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I have some questions about Theorem 4.17 of Rudin's Real and Complex Analysis, attached below.

  1. Inequality 4.14(5) is $$\sum_{\alpha\in F} |\hat x(\alpha)|^2 \le \|x\|^2$$ How does the above inequality holding for every finite set $F\subset A$ guarantee Bessel's inequality (1)? I think we can extend the inequality to countable sets by taking the limit, but what about uncountable sets? I saw here that the uncountable case is actually not possible, but I do not quite understand the reasoning. In Rudin's book, it is shown that if $\varphi\in \ell^2(A)$ then $\{\alpha\in A: \varphi(\alpha)\ne 0\}$ is at most countable. Is this related?
  2. Just want to confirm, $\|x\|^2$ cannot possibly be infinite, right? I guess that is how the author concludes that $\hat x \in \ell^2(A)$ (using Bessel's inequality).
  3. How does Theorem 4.14(a) show that $f$ is an isometry of $P$ onto the dense subspace of $\ell^2(A)$ consisting of those functions whose support is a finite subset of $A$?

Attached for reference:
Theorem 4.17: enter image description here Theorem 4.14(a): enter image description here

stoic-santiago
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  1. Take a look at 4.15 in Rudin. The sum in question on the LHS is defined as the supremum of all finite sums. So, if the inequality holds for all finite sums, it holds for the supremum as well.
  2. You're right, $\|x\|$ and hence $\|x\|^2$ are finite. So, $\sum_{\alpha\in A}|\hat{x}(\alpha)|^2\leq \|x\|^2<\infty$ and thus $\hat{x}\in \ell^2(A)$.
  3. Unwind the definitions.
peek-a-boo
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  • About 1: How do we rule out the possibility of $A$ being uncountable? The argument you present only works for countable sums. About 3: Could I get some help? It's why I posted the question. – stoic-santiago Jun 06 '21 at 09:49
  • @epsilon-emperor no. reread 4.15 in Rudin carefully. – peek-a-boo Jun 06 '21 at 09:49
  • Alright, I see. You're right. The question linked to this is incorrect then, right? (https://math.stackexchange.com/questions/3307246/bessels-inequality-in-rudins-rca?noredirect=1&lq=1) – stoic-santiago Jun 06 '21 at 09:51
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    @epsilon-emperor when we are writing $\sum_{i\in I}t_i$ for non-negative numbers $t_i$ and $I$ is uncountable, then by definition we mean that $\sum_{i\in I}t_i:=\sup_{F\subset I}\sum_{i\in F}t_i$, where $F$ ranges over the finite subsets of $I$. therefore, if we have $\sum_{i\in F}t_i\leq M$, where $M$ is a constant independent of $F$, then we can take supremum w.r.t $F$ and conclude that $\sum_{i\in I}t_i\leq M$. – Just dropped in Jun 06 '21 at 09:51
  • @epsilon-emperor which part of the link are you claiming is incorrect? – peek-a-boo Jun 06 '21 at 09:54
  • Just the fact that OP (of the other post) said uncountability of $A$ could be a problem. @peek-a-boo – stoic-santiago Jun 06 '21 at 09:57
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    @epsilon-emperor cardinality of $A$ is not an issue – peek-a-boo Jun 06 '21 at 09:58
  • @peek-a-boo I already know that the subspace of $\ell^2(A)$ of functions with finite support $F\subset A$ is dense in $\ell^2(A)$ (same as saying that simple functions are dense in $L^p(\mu)$). I need to show two things: (1) that $f$ is an isometry, i.e. $|f(y) - f(x)|_2 = |y-x|$ for all $x,y\in H$. We already have $|f(y) - f(x)|_2 \le |y-x|$, so we need to show $\ge$ direction of the inequality. (2) that $f(x) = \hat x$ has finite support for every $x\in H$. How do I do (1) and (2)? – stoic-santiago Jun 06 '21 at 10:09
  • @epsilon-emperor what is the definition of $P$ and what does theorem 4.14 say? – peek-a-boo Jun 06 '21 at 10:17
  • $P$ is the space of all finite linear combinations of vectors $u_\alpha$. Right, that helps me see that the support of every function in $f(P)$ is finite. I've linked Theorem 4.14(a) in my answer, and I still do not see why $f$ is an isometry. @peek-a-boo – stoic-santiago Jun 06 '21 at 10:21
