Dual space of $\ell_{1}$ is $\ell_{\infty}$

Question

This question has been asked here (see below link) but I still have some questions after reading the answers and following the books content. I reference the following question because the OP has added a picture of the solution presented by Kreyszig.

same_question_asked

1. Do we have to show that there is an isomorphism between $l_{1}^*$ and $l_{\infty}$ or show that $l_{1}^* = l_{\infty}$, i.e. one is a subset of the other?

I just assume that Kreyszig does the former. In other words, we must show that $\phi(f) = (f(e_{1}), \ldots)$ is a homomorphism between vector spaces, is bijective and that the norms are the same i.e. $\|f\| = \|x\|_{ \infty}$.

2. For showing homomorphism I have the following questions:

I don't see where Kreyszig shows that $\phi$ is linear, i.e $\phi(\alpha f + \beta g) = \alpha \phi(f) + \beta \phi(g)$; this should be shown by $\phi(\alpha f + \beta g) = (\alpha f(e_{k}) + \beta g(e_{k}), \ldots ) = \alpha \phi(f) + \beta \phi(g)$ by the definition of addition and scalar multiplication in sequence space.

3. For showing bijective I have the following questions:

Where does the author show injectivity? Shouln't he show $\phi(f_{1}) = \phi(f_{2}) \iff (f_{1}(e_{1}), \ldots ) = (f_{2}(e_{1}), \ldots ) \iff f_{1}(e_{1}) = f_{2}(e_{1}), \ldots \iff f_{1} = f_{2}$?

Does he show surjectivity after (7) in above attached math stack question? If so, shouldn't he show that for all $b \in l_{\infty}$ there is a $g \in l_{1}^{'}$ such that $b = g$, i.e. $g$ should be a sequence? Also I do not understand the identities where he shows that $g$ is bounded.

4. For showing that the norms are equal I have the following questions:

I do not understand how $|f(e_{k})| \leq \|f\| \|x\|$ in (7) in above attached math stack question.

Also is he assuming that each $c \in l_{\infty}$ is equal to $\gamma_{k} = f(e_{k})$? How is this true? Shouldn't we use an arbitrary $c \in l_{\infty}$?

I also do not understand the entire line of identity and inequality in 7(a) in above attached math stack question.

Answer 1

For your first question, the statement $(\ell^1)^* = \ell^\infty$ is the statement they are isomorphic, meaning there is a linear bicontinuous bijection $T: (\ell^1)^* \rightarrow \ell^\infty$. (Bicontinuous meaning $T$ and $T^{-1}$ are continuous.) The requirement of linearity (which does not appear in the definition of isomorphic metric spaces) relates to the fact a Banach space is a (normed) vector space. I think at this point some authors define an isomorphism to demand $T$ also be an isometry, which relates to your statement we must show $||f|| = ||\phi(f)||$, but normally I say in this case they are isometrically isomorphic. Of course, if indeed $T$ is an isometry then $T$ is automatically bounded, and a bounded linear bijection by the open mapping theorem is automatically an isomorphism.

For your second and third questions about showing $\phi$ is a homomorphism (linearity): it might be omitted because it is presumably 'trivial' to check $\phi$ is linear. The same goes for showing $\phi$ is injective. (As Nate said.)

The author does show surjectivity after (7). You are slightly off in your definition of surjectivity; we must show for all sequences $b = (\beta_k) \in \ell^\infty$ there exists a linear functional, not a sequence, $g \in (\ell^1)^*$ such that $b = \phi(g)$, not $b = g$. But (6) establishes every linear function $g$ is necessarily of the form

$$g(x) = \sum_{k=1}^\infty \xi_k g(e_k),$$

for $x = (\xi_k) \in \ell^1$. Given $b = (\beta_k)$, we define $g(e_k) = \beta_k$, so that

$$g(x) = \sum_{k=1}^\infty \xi_k \beta_k.$$

Then this guarantees $\phi(g) = (g(e_k)) = (\beta_k) = b$, as required. This only says $g$ is a linear operator; to show boundedness, we need to show $|g(x)| \leq C||x||$ for some constant $C > 0$.

The first part is $$|g(x)| \leq \sum |\xi_k\beta_k|$$ is the triangle inequality $||\sum a_i|| \leq \sum ||a_i||$.

The second part is $$ \sum |\xi_k\beta_k| \leq \sup_i |\beta_i| \sum |\xi_i|$$ is because $|\xi_k\beta_k| = |\beta_k||\xi_k| \leq (\sup_i |\beta_i|)|\xi_k|$. That the supremum even exists (i.e. is finite) is because $\beta \in \ell^\infty$.

The last is by definition of $||x|| = \sum |\xi_k|$.

Your last (fourth) set of questions pertain to why the norms are equal. First, your question has a typo; in (7), they write

$$|f(e_k)| \leq ||f|| ||e_k||$$

which is by definition of $f \in (\ell^1)^*$; it is a bounded linear operator, and so by definition it's norm $||f||$ satisfies the inequality $|f(x)| \leq ||f|| ||x||$ for all $x \in \ell^1$.

He doesn't need to assume $c \in \ell^\infty$ is arbitrary; the first reason is that we have already assumed $\phi$ is bijective, so every $c \in \ell^\infty$ is indeed the image $\phi(f)$ of some $f$ with $f(e_k) = \gamma_k$. The second reason is all you're trying to show is that $||f|| = ||\phi(f)|| = ||c||$.

Lastly, in 7(a), this follows from the same reasoning as why $g$ is bounded. (Triangle inequality; then $|\gamma_k| \leq \sup_i |\gamma_i|$, then definition of $||x||$.)