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Can you tell me if I got the following homework right? Nitpicking is welcome.

a) Recall that $$ c_0 (\mathbb{N}) = \{ f: \mathbb{N} \rightarrow \mathbb{C} \mid \lim_{n \rightarrow \infty } f(n) = 0\} \subset l^\infty ( \mathbb{N}) $$ is a Banach space with respect to the supremum norm $\| . \|_\infty $. Show that $(c_0(\mathbb{N}))^\ast \cong l^1(\mathbb{N})$ where the dual pairing is given by $$ \langle f, g \rangle = \sum_{n = 0}^\infty f(n) g(n)$$ for $f \in c_0(\mathbb{N})$ and $g \in l^1(\mathbb{N})$.

b) Show that $(l^1(\mathbb{N}))^\ast \cong l^\infty(\mathbb{N})$

c) Compute the dual of $c(\mathbb{N}) = \{ f: \mathbb{N} \rightarrow \mathbb{C} | \lim_{n \rightarrow \infty} f(n) $ exists $\}$

d) Show that $l^1$ is not reflexive by showing that a Banach limit does not come from a pairing as above with an element of $l^1$

My answers:

a) We want to show that $\varphi : l^1(\mathbb{N}) \rightarrow (c_0 (\mathbb{N}))^\ast$ defined by $g \mapsto \langle ., g\rangle$ is an isomorphism. First, I think, we need to verify that it maps into $(c_0 (\mathbb{N}))^\ast$. So let $f \in c_0(\mathbb{N})$. Then $$ | \langle f,g \rangle | = \Big | \sum_{n=0}^\infty f(n) g(n) \Big | \leq \sum_{n=0}^\infty |f(n)| |g(n)| \leq \| f \|_\infty \sum_{n=0}^\infty | g(n)| < \infty$$

Next we need to verify that it's a vector space homomorphism, that is, it's linear. For this, let $\alpha \in \mathbb{C}, g \in l^1(\mathbb{N})$ and $f_1 , f_2 \in c_0(\mathbb{N})$. Then $$ \langle \alpha (f_1 + f_2), g \rangle = \sum_{n=0}^\infty \alpha (f_1 + f_2)(n) g(n) = \alpha \sum_{n=0}^\infty f_1(n)g(n) + \alpha \sum_{n=0}^\infty f_2(n)g(n) = \alpha \langle f_1 , g \rangle + \alpha \langle f_2 , g \rangle$$

Next we need to show that $\varphi$ is injective. For this let $g \in l^1(\mathbb{N})$ such that $\langle f,g \rangle = 0$ for all $f$ in $c_0(\mathbb{N})$. Then in particular, $\langle f_N,g \rangle = 0$ for $$ f_N (n) := \begin{cases} 1 & n = N \\ 0 & otherwise \end{cases}$$ So $g(n) = 0$ for all $n$ and so $\ker \varphi = \{ 0 \}$.

The last thing to verify is that $\varphi$ is surjective. Consider any $\lambda \in c_0(\mathbb{N})^\ast$. For the next step $\mathbb{N}$ should be locally compact and Hausdorff. This would be the case if $\mathbb{N}$ had the discrete topology but the topology is not specified in the homework so I'm not so sure about the following:

By Riesz-Markov there exists a unique regular countably additive complex Borel measure $\mu$ on $\mathbb{N}$ such that $\varphi(f) = \int_X f(x) d \mu$. Set $g(n) := \mu(n)$ then $\varphi(f) = \int_X f(x) d \mu = \sum_{n=0}^\infty f(n) g(n)$. $g(n) \in l^1$ because $\mu$ is regular and therefore has finite measure on each compact set, in particular $\{ n \}$.