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    @epsilon-emperor theorem 4.14 says $f|P$ is an isometry. Given $x\in P$, by definition there is a finite set $F\subset A$ and coefficients $c{\alpha}$ such that $x=\sum_{\alpha\in F}c_{\alpha}u_{\alpha}$. In other words, (by definition) there is a function $c:A\to\Bbb{C}$ which vanishes outside $F$ such that $x=\sum_{\alpha\in F}c_{\alpha}u_{\alpha}$. Now, $f(x)=\hat{x}$ is precisely the same function as $c$. i.e $x=\sum_{\alpha\in F}\hat{x}(\alpha)\cdot u_{\alpha}$, and 4.14 says $|x|^2=\sum_{\alpha\in F}|c_{\alpha}|^2=|\hat{x}|^2=|f(x)|^2$. Taking square roots completes the "proof". – peek-a-boo Jun 06 '21 at 10:26
  • @peek-a-boo Can you please clarify why $|x|^2<\infty$ ... I also found this related question, but I am not clear about the answer. https://math.stackexchange.com/questions/23744/is-the-norm-on-a-hilbert-space-always-finite – texmex Aug 08 '23 at 14:09
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    @texmex a norm $|\cdot|$ is by definition a map from the vector space into $[0,\infty)$ satisfying a bunch of conditions (homogeneity, triangle inequality). So, the answer is that a norm is by definition finite. So, if you now square something finite, you again get something finite. – peek-a-boo Aug 08 '23 at 14:50
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    For a Hilbert space, the inner product is by definition (bilinear/sesquilinear) map $H\times H\to\Bbb{C}$ satisfying a bunch of conditions. Then, you define $|x|^2=\langle x,x\rangle$, so this is an element of $\Bbb{C}$ (actually the axioms tell us it is a very particular element of $\Bbb{C}$, namely it belongs to $[0,\infty)$). So really, it is by definition. If this is not clear, you need to re-read the definitions given in Rudin for norms and inner products. There is nothing to prove here. – peek-a-boo Aug 08 '23 at 14:52
  • @peek-a-boo Thanks, Much appreciated. It is good to know that the norm is always finite. – texmex Aug 10 '23 at 12:56
  • @peek-a-boo In RCA Definition 4.13: The definition of $\hat{x}$ is given as a complex function defined on the index set $A$ for every $x \in H$ such that $\hat{x}(\alpha)=(x,u_{\alpha})$. How are we to interpret this definition of $\hat{x}$ as used in Theorem 4.17. Specifically referring to the third part of the question you answered, we have to show that ${|y_{1}-y_{2}|}$=${|\hat{y_{1}}-\hat{y_{2}}|}$ where $y_{1}, y_{2}, \in H$ We can take $y_{1}, y_{2}, \in P$ or as linear combinations of the orthonormal vectors. But what about $\hat{y_{1}},\hat{y_{2}}$ – texmex Aug 21 '23 at 15:42
  • Please let me know if the above question is unclear. Thanks – texmex Aug 21 '23 at 15:45
  • @texmex you really should ask a separate question. – peek-a-boo Aug 21 '23 at 16:48
  • @peek-a-boo Indeed, that makes sense. Was thinking of doing that. The related question is here: https://math.stackexchange.com/questions/4756698/rudin-rca-theorem-4-17-questions . Also, I am not sure how the definitions in Rudin RCA ensure that the Norm is finite. I have added that as a second question as well. . Please let me know your thoughts. – texmex Aug 22 '23 at 05:21
  • please read the definition of inner product. It is a complex number. $\infty$ is not a complex number. The norm is $|x|=\sqrt{\langle x,x\rangle}$, the square root of a non-negative real number, i.e the square root of something in $[0,\infty)$, so it belongs to $[0,\infty)$, by definition. – peek-a-boo Aug 22 '23 at 05:23
  • whoever/wherever you got the idea that $\infty$ is a real or complex number, you need to forget that immediately. – peek-a-boo Aug 22 '23 at 05:26
  • @peek-a-boo Could you please address the first part of the question here as well? https://math.stackexchange.com/questions/4756698/rudin-rca-theorem-4-17-questions?noredirect=1&lq=1 – texmex Aug 24 '23 at 02:35