Orthonormal Bases on Hilbert Spaces

[Euge]

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.

 

[Office_Shredder]

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.

Spoiler

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.

 

[Paul Colby]

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

Spoiler

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.

 

[Office_Shredder]

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]

I was right! I did miss something simple

 

[mathman]

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 $y_n=0$?

 

[Paul Colby]

Some $y_n=0$?

$y=0$ isn’t normal

 

[Paul Colby]

I think I have a partial solution,

Spoiler

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 $H$.