Orthonormal Bases on Hilbert Spaces

In summary, an orthonormal basis on a Hilbert space is a set of vectors that are both orthogonal and normalized. It can be constructed using the Gram-Schmidt process or by finding eigenvalues and eigenvectors of a self-adjoint operator. Orthonormal bases have important applications in mathematics and physics, providing a way to represent any vector in a Hilbert space in terms of a finite number of basis vectors. The Gram-Schmidt process involves a series of orthogonal projections and normalizations to create an orthonormal basis. It can be infinite, but for practical purposes, a finite basis is often used.
  • #1
Euge
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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.
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
No, @Paul Colby ##||y_n-x_n||^2=2## for all ##n##! Try a finite dimensional example (and ignore the tail) to check.
 
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  • #5
I was right! I did miss something simple
 
  • #6
Euge said:
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##?
 
  • #7
mathman said:
Some ##y_n=0##?
##y=0## isn’t normal
 
  • #8
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##.
 

Related to Orthonormal Bases on Hilbert Spaces

1. What is an orthonormal basis on a Hilbert space?

An orthonormal basis on a Hilbert space is a set of vectors that are both orthogonal (perpendicular) to each other and have a unit length. These vectors form a basis for the Hilbert space, meaning that any vector in the space can be expressed as a linear combination of the basis vectors.

2. Why are orthonormal bases important in Hilbert spaces?

Orthonormal bases are important in Hilbert spaces because they allow for the representation of any vector in the space using a finite number of basis vectors. This makes it easier to work with and analyze vectors in the space, as well as perform calculations and solve problems.

3. How do you find an orthonormal basis on a Hilbert space?

To find an orthonormal basis on a Hilbert space, you can use the Gram-Schmidt process. This involves taking a set of linearly independent vectors and orthogonalizing them by projecting each vector onto the orthogonal complement of the previous vectors. The resulting set of orthogonal vectors can then be normalized to become an orthonormal basis.

4. Can an orthonormal basis on a Hilbert space be infinite?

Yes, an orthonormal basis on a Hilbert space can be infinite. In fact, many Hilbert spaces have an infinite number of orthonormal basis vectors. For example, the set of all polynomials of degree n or less forms an infinite orthonormal basis for the space of all polynomials.

5. What are some applications of orthonormal bases on Hilbert spaces?

Orthonormal bases on Hilbert spaces have various applications in mathematics, physics, and engineering. They are used in Fourier analysis, signal processing, quantum mechanics, and many other fields. They also have practical applications in data compression and image processing, where they can be used to reduce the amount of data needed to represent a signal or image.

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