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Can you please help to understand what exactly the definitions of these spaces, actually I'm so confused.

Firstly, the Hilbert space $\ell^2 =\{ \{a_n\}_0^\infty: \sum_0^\infty |a_n|^2 <\infty$ }. Then can I conclude that if $(a_0, a_1, a_2, ...) \in \ell^2 \implies a_0, a_1, a_2, ... \in \mathbb{D}$ $(\mathbb{D}=\{ z \in \mathbb{C}: |z|<1 \})$??

Secondly, the Hilbert space $L^2 = \{ f \in S^1: \frac{1}{2\pi} \int_0^{2\pi} |f|^2 < \infty \}$ (all lebesgue integrable functions with finite square-integral), where $S^1 = \{ z \in \mathbb{C}: |z|=1 \}$. Did it implies that every analytic function $f : S^1 \to S^1$ is in $L^2$,? How can I write a function $f \in L^2$ ?

Thirdly, the Hardy-Hilbert space $ H^2 = \{ f: f(z) = \sum_0 ^ \infty a_n z^n \; and \; \sum_0^\infty |a_n|^2 < \infty \} $. Can I conclude that every function $f \in H^2 \implies f$ is analytic and $f: \mathbb{D} \to \mathbb{D}$ ??

Is true that $\ell^2 \subset H^2 \subset L^2$?? If does, how can it be?

I'm so sorry for this long question, but I'm really confused.

HallaSurvivor
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hala sami
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  • From a purely set theoretic standpoint, surely $\ell^2$ is not a subset of $H^2$. In fact they are disjoint sets. – Lee Mosher Jul 18 '22 at 21:26

1 Answers1

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  1. *can I conclude that... * No, why should $|a_n| < 1$? This being a vector space, every scalar multiple of a member of the space is also in the space. So the possible values can't be bounded.

  2. No, not $f \in \mathbb S^1$, rather complex-valued measurable functions defined on $\mathbb S^1$.

2a) Did it implies that... By "analytic function $f: S^1 \to S^1$ I suppose you mean a function analytic in a neighbourhood of $S^1$, and whose values on $S^1$ happen to fall in $S^1$. It's a nice complex-analysis exercise to characterize what these are: there are probably fewer than you realize. Anyway, since the values are in $S^1$ they are bounded and they are certainly measurable, so yes they are in $L^2$.

2b) How can I write a function $f \in L^2$? I don't understand the question. Write it as you would write any function.

  1. Can I conclude that... Since the power series converges absolutely for $|z| < 1$, it is analytic on $\mathbb D$, but again there's no reason to think the values are in $\mathbb D$.

  2. Is it true that $\ell^2 \subset H^2 \subset L^2$? They are different sorts of things. A member of $\ell^2$ is a sequence, not a function. A member of $H^2$ is a function on $\mathbb D$, while a member of $L^2$ is a function on $\mathbb S^1$. However, a function $f$ in $H^2$ does have radial limits almost everywhere on $\mathbb S^1$; the map taking $f$ to the function $f^*$ given by $f^*(s) = \lim_{r \to 1-} f(rs)$ turns out to be an isometry from $H^2$ to a subspace of $L^2(\mathbb S^1)$. In other words, $H^2$ can be identified in a natural way as a subspace of $L^2$.

Robert Israel
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  • For 1: isn't it true that $|a_n|<1$ eventually, i.e. for large enough $n$? – Zeekless Jul 18 '22 at 21:48
  • If OP is talking about the function being a functional, you can use Riesz representation theorem to represent it as the inner product with a fixed $L^2$ function, but doubt that's helpful :) – Alan Jul 18 '22 at 23:07
  • @Zeekless Yes, but that's a different question. – Robert Israel Jul 19 '22 at 15:32
  • @RobertIsrael I think I understand what you mean, but for Shure if $f \in L^2 $, then its domain in $\mathbb{S^1}$ and the rang may or may not be in $\mathbb{S^1}$. And if $f \in H^2 $, then its domain in $\mathbb{D}$ and the rang may or may not be in $\mathbb{D}$. Can I ask a last question, Is there a function $f$ Defined on the half-upper complex plane belong to $H^2, $ or a function defined on the complex plane and belong to $L^2$ ?? – hala sami Jul 20 '22 at 11:29
  • a) Not the same $H^2$. There is, however, a version of $H^2$ for the upper half plane. b) Of course, e.g. the constant function $0$. – Robert Israel Jul 21 '22 at 02:11