Properties of dual spaces of sequence spaces

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!

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.