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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.

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  • $\begingroup$ It'd be nice if you could organize your questions a little better - perhaps numbering them. Currently it's a bit overwhelming to try to write an answer. $\endgroup$ Jan 10, 2020 at 2:33
  • $\begingroup$ You're correct that Kreyszig doesn't explicitly show the linearity and injectivity; he presumably considers them obvious (as would I). But you are of course welcome to write out the details if you find it helpful. $\endgroup$ Jan 10, 2020 at 2:37
  • $\begingroup$ @NateEldredge I edited and highlighted accordingly. Hopefully this makes it easier. $\endgroup$
    – user740738
    Jan 10, 2020 at 2:44
  • $\begingroup$ My edit was for typos in Q 1., where you had $l_1$ but meant $l_1^*$.... BTW when a subscript or supscript is a single keystroke, you don't need braces. E.g. l_1^* instead of l_{1}^{*}... This also applies elsewhere, e.g. \sin x and \sqrt x. But NOT for \frac nor \binom $\endgroup$ Jan 10, 2020 at 8:28

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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 $||x||$.)

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  • $\begingroup$ Great answer! I seem to understand by first reading. Will follow through the details and check by myself and if questions remain, then I will ask. Thank you. $\endgroup$
    – user740738
    Jan 10, 2020 at 3:00
  • $\begingroup$ Indeed, Kreyszig demands for isomorphic normed spaces that there is a bijective linear operator such that it preserves the norm i.e. $||Tx|| = ||x||$. $\endgroup$
    – user740738
    Jan 10, 2020 at 3:14
  • $\begingroup$ I think it's fine for most intents and purposes to demand an isometry, but isomorphic but not isometrically isomorphic spaces exist, see here. $\endgroup$
    – Riley
    Jan 10, 2020 at 3:21
  • $\begingroup$ I think that saying isometrically isomorphic is much more accurate since we use isomorphism when dealing with algebraic structures (without any notion of metric and norm) and we use isometric in the theory of metric spaces. Thank you for the useful link. $\endgroup$
    – user740738
    Jan 10, 2020 at 3:25
  • $\begingroup$ In fact, what really matters for most purposes is not just that the spaces are isomorphic, nor even that they are isometrically isomorphic, but that a specific natural map between them (the map $\phi$ defined here) is an isometric isomorphism. $\endgroup$ Jan 10, 2020 at 4:08

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