Meredith's Math Blog

Note: this was originally a thread on my twitter . Last time I talked about subspaces of the Hardy space that are invariant under the shift operator . Now let's talk about the backward shift! (Most of the discussion here comes from the excellent "Introduction to Model Spaces and their Operators" by Garcia, Mashreghi, and Ross, which I cannot recommend highly enough.) Our shift invariant subspaces are of the form \(uH^2\) where u is an inner function The backward shift is the adjoint of the shift, so in some sense, invariant subspaces for the backward shift should be "opposite" of those of the shift. In fact, we define "model spaces" as the orthogonal complement in \(H^2\) of \(uH^2\)! As a reminder of your linear algebra, the orthogonal complement of a subspace is everything orthogonal to every element of that subspace. That is, if you take the inner product of anything in a subspace V with anything in the orthogonal complement V-perp, you get zero   I...

 Note: this was originally a thread on my twitter Ok, I need to compile some knowledge in my brain, so let's talk about shift invariant subspaces! First, we need to recall the Hardy space on the disk. This is a Hilbert space of analytic functions with square summable Taylor coefficients. The Hardy space can be thought of either as functions analytic on the disk, or as a subspace of \(L^2(\mathbb{T})\) consisting of functions with all negative coefficients zero. Either way, power series are life, and we care about the coefficients.   There are lots of interesting bounded linear operators (or "operators" if you're me) on \(H^2\), but here I'll talk about two of the most important: the shift and backward shift. The shift is multiplication by z. It gets its name from what that would do to the Taylor coefficients: it shifts them over by one. The shift is an isometry! Putting more zeros at the front isn't going to change the square sum of a sequence.   The backwar...