# Thursday, January 27

:::{.remark}
Notes:

- Bessel:
  \[
  \sum_n \abs{ \inp{f}{\phi_n} } \leq \norm{f}^2,\qquad \norm{f} \da \sqrt{\inp{f}{f}}
  .\]
  - Prove using the fact that
  \[
  0\leq \norm{f - \sum_{n\leq m} \inp{f}{\phi_n} \phi_n }^2 = \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2
  .\]

- Best fit: 
\[
\norm{f - \sum_{n\leq m} c_n \phi_n } \geq \norm{f - \sum_{n\leq m} \inp{f}{\phi_n} \phi_n }
,\]
so define projections $P_M(f) \da \sum \inp{f}{\phi_n} \phi_n$ for $\spanof\ts{\phi_n} = M$.
  - Prove using 
  \[
  \norm{f - \sum_{n\leq m} c_n \phi_n } 
  &= \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2 + \sum_{n\leq m} \abs{\inp{f}{\phi_n} - c_n }^2 \\
  &\geq \norm{f}^2 - \sum_{n\leq m} \abs{\inp{f}{\phi_n}}^2
  .\]
- Hilbert spaces are separable: have countable dense subsets.
  - $\ell^\infty(\ZZ)$ is not separable.
- For $\ts{\phi_n}$ orthonormal and $\ts{c_n}$ scalars, $\sum c_n \phi_n$ is convergent iff $\ts{a_n}\in \ell^2(\ZZ)$, so $\sum\abs{c_n}^2<\infty$.
  If it converges, it can be rearranged, and 
  \[
  \norm{\sum c_n \phi_n}^2 = \norm{\ts{c_n}}_{\ell^2(\ZZ)}^2 = \sum \abs{c_n}^2
  .\]
  - To prove, use that $\norm{ \sum_{i\leq n\leq j}c_n \phi_n}^2 = \sum_{i\leq n\leq j}\abs{c_n}^2$, so the sequence $\ts{S_m}_{m\geq 0}$ where $S_m \da \sum_{n\leq m} c_n\phi_n$ is Cauchy since $\sum\abs{c_n}^2$ converges.
  - If $g = \sum c_n \phi_n$ and $f = \sum c_{m_n} \phi_{m_n}$ is a rearrangement, if $\ts{c_n}\in \ell^2$ then $\norm{f}^2 = \norm{g}^2 = \sum \abs{c_n}^2$.
  Then $\norm{f-g}^2 = \norm{f}^2 + \norm{g}^2 - 2\Re\inp{f}{g}$, but $\inp{f}{g} = \sum \abs{c_n}^2$, so $\norm{f-g} = 0$.
- If $K = \ts{\phi_n} \subseteq \mch$ is a proper subset, so $\spanof\ts{\phi_n}\neq \mch$, write $P_K(f) = \sum \inp{f}{\phi_n}\phi_n$.
  Then $P_K(f) = 0$ if $f\in \cl(K)^\perp$, $P_K(f) = f$ if $f\in \cl(K)$, and if $f\not \in K$, since $\cl(K)\leq \mch$ is closed then there exists a $g\in \cl(K)$ where $\norm{f-g} = \dist(f, K)$.
  Write $f = h + (f-g) \in \cl(K) \oplus \cl(K)^\perp$, so $f = P_K f + (I-P_K)f$.
- Recall complete subspaces of Banach spaces are closed.
- Next theorem: every separable Hilbert space admits an orthonormal basis.
:::

:::{.theorem title="?"}
If $\mch$ is a separable Hilbert spaces and $K = \ts{\phi_n}$ is an orthonormal set, then TFAE

- $K$ is complete, i.e. $K^\perp = 0$ (taking the closure is not needed),
- $\cl{\spanof K} = \mch$,
- $K$ is an orthonormal *basis*,
- For all $f\in \mch$, $\norm{f} = \sum_{n\geq 0} \abs{\inp{f}{\phi_n}}^2$ (Parseval).
:::

:::{.remark}
Note $\inp{f}{\phi_n}$ is the $n$th Fourier coefficient $\hat{f}(\xi) = \sum \inp{f}{\phi_n}\phi_n(\xi)$, and Parseval says $\norm{f}^2 = \norm{\hat{f}}^2$.
:::

:::{.proof title="?"}
$1\implies 2$: 
Let $f\in \mch \sm \cl\spanof(K)$ and project, so $f= g+h$ with $g,h\neq 0$.
But $g\in \cl\spanof(K)$ and $h\in \cl\spanof(K)^\perp = 0$, forcing $K^\perp = \cl(K)^\perp \neq 0$. $\contradiction$

$2\implies 3$:
Follows directly from previous lemma that $f = P_Kf + (I-P_K)f$.

$3\implies 4$:
Write $f\in \mch$ as $f = \sum \inp{f}{\phi_n} \phi_n$ by sending $m\to \infty$ in the previous lemma.

$4\implies 1$:
Toward a contradiction, suppose $f\neq 0\in K^\perp$.
Then $\norm{f} \neq 0$ but $\inp{f}{\phi_n}=0$ for all $n$, contradicting Parseval. $\contradiction$
:::