b) Here we need to verify that $\varphi^\prime: g \mapsto \langle ., g \rangle$ is an isomorphism where $g \in l^\infty(\mathbb{N})$. By the same arguments as in a) I showed that it maps into $(l^1 (\mathbb{N}))^\ast$, that it's a homomorphism and that it's injective. To verify that it's surjective, consider any $\lambda : l^1(\mathbb{N}) \rightarrow \mathbb{C}$. Then $\lambda$ can be split into positive real functions as follows: $\lambda = \Re (\lambda) + i \Im(\lambda) = \Re (\lambda)^+ - \Re (\lambda)^- + i \Im(\lambda)^+ - i \Im(\lambda)^-$. Then for each part separately, there exists a unique regular measure by the same argument as in a), so $\varphi^\prime$ is surjective.

c) To compute $(c(\mathbb{N}))^\ast$ consider an $s \in c(\mathbb{N})$. Then $\lim_{n \rightarrow \infty} s_n = k$ for some $k \in \mathbb{C}$ and $\lim_{n \rightarrow \infty}(s_n - k) = 0$ so $(s_n - k)$ is an element of $c_0(\mathbb{N})$. Now I'm not sure about my next step. I suspected that if $V, V^\prime$ are isomorphic vector spaces then their duals $V^\ast$ and $(V^\prime)^\ast$ are isomorphic. So I constructed a bijective linear function from $c(\mathbb{N})$ into $c_0(\mathbb{N}) \times \mathbb{C}$ as follows: $s_n \mapsto (s_n - k, k)$ and verified that it's bijective and linear.

d) To show that $l^1$ is not reflexive I used Kakutani's theorem which says that $l^1$ is reflexive if and only if the closed unit ball is weakly compact. To find an open cover of the unit ball that doesn't contain a finite subcover pick $B_1 (f) $, the ball of radius $1$ around $f$ for all $f$ with $\| f \| = 1$. Then any two such $f$ have distance $2$ and therefore no $f$ is in any $B_1 (f^\prime) $ for any $f^\prime$ with $\| f^\prime \| = 1$. There is an infinite number of $f$'s with $\| f \| = 1$ so any finite collection of balls $B_1 (f) $ leave an infinite number of $f$'s uncovered. So $l^1$ is not reflexive.

Thanks for your help!

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First a few comments on a): in the verification that $\varphi$ maps into $l^1$ you dropped some absolute value signs. In the last step, of course $\mathbb{N}$ has the discrete topology. But appealing to Riesz-Markov is serious overkill: evaluate a linear functional at the standard basis vectors (which you denoted $f_n$, here $e_n$ would be more standard), get a sequence $g = (g_n)_{n=0}^{\infty}$. Show that it is summable by pairing it with finite sequences $h_{N}$ adjusting the signs of the $g_n$. Finally, congratulations on how you changed your writing, this is a pleasant surprise! –  t.b. Nov 20 '11 at 12:46
    
@t.b.: I corrected the typos. Why 'of course'? It's not obvious to me. Thank you, I like my writing much better now than before you helped me change it. –  Matt N. Nov 20 '11 at 12:54
    
(I'll address your questions in a minute) A few LaTeX tips first: For writing sets $\{x \in X \mid x \text{ is blah}\}$ it looks a bit nicer to use \mid instead of |: compare the previous set with $\{x \in X | x \text{ is blah}\}$. For the dot as a placeholder it's more standard to use \cdot instead of a plain period. To insert text into a formula, use the \text{my explanation} (note that you can adjust spacing by starting/ending it with a blank, so e.g. \text{ my explanation }. –  t.b. Nov 20 '11 at 12:57
    
"of course" because it's the natural topology and as was pointed out to you elsewhere $c_0$ is the space of continuous functions vanishing at infinity on $\mathbb{N}$ with the discrete topology. This wasn't mentioned in the exercise because it is not intended that you use any heavy machinery. Evaluating a functional $\varphi$ at the standard basis functions $e_n$ gives you a sequence $(\varphi(e_n))_{n=0}^\infty$ for which you can then verify directly that it does what you want. This applies to b), too, where I don't quite understand where you get the measures from - you're no longer working.. –  t.b. Nov 20 '11 at 13:06
    
