Dr. Matthieu R Bloch
Monday September 13, 2021
Last time: Hilbert spaces
Key take-way: we can avoid problems in infinite dimensions
Key concepts: completeness is the property that we want
Questions?
Today:
Wednesday September 15, 2021: lots of examples
Reading: Romberg, lecture notes 5 and 6
Let \(\calH\) be a Hilbert space with induced norm \(\dotp{\cdot}{\cdot}\) and induced norm \(\norm{\cdot}\) ; let \(\calT\) be subspace of \(\calH\)
For \(x\in\calH\), what is the closest point of \(\hat{x}\in\calT\)? How do we solve \(\min_{y\in\calT}\norm{x-y}\)?
This problem has a unique solution given by the orthogonality principle
Let \(\calX\) be a pre-Hilbert space, \(\calT\) be a subspace of \(\calX\), and \(x\in\calX\).
If there exists a vector \(m^*\in\calT\) such that \(\forall m\in\calT\) \(\norm{x-m^*}\leq \norm{x-m}\), then \(m^*\) is unique.
\(m^*\in\calT\) is a unique minimizer if and only if the error \(x-m^*\) be orthogonal to \(\calT\).This doesn’t say that \(m^*\) exists!
Let \(\calH\) be a Hilbert space, \(\calT\) be a closed subspace of \(\calX\), and \(x\in\calX\).
There exists a unique vector \(m^*\in\calT\) such that \(\forall m\in\calT\) \(\norm{x-m^*}\leq \norm{x-m}\).
\(m^*\in\calT\) is a unique minimizer if and only if the error \(x-m^*\) be orthogonal to \(\calT\)The orthogonality principle gives us a procedure for computing the closest point