# Tuesday, February 01


:::{.remark}
Notes:

- Assume $\mch$ is a **separable** Hilbert space: there exists a countable set of vectors $\ts{v_i}_{i\in \ZZ}$ which span a subspace that is dense in $\mch$, so $\cl(\spanof \ts{v_i}) = \mch$.
- Complete subspaces: $M\leq \mch$ with $M^\perp = 0$. 
- Show that for any $S \subseteq \mch$, $S^\perp$ is closed in $\mch$, $S^\perp = \qty{\cl\spanof S}^\perp$, and $f\in S \intersect S^\perp \implies f=0$.
- If $K = \ts{\phi_k}_{k\in \ZZ}$ is an orthonormal set in $\mch$, then TFAE
  - $K \leq \mch$ is a complete subspace, so $K^\perp = 0$, i.e. $\inp{f}{\phi_k} = 0$ for all $k$ implies $f=0$.
  - $\cl\spanof K = \mch$, so every $f\in \mch$ is the limit of a sequence of vectors from $\spanof K$.
  - $K$ is an orthonormal *basis*
  - Parseval: equality in Bessel, i.e. $\norm{f} = \sum \abs{ \inp{f}{\phi_k} }^2$
- Lemma: if $M, N\leq \mch$ with $\dim M < \dim N$, then $M^\perp \intersect N \neq 0$.
  - Without loss of generality assume $\dim N = n + 1$ where $n\da \dim M$, take a basis $\ts{\psi_k}_{k\leq n+1}$ for $N$.
  - Try to find $f\in N$ with $f\perp M$, i.e. coefficients $\ts{b_i}_{i\leq n}$ with $\sum b_i \psi_i \perp \phi_k$ for every $\phi_k$ basis elements of $M$.
  - This is a linear system of $n$ equations in $n+1$ unknowns, so it has a nontrivial solution.
- Theorem, orthonormal bases are stable: if $\ts{\phi_k}$ is an orthonormal basis and $\ts{\psi_k}$ is an orthonormal *system*, if $\sum \norm{\phi_k - \psi_k}^2 < \infty$ then $\ts{\psi_k}$ is a basis.
  - Assume note, then find a $\psi_0 \in \mch$ with $\norm{\psi_0} = 1$ and $\inp{\psi_0}{\psi_j} = 0$ for all $j$.
  - Choose $N\gg 1$ so that $\sum_{k\geq N} \norm{\psi_k - \phi_k} < 1$.
  - Use previous lemma to produce $w\in \spanof\ts{\psi_0, \psi_1,\cdots, \psi_N}$ with $w\neq 0$ and $w\perp \phi_j$ for all $j\leq N$.
  - Note $w\perp \spanof\ts{\psi_n}_{n > N}$.
  - Apply Parseval: 
  \[
  0 
  &< \norm{w}^2 \\
  &= \sum_{n\geq 1} \abs{\inp{w}{\phi_n}}^2 \\
  &= \sum_{n\geq N+1}\abs{\inp{w}{\phi_n}}^2 \\
  &= \sum_{n\geq N+1} \abs{\inp{w}{\phi_n - \psi_n}}^2 \\
  &\leq \norm{w}^2 \sum_{n\geq N+1}\norm{\phi_n - \psi_n}^2 \\
  &< \norm{w}^2 \cdot 1
  ,\]
  where we've used that $\inp{w}{\psi_n} = 0$ for $n\geq N$.
  $\contradiction$
- $\mch$ admits a countable orthonormal basis iff $\mch$ is separable.
  - $\implies$: clear, since the basis is countable, and every element is a limit of partial sums against the basis.
  - $\impliedby$: Gram-Schmidt.
    - $h_1 = \psi_1$ and $\phi_1 = h_1/\norm{h_1}$
    - $h_n = \psi_n - \sum_{1\leq k\leq n-1} \inp{\psi_k}{\phi_k} \phi_k$ and normalize.
- Exercise: a closed subspace of a separable Hilbert space is separable.
- Linearly isometric inner product spaces: $E\sim F$ iff there is a map $A: E\surjects F$ with 
  - $A(au + bv) = aAu + bAv$
  - $\norm{Au}_F = \norm{u}_E$
- Theorem: if $\mch_1, \mch_2$ are infinite dimensional separable Hilbert spaces, then $\mch_1 \sim \mch_2$. 
  Thus for any Hilbert space $\mch$ over $\CC$, $\mch \sim \ell^2(\CC)$.
  - Pick orthonormal bases $\ts{\phi_k} \subseteq \mch_1, \ts{\psi_k} \subseteq \mch_2$.
  - For $u\in \mch_1$, define $Au \da \sum \inp{u}{\phi_k}\psi_k$, which converges -- this will be the linear isometry, and satisfies condition (i).
  - Check $\norm{Au}_{\mch_2}^2 = \sum_{k\geq 1}\abs{\inp{u}{\phi_k}}^2 = \norm{u}_{\mch_2}$, which is condition (ii).
  - Check $A$ is surjective: for $y\in \mch_2$, write $y = \sum_{k} \inp{y}{\psi_k} \psi_k = Av$ for $v\da \sum_k \inp{y}{\psi_k}\phi_k \in \mch_1$.
- Non-separable spaces: look at *almost-periodic functions*.
  - E.g. $\sum_{k\leq n} c_k \exp(i\lambda_k t)$ for $\lambda_k \in \RR$.
:::