..with a space of continuous functions, but rather with a space of integrable functions. Further comments will follow in an answer. –  t.b. Nov 20 '11 at 13:06
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1 Answer

up vote 7 down vote accepted

The work on a) is perfectly fine up to the point where you verify that $\varphi: \ell^1(\mathbb N) \to c_0(\mathbb N)$ is surjective. Of course, you can argue with Riesz-Markov, as you did, but that's cracking a nut with a sledgehammer. It can be done in a completely straightforward way:

Let $\lambda: c_0(\mathbb N) \to \mathbb{C}$ be a continuous linear functional. Put $g_n = \lambda(e_n)$, where $$e_{n}(k) = \begin{cases} 1, & \text{if } n = k, \\ 0 & \text{otherwise.} \end{cases}$$ Then $g = (g_n)_{n=0}^{\infty}$ is a sequence and you want to show it to be summable, hence $\lambda = \varphi(g)$ using the pairing you describe. To this end, let $\alpha_{n} = \frac{\overline{g}_n}{|g_n|}$ if $g_n \neq 0$, otherwise let $\alpha_n = 0$. The sequence $f_N = (\alpha_0, \alpha_1,\ldots,\alpha_N,0,0,\ldots)$ is in $c_0(\mathbb N)$, has norm $\|f_N\|_\infty \leq 1$ and $$ \sum_{n = 0}^N |g_n| = \lambda(f_N) \leq \|\lambda\|, $$ so the sequence $g$ is indeed in $l^1(\mathbb{N})$ and as mentioned before $\lambda = \varphi(g)$.

Added: Next, you should show that $\varphi: l^1(\mathbb{N}) \to (c_0)^\ast$ is an isometric isomorphism. To prove this, argue that $\varphi$ is a bijective isometry: you've shown that $\varphi$ is of norm $\leq 1$ in your very first argument and combing your work with my argument above shows that it is bijective. To see that it is isometric, use the pairing as I did above to see that $\|g \| \geq \|\varphi(g)\| \geq \langle g,f_N \rangle \geq \|g\| - \varepsilon$ for $N$ large enough. (No need to appeal to the open mapping theorem here.)


In b) you appeal to the Riesz-Markov theorem again. I don't understand how you do that, since $l^1(\mathbb{N})$ is not a space of continuous functions with the $\sup$-norm. In the spirit of your solution of a) you could appeal to the duality theory of $L^p$-spaces by identifying $l^1(\mathbb N) = L^1(\mathbb{N},\mathfrak{P}(\mathbb{N}),\#)$, where $\#$ is counting measure on the power set $\mathfrak{P}(\mathbb{N})$ of $\mathbb{N}$, but again this is serious overkill. I suggest that you try to mimic the argument I gave for a).


Your argument for c) works insofar as you can indeed show that isomorphic Banach spaces have isomorphic duals (try to do that!). However, the isomorphism you write down is not an isometric isomorphism, and indeed $c(\mathbb{N})$ and $c_0(\mathbb{N})$ are not isometrically isomorphic, as David explained in this thread.

However, the exercise asks you to identify the dual space of $c(\mathbb{N})$ and your argument only gives you that it is isomorphic to $l^1(\mathbb{N})$, but it doesn't give an explicit identification. Try to use the pairing $$ \langle f,g \rangle_{l^1,c} = f_0 \cdot (\lim_{n\to\infty} g_n) + \sum_{n=1}^{\infty} f_n \,g_n \qquad f \in l^1(\mathbb{N}), \; g \in c(\mathbb{N}) $$ and again adapt the argument I gave for a).


Concerning d): You didn't do what was asked. Using the Kakutani theorem seems to be a bit laborious(1) (also the argument using Banach limits is a bit roundabout: if $l_1 = (c_0)^\ast$ were reflexive then so were $c_0$, but $(c_0)^{\ast\ast} \cong l^\infty$ and the canonical embedding $c_0 \to (c_0)^{\ast\ast} \cong l^\infty$ is simply the inclusion, which obviously isn't surjective).

