Prof. Matthieu Bloch
Wednesday, September 11, 2024 (v1.1)
Any separable Hilbert space is isomorphic to \(\ell_2\); the isomorphism preserves the norm and inner product.
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 \[ A\norm[2]{\bfalpha}^2 \leq \norm[\calH]{x}^2\leq B\norm[2]{\bfalpha}^2 \]
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 \(\sum_{i\geq 1}\alpha_i^2<\infty\).
Computing expansion on Riesz basis not as simple in infinite dimension: Gram matrix is "infinite"
The Gramian 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!
A fundamental problem in supervised machine learning can be cast as follows \[ \textsf{Given dataset }\calD\eqdef\{(\vecx_i,y_i)\}_{i=1}^n\textsf{, how do we find $f$ such that $f(\bfx_i)\approx y_i$ for all }i? \]
We need to introduce several ingredients to make the question well defined
We can then formulate the question as \[ \min_{f\in\calF}\sum_{i=1}^n\ell(f(\bfx_i),y_i) \]
We will focus quite a bit on the square loss \(\ell(u,v)\eqdef (u-v)^2\), called least-square regression
A common choice of \(\calF\) is the set of continuous linear functions.
\(f:\bbR^d\to\bbR\) is linear iff \[ \forall \bfx,\bfy\in\bbR^d,\lambda,\mu\in\bbR\quad f(\lambda\bfx + \mu\bfy) = \lambda f(\bfx)+\mu f(\bfy) \]
We will see that every continuous linear function on \(\bbR^d\) is actually an inner product, i.e., \[ \exists \bftheta_f\in\bbR^d\textsf{ s.t. } f(\bfx)=\bftheta_f^\intercal\bfx \quad\forall \bfx\in\bbR^d \]
Canonical form I
Canonical form II
Allow for affine functions (not just linear)
Add a 1 to every \(\vecx_i\) \[ \min_{\bftheta\in\bbR^{d+1}} \norm[2]{\bfy-\matX\bftheta}^2\textsf{ with } \matX\eqdef\mat{c}{1-\vecx_1^\intercal-\\\vdots\\1-\vecx_n^\intercal-} \]
Let \(\calF\) be an \(d\)-dimensional subspace of a vector space with basis \(\set{\psi_i}_{i=1}^d\)
The problem becomes \[ \min_{\bftheta\in\bbR^d}\norm[2]{\bfy-\boldsymbol{\Psi}\bftheta}^2\textsf{ with }\boldsymbol{\Psi}\eqdef \mat{c}{-\psi(\bfx_1)^\intercal-\\\vdots\\-\psi(\bfx_n)^\intercal-}\eqdef\mat{cccc}{\psi_1(\bfx_1)&\psi_2(\bfx_1)&\cdots&\psi_d(\bfx_1)\\ \vdots&\vdots&\vdots&\vdots\\ \psi_1(\bfx_n)&\psi_2(\bfx_n)&\cdots&\psi_d(\bfx_n) } \]
We are recovering a nonlinear function of a continuous variable