Dr. Matthieu R Bloch

Monday, September 27, 2021

**Assignments 2/3:**Grading ongoing**Midterm 1**- Grading ongoing
- General logistics of exams

**Assignment 4**assigned today or tomorrow

**Last time**: Orthobases- Things work in infinite dimension too with orthobases (without thinking too much)
- Last little bit to prove: isomorphism to \(\ell_2\)

**Today:**What happens with non-orthogonal bases**Wednesday September 29, 2021**: Least-square regression**Reading:**Romberg, lecture notes 7

The following properties of an orthonormal set \(\set{e_i}_{i\geq 1}\) are equivalent:

- Finite linear combinations of elements in \(\calH\) are dense in \(\calH\);
- If \(v\in\calH\) and \(\forall j\geq 1\) \(\dotp{v}{e_j}=0\) then \(v=0\);
- If \(v\in\calH\) then \(\sum_{i=1}^n\dotp{v}{e_i}e_i\) converges to \(v\) (in the norm \(\norm{\cdot}\)) as \(n\to\infty\);
- If \(v\in\calH\) and \(\forall k\) \(a_k\eqdef \dotp{v}{e_k}\) then \(\norm{v}^2=\sum_{k\geq 1}\abs{a_k}^2\).

This does not say anything about the existence of such nice orthornormal set

- A space is separable if it contains a
*countable*dense subset. Separability is the key property to deal with sequences instead of collections

- Any separable Hilbert space has an orthonormal basis.
Most useful Hilbert spaces are separable! We won’t worry about non-separable Hilbert spaces

- Any separable Hilbert space is isomorphic to \(\ell_2\)

- Let \(\set{v_i}_{i=1}^n\) be a linearly independent set in a Hilbert space \(\calH\) of dimension \(n\). Then, for any \(x\in\calH\), \(x=\sum_{i=1}^n\alpha_iv_i\) for some \(\bfalpha\in\bbR^n\). In addition, there exists \(A,B>0\) such that \[ A\norm[2]{\bfalpha}^2 \leq \norm[\calH]{x}^2\leq B\norm[2]{\bfalpha}^2 \]
Inequality is tight on both sides

For orthobases, \(A=B=1\)

**Interpretation**:- The values of \(A\) and \(B\) govern the
*stability*of the representation

- The values of \(A\) and \(B\) govern the
**Examples**

**Recall from orthobases**:- perfectly stable representation \(A=B=1\)
- Efficient computation of representations: \(\alpha_i=\dotp{x}{v_i}\)

- For any \(x\in\calH\) with basis \(\set{v_i}_{i=1}^n\)we have \[ x=\sum_{i=1}^n\alpha_iv_i\qquad\text{with}\qquad\bfalpha = \matG^{-1}\mat{c}{\dotp{x}{v_1}\\\dotp{x}{v_2}\\\vdots\\\dotp{x}{v_n}} \] There also exists a basis \(\set{\tilde{v}_i}_{i=1^n}\) such that \(\alpha_i=\dotp{x}{\tilde{v}_i}\)

- \(\set{v_i}_{i=1}^\infty\) is a
**Riesz basis**for Hilbert space \(\calH\) if \(\text{cl}(\text{span}(\set{v_i}_{i=1}^\infty))=\calH\) and there exists \(A,B>0\) such that \[ A\sum_{i=1}^\infty\alpha_i^2\leq \norm[\calH]{\sum_{i=1}^n\alpha_iv_i}^2\leq B\sum_{i=1}^\infty\alpha_i^2 \]**uniformly**for all sequences \(\set{\alpha_i}_{i\geq 1}\) with \(\sum_{i\geq 1}\alpha_i^2<\infty\). In infinite dimension, the existence of \(A,B>0\) is

**not**automatic.**Examples**

Computing expansion on Riesz basis not as simple in infinite dimension: Gram matrix is “infinite”

The Grammiam is a

**linear operator**\[ \calG:\ell_2(\bbZ)\to\ell_2(\bbZ): \bfx\mapsto \bfy\text{ with }[\calG(\bfx)]_n\eqdef y_n=\sum_{\ell=-\infty^\infty}\dotp{v_\ell}{v_n}x_\ell \]**Fact:**there exists another linear operator \(\calH:\ell_2(\bbZ)\to\ell_2(\bbZ)\) such that\[ \calH(\calG(\bfx)) = \bfx \] We can replicate what we did in finite dimension.