The exercise asks you to show that a Banach limit $\ell: l^\infty(\mathbb{N}) \to \mathbb{C}$ is not of the form $\ell (f) = \langle g, f \rangle_{l^1,l^\infty}$ for any $g \in l^1(\mathbb{N})$. To this end, notice that $\ell$ extends the limit functional on $c(\mathbb{N})$, hence $\ell(e_n) = 0$ for all $n$ and that would imply that $g = 0$, which it can't be because $\ell \neq 0$.

(1) Added: The problem with your argument for d) is that the norm-balls are not weakly open, so you don't get an open cover of the unit ball, so you cannot conclude this way. I do not see an easy way to fix this, that's why I said that using that theorem seems a bit laborious.


Finally, I'm seriously impressed by the improvement of your writing style. This is much more than what I expected to see. I'll add a few remarks on that later on.

Added: Nothing to add, in fact :)

The only point I don't really like is that you use the prime in your notation and this can be a bit confusing because the prime is often used for the dual space and the adjoint morphism.

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Why do I want to show that $\varphi$ is isometric in a)? –  Matt N. Nov 21 '11 at 14:23
    
1) This seems to be the easiest way to show that $\varphi$ actually is an isomorphism of Banach spaces (this way you avoid the open mapping theorem). 2) Given a Banach space $X$ its dual space $X^\ast$ comes along with a specific norm, namely $\|f\|_{X^\ast} = \sup_{\|x\| \leq 1} |f(x)|$. So, if you only exhibit $X^\ast$ up to isomorphism of Banach spaces (meaning: continuous linear map with continuous [linear] inverse) the information is incomplete. 3) If you're able to prove a more precise statement without any effort, do it! :) –  t.b. Nov 21 '11 at 14:27
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Sorry for the confusion. We are talking about two definitions of "isomorphism of Banach spaces" here. We know that $\varphi$ is bijective already, so the open mapping theorem would let you conclude that $\varphi$ has a bounded inverse (that's what David suggested in a comment above). But the open mapping theorem's much more difficult to prove and yields a weaker result than just verifying that $\varphi$ is an isometric bijection, so that $\varphi^{-1}$ must be isometric, too (in particular it is bounded) by a completely elementary argument. –  t.b. Nov 21 '11 at 16:44
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@caligula: You're right insofar as we made a somewhat arbitrary but natural choice (what you call fixing an element): We have a special linear functional $f = \lim: c \to \mathbb{C}$. The adjoint $f^\ast: \mathbb{C}^\ast \to c^\ast$ gives us a particular element $f^\ast(1) \in (c)^\ast$ (here $1$ is $1 \in \mathbb{C}$). This $f^\ast(1) \in c^\ast$ is the element we're "fixing". You should really think $c = c_0 \oplus \mathbb{C}$ and $\ell^1 \cong \ell^1 \oplus \mathbb{C} = (c_0)^\ast \oplus (\mathbb{C})^\ast = (c_0 \oplus \mathbb{C})^\ast$. The "fixing" happens where I wrote $\cong$. –  t.b. Nov 22 '11 at 1:31
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@t.b.: Isomorphic Banach spaces have isomorphic duals: Let $\varphi : V \rightarrow V^\prime$ be a linear bijective map between two Banach spaces (prime is not meant to denote dual) with $\varphi^{-1}: V^\prime \rightarrow V$ bounded and linear. Let $\lambda \in {V^\prime}^\ast$ be a linear functional. Define $\varphi^\ast: {V^\prime}^\ast \rightarrow V^\ast$ as $\lambda \mapsto \lambda \circ \varphi^{-1}$. Then $\varphi^\ast$ is an isomorphism: (i) linear, injective and surjective are clear (ii) ${\varphi^\ast}^{-1}$ is bounded follows using the open mapping thm. and surjectivity from (i) –  Matt N. Nov 24 '11 at 12:47
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