Unsure how $(\ell^1)^*\cong\ell^\infty$

Question

I do not really understand how the dual space of $\ell^1$ is isomorphic $\ell^\infty$ in that $\ell^\infty$ is a set of sequences and a dual space is a set of functions. I know I am most likely not understanding something relatively simple but I cannot seem to wrap my head around why this is the case.

Answer 1

The way you show one vector space $W$ over $\mathbf R$ (say) can be viewed as the dual of another vector space $V$ over $\mathbf R$ is to figure out a way to make each $w \in W$ act as an $\mathbf R$-linear map $f_w \colon V \to \mathbf R$ that depends linearly on $w$ ($f_{w + w'} = f_w + f_{w'}$, $f_{cw} = cf_{w}$ in $V^*$), and make sure each $f_w$ is a continuous map on $V$ if $V$ and $W$ are Banach spaces (since for those, the dual space is defined to be continuous linear functionals, not all linear functionals).

Example 1. If $V = \mathbf R^n$ then we can make $V$ look like its own dual space by associating to each $\mathbf v$ the $\mathbf R$-linear map $f_\mathbf v \colon V \to \mathbf R$ where $f_{\mathbf v}(\mathbf x) = \mathbf v \cdot \mathbf x$ for all $\mathbf x \in V$. Each $f_\mathbf v$ is in $V^*$ and the map $V \to V^*$ where $\mathbf v \mapsto f_\mathbf v$ is linear ($f_{\mathbf v + \mathbf v'} = f_\mathbf v + f_{\mathbf v'}$, $f_{c\mathbf v} = cf_{\mathbf v}$ in $V^*$) and injective (if $f_\mathbf v = 0$ in $V^*$ then $f_\mathbf v(\mathbf x) = 0$ for all $\mathbf x$ in $V$, so in particular $0 = f_\mathbf v(\mathbf v) = \mathbf v \cdot \mathbf v$, implying $\mathbf v = \mathbf 0$). Since $V$ and $V^*$ have the same (finite) dimension, a linear injective map between them is surjective and thus an isomorphism, so every element of $V^*$ has the form $f_\mathbf v$ for a unique $\mathbf v$ in $V$. By turning each $\mathbf v$ in $V$ into $f_\mathbf v$ in $V^*$, $V$ turns into its own dual space.

Example 2. (This is the one you asked about.) For a bounded sequence $\mathbf b \in \ell^\infty$ and an $\ell^1$-sequence $\mathbf x \in \ell^1$, the "inner product" $$ \mathbf b \cdot \mathbf x := \sum_{n \geq 1} b_nx_n $$ is absolutely convergent. Write this series as $f_{\mathbf b}(\mathbf x)$, so $f_{\mathbf b} \colon \ell^1 \to \mathbf R$ is $\mathbf R$-linear. Set $b = |\mathbf b|_\infty = \max_{n \geq 1} |b_n|$. Then $$ |f_{\mathbf b}(\mathbf x)| \leq \sum_{n \geq 1} |b_n||x_n| \leq b\sum_{n \geq 1} |x_n| = b|\mathbf x|_1 = |\mathbf b|_\infty|\mathbf x|_1, $$ so $f_{\mathbf b} \colon \ell^1 \to \mathbf R$ is a bounded linear functional on $\ell^1$ and thus it is continuous (a linear map between normed vector spaces is continuous if and only if it is bounded).

Thus to each $\mathbf b \in \ell^\infty$ we have constructed a continuous linear functional $f_{\mathbf b} \in (\ell^1)^*$, so we get a function $\ell^\infty \to (\ell^1)^*$ by $\mathbf b \mapsto f_{\mathbf b}$. That is how we make elements of $\ell^\infty$ into elements of $(\ell^1)^*$. Check this mapping $\ell^\infty \to (\ell^1)^*$ is $\mathbf R$-linear and continuous. Then its injectivity is easy: show the kernel is $\{\mathbf 0\}$ by applying $f_{\mathbf b}$ to well-chosen elements of $\ell^1$. Surjectivity of $\ell^\infty \to (\ell^1)^*$ is more subtle (the only "finicky" part, in your lecturer's terminology), and it depends in an essential way on all of the elements of $(\ell^1)^*$ being continuous linear functionals. See here.

Warning. By switching the roles of $\ell^\infty$ and $\ell^1$, we get a natural mapping $\ell^1 \to (\ell^\infty)^*$, say $\mathbf x \mapsto g_{\mathbf x}$, where $g_{\mathbf x}(\mathbf b) = \mathbf b \cdot \mathbf x$. This is easily seen to be $\mathbf R$-linear, continuous, and injective, so you might think it leads to $(\ell^\infty)^* \cong \ell^1$. But that is false! The mapping $\ell^1 \to (\ell^\infty)^*$, where $\mathbf x \mapsto g_{\mathbf x}$ is not surjective. And in fact $(\ell^\infty)^*$ is not isomorphic to $\ell^1$ at all (not even by some mapping other than the one we tried to use). Unfortunately there is no concrete example of an element of $(\ell^\infty)^*$ that's not of the form $g_{\mathbf x}$: that $(\ell^\infty)^* \not\cong \ell^1$ is due to non-constructive reasons related to concepts like Banach limits, nonprincipal ultrafilters, and Stone-Cech compactifications. See here.

Answer 2

Let $L$ be a linear functional on $\ell^1$, we have $L:\ell^1 \rightarrow \mathbb{R}$. $L$ is defined by its action on a basis, $\{L(e_i)\}$ and can be written as $L(a) = L(\sum_i a_i e_i) = \sum_i a_i L(e_i)$. So lets form the isomorphism $\phi(L) = \{L(e_i)\}$. If $sup_i |L(e_i)| < \infty$ then $|\sum_i a_i L(e_i)| \leq \sum_i |a_i L(e_i)| \leq \sup_i |L(e_i)| \times \sum_i |a_i| < \infty$. Hence $\ell^{\infty} \subset \phi(\ell^1)$, Now if $sup_i |L(e_i)| = \infty$, then there exits a subsequence $\{L(e_{n_i}) : |L(e_{n_i})| > N_M, \forall i \geq M \}$. Let $M_N = \inf\{n_p: |L(e_{n_i})| > N, \forall i \geq n_p\} + 1$. Now choose the $\ell_1$ sequence $x_{M_N} = sign(L(e_{M_N})) \times \frac{1}{N^2}$ and other $x_i$ are equal to $0$. Now $L(x) = \sum_{N} x_{M_N} L(e_{M_N}) \geq \sum_{N} \frac{1}{N} $ which is not in $\ell^1$. Hence $\phi(\ell^1) = \ell^{\infty}$.

EDIT: In the above I assumed $(\ell^1)^*$ is a space of linear functionals on $\ell^1$ of bounded norm i.e., $\sup_x |Lx|/||x||_{\ell^1} < \infty $ for $L \in (\ell^1)^*$. The insight is linear functional which is a linear map of bounded norm is defined by its action on the basis elements and hence can be mapped to a fixed sequence and since the linear functional maps a sequence to its inner product by a fixed sequence and since we want this inner product to converge in norm this forces the sequence to which the linear functional is mapped to have certain properties w.r.t norm. Here the norm forced upon is $\ell^{\infty}$. In general it's the holder's inequality which dictates what norm is forced upon in the sequence to which linear functional is mapped to so that the inner product converges. Goes to say about beauty of holder's inequality.