do we find a contradiction? $\endgroup$ Jul 13, 2021 at 20:54
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    $\begingroup$ @epsilon-emperor: yes, those pesky typos... I just fixed it. The most interesting part of this problem is that this types of examples can only be realize with the help of Hand-Banach's theorem, which is a weak version (in a way) of the axiom of choice. $\endgroup$ Jul 13, 2021 at 20:54
  • $\begingroup$ The Hahn Banach Theorem is a weak version of AOC? I won't be able to sleep tonight. I'm curious. $\endgroup$ Jul 13, 2021 at 20:55
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    $\begingroup$ Just take a look at the link that I sent. There is an answer by GEdgard that gives a short overview of the fact. TO get into all the details is to go into the gusts of Set Theory and, unless you want to pursue research in Logic, just known the fact is enough. Later on, if you ever study Logic and Set theory, you can take a look at those things, $\endgroup$ Jul 13, 2021 at 20:59
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    $\begingroup$ @epsilon-emperor: Yes, that is basically the end of the story. One can actually show that the extension $\Lambda$ can be taken to be nonnegatoive, that is $\Lambda x\geq0$ if $x\in\ell^+_\infty$. There is a big theorem that gives a general answer due to Kantorovich. A more elementary treatment for this fact (positivity) can be found in Conway's book on functional analysis (under Banach limits) $\endgroup$ Jul 13, 2021 at 21:02
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Concerning part 3 (following what's mentioned in the comments), the form of the question indicates that you can use Hahn--Banach somewhere. Think of extending a function which is $0$ on $c_0$ and is bounded above by the functional $\limsup_n x$ (which is sublinear).

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