A few problems I thought about while working.

$\ell^p(A)$ is always complete

If $A$ is empty, then let’s agree to ignore this irrelevant case.

Otherwise, suppose $\{f_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence. We would like to define $f = \lim_{n\to\infty}f_n$, where the limit is pointwise. The nice thing about $\ell^p$ is that every point has mass (formally, every $\alpha\in A$ is an atom of the counting measure), so this idea actually works.

We first show that $\lim_{n\to\infty}f_n(\alpha)$ exists for each $\alpha\in A$. Note that \(|f_n(\alpha) - f_m(\alpha)|^p \leq \sum_{\beta\in A}|f_n(\beta) - f_m(\beta)|^p \to 0, \text{as }n,m\to\infty.\) Thus, $\{f_n\}_{n\in\mathbb{N}}$ is pointwise Cauchy, and therefore $f=\lim_{n\to\infty}f_n$ is well-defined.

It’s pretty clear that $f\in\ell^p(A)$, since \(\|f\|_p \leq \|f_n\|_p + \|f_n-f\|_p < \infty.\)

The proof for when $p=\infty$ is largely the same.

This is not interesting since $L^p$ spaces are always complete, but it is an example of how counting measure lets us write pointwise estimates in terms of the $\ell^p$ norm.

Every Orthonormal Basis Has the Same Size

Interestingly, this proof actually requires two different proofs in the finite and infinite dimensional cases.

Finite Dimensions

This is basically the star result of finite-dimensional linear algebra, in particular the Steinitz Exchange Lemma.

Infinite Dimensions

We will assume we are working in an infinite-dimensional Hilbert space $H$. Let $\{v_\alpha\}_{\alpha\in A}$ and $\{w_\beta\}_{\beta\in B}$ are two bases of the space. Now, define \(E_\beta = \{ \alpha\in A : \langle w_\beta, v_\alpha\rangle \neq 0 \}.\) Since $|w_\beta|^2 = \sum_{\alpha\in A} |\langle w_\beta, v_\alpha\rangle |$, we know that $E_\beta$ is countable. Moreover, $A = \bigcup_{\beta\in B}E_\beta$. If not, then there would exist $\alpha$ such that $\langle w_\beta, v_\alpha\rangle = 0$ for all $\beta$, implying that $v_\alpha = 0$ by Parseval’s Theorem. This clearly cannot happen.

Thus, \(|A| = \left| \bigcup_{\beta\in B} E_{\beta}\right| \leq |\mathbb{N}\times B | = |B|.\) We are using the fact if $B$ is infinite then $|\mathbb{N} \times B| = |B|$ and the fact that each $E_\beta$ is at most countable. However, nothing here was special about the role of $A$ and $B$, so reversing the argument shows that $|A| \leq |B|$ and $|B|\leq |A|$. Applying the Cantor-Schröder-Bernstein theorem completes the result.

Corollary: $\ell^2(A) \cong \ell^2(B)$ if and only if $|A| = |B|$

This follows from noting that $\{\chi_{\{\alpha\}}\}_{\alpha\in A}$ and $\{\chi_{\{\beta\}}\}_{\beta\in B}$ form orthonormal bases of their respective spaces.

Von Neumann’s Mean Ergodic Theorem