# real analysis – Proof of (ii)\$implies\$ (iii) in Theorem \$4.18\$, Rudin’s RCA

Rudin says that (ii)$$implies$$(iii) of Theorem $$4.18$$ follows from Theorem $$4.17$$. I do not see how, although I have tried two different approaches. I have described my work below and attached the theorems for reference below at the end of the post.

Approach 1: Assume (ii), i.e. $$P$$ is dense in $$H$$. Then for every $$epsilon > 0$$ and $$xin H$$, there exists a finite $$Fsubset A$$ and $$c_alpha$$‘s such that $$left|x – sum_{alpha in F} c_alpha u_alpharight| < epsilon$$
We have $$sum_{alphain A} |hat x(alpha)|^2 le |x|^2$$ from Theorem $$4.17$$ already. If we can show $$sum_{alphain A} |hat x(alpha)|^2 ge |x|^2$$, we are done. I observed that
$$|x| le left|x – sum_{alpha in F} c_alpha u_alpharight| + left|sum_{alphain F}c_alpha u_alpha right| < epsilon + sum_{alphain F}|c_alpha|^2$$
which doesn’t really help.

Approach 2: The proof of Theorem $$4.17$$ says that $$f$$ is an isometry of $$P$$ onto the dense subspace of $$ell^2(A)$$ consisting of finitely supported functions. Since we assume $$P$$ is dense in $$H$$, there is a sequence $$(x_n)$$ of elements in $$P$$, such that $$x_nto xin H$$. Since $$0in P$$ and $$f$$ is an isometry of $$P$$, we see that
$$|x_n|^2 = |f(x_n)|^2$$
$$|x_n|^2 = |hat x_n|^2 = sum_{alphain A}|hat x_n(alpha)|^2$$
Taking limits,
$$lim_{ntoinfty}|x_n|^2 = lim_{ntoinfty} sum_{alphain A}|hat x_n(alpha)|^2$$
Rudin defines $$sum_{alphain A} phi(alpha)$$ as $$sup_{Fsubset A, |F| < infty} sum_{alphain F}phi(alpha)$$ where $$F$$ is a finite subset of $$A$$, i.e. we take supremum over all finite sums.
Then,
$$lim_{ntoinfty}|x_n|^2 = lim_{ntoinfty} sup_{Fsubset A} sum_{alphain F, |F| < infty}|hat x_n(alpha)|^2$$
If we can switch $$lim$$ and $$sup$$ in the expression above, we are essentially done – but I’m not sure if the swapping is allowed.

Attached for reference:
Theorem 4.18:

Theorem 4.17:

Thank you!