# Orthonormal Bases on Hilbert Spaces

• POTW
Gold Member
MHB
POTW Director
Let ##H## be a Hilbert space with an orthonormal basis ##\{x_n\}_{n\in \mathbb{N}}##. Suppose ##\{y_n\}_{n\in \mathbb{N}}## is an orthonormal set in ##H## such that $$\sum_{n = 1}^\infty \|x_n - y_n\|^2 < \infty$$ Show that ##\{y_n\}_{n\in \mathbb{N}}## must also be an orthonormal basis.

topsquark

Staff Emeritus
Gold Member
Here's a brief outline of an idea. I might have time to flesh it out this weekend, or maybe an enterprising student who wants to take a crack at this but isn't sure where to start is inspired to go through in detail and see if it turns into a proof.

since terns of the sum go to zero, you know that ##x_n\approx y_n## for large ##n##. So there's some ##N## for which every element of ##\span(x_{N+1},...)## is close to an element of ##\span(y_{N+1},...)## and vice versa. This might require using the convergence of the sum, in some additional way.

##x_1,...,x_N## and ##y_1,...,y_N## span ##N## dimensional spaces ##X## and ##Y##. Every element of ##Y## is in the span of all the x's and is almost orthogonal to the infinite dimensional space in the last paragraph, so is well approximated by an element in ##X##. Since ##X## and ##Y## have the same dimension, every element of ##X## is well approximated by an element of ##Y## as well.

If the y's do not span the space, there is some vector orthogonal to them. But that vector cannot be both almost orthogonal to ##X## and also almost orthogonal to the span of the other x's, which means it cannot be orthogonal to both ##Y## and the span of the other y's.

topsquark
Gold Member
I've given a counter example. Could be missing something simple (as is usual for me).

Okay, I've edited this. It's more in line with what I was intending. It's still wrong because ##y_1 \notin H##.

The statement is false as the following counter example shows. Let ##z_n## be an orthonormal set of vectors where ##n\in\mathbb{N}##. Let, $$x_n = z_1,z_3,z_4,\cdots$$ Then let $$y_n = z_2, z_3,z_4, \cdots$$ Clearly, both sets ##x_n## and ##y_n## define orthogonal sets. ##\sum_1^\infty \|| x_n - y_n||^2 = 2 < \infty## since every term is 0 except for the first term which is 2.

Last edited:
Staff Emeritus
Gold Member
No, @Paul Colby ##||y_n-x_n||^2=2## for all ##n##! Try a finite dimensional example (and ignore the tail) to check.

Paul Colby
Gold Member
I was right! I did miss something simple

Let ##H## be a Hilbert space with an orthonormal basis ##\{x_n\}_{n\in \mathbb{N}}##. Suppose ##\{y_n\}_{n\in \mathbb{N}}## is an orthonormal set in ##H## such that $$\sum_{n = 1}^\infty \|x_n - y_n\|^2 < \infty$$ Show that ##\{y_n\}_{n\in \mathbb{N}}## must also be an orthonormal basis.
Some ##y_n=0##?

Gold Member
Some ##y_n=0##?
##y=0## isn’t normal

Gold Member
I think I have a partial solution,

we may write an operator $$U=\sum_{n=1}^\infty |y_n\rangle\langle x_n|$$ which is defined on all of ##H## and is norm preserving. Further, $$(1-U)|x_n\rangle=|x_n\rangle-|y_n\rangle.$$ Clearly, $$\|1-U\|^2 \le M = \sum_{n=1}^\infty \|x_n-y_n\|^2$$

For the case ##M < 1##, ##U## has an inverse on ##H##. Let ##X=1-U##. The sum, $$\sum_{n=0}^\infty X^n = \frac{1}{1-X} = U^{-1},$$ converges and defines the inverse operator on all of ##H##. The ##|y_n\rangle## therefore span ##H##.