What Distinguishes Hilbert Spaces from Euclidean Spaces?

[ajayguhan]

I know that hilbert space is infinite dimension space whereas eucledian is Finite n dimensional space, but what are all other differences between them?

 

[mathman]

There are theorems true in Euclidean space that are not valid in Hilbert space because of the infinite dimension.

For example: In Euclidean space if a sequence of vectors {x_n} converge pointwise to x, then
||x_n-x|| -> 0.

This is not necessarily true in Hilbert space.
Example: kth component of x_n = 0, for k≠n, = 1, for k=n. x_n -> 0 vector pointwise, but ||x_n|| = 1.

 

[R136a1]

A Hilbert space is not necessarily infinite dimensional. Some Hilbert spaces are finite dimensional and are isomorphic to Euclidean space.

So Euclidean space is a special case of a Hilbert space. But of course, there are many Hilbert spaces which are not Euclidean spaces.

 

[johnqwertyful]

Hilbert spaces are not necessarily infinite dimensional, I don't know where you heard that. Euclidean space IS a Hilbert space, in any dimension or even infinite dimensional.

A Hilbert space is a complete inner product space. An inner product space is a vector space with an inner product defined on it. An inner product induces a norm, which induces a metric. A metric space is called complete if every Cauchy sequence is convergent.

 

[R136a1]

Hilbert spaces are not necessarily infinite dimensional, I don't know where you heard that.

I guess he's taking a QM course. I have seen QM courses being taught by defining Hilbert space as infinite dimensional. Some even define Hilbert space as $L^2(\mathbb{R})$ or $\ell^2$. I don't know why they don't teach it the right way though.

 

[ajayguhan]

If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?

Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.

 

[johnqwertyful]

If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?

Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.

Euclidean space is a specific example of a Hilbert space. Euclidean space is the set of all n lists of numbers (n is is the dimension, and may be infinite dimensional), over real numbers, with the dot product as the inner product.

 

[ajayguhan]

An example space which is hilbert but not eucledian space can be set of all continuous function in a given interval..correct ?

 

[Jorriss]

I guess he's taking a QM course. I have seen QM courses being taught by defining Hilbert space as infinite dimensional. Some even define Hilbert space as $L^2(\mathbb{R})$ or $\ell^2$. I don't know why they don't teach it the right way though.

I don't know who would describe a hilbert space as infinite dimensional only as one of the most important hilbert spaces in QM is two-dimensional. It's hard to talk about Hilbert spaces properly when a very good portion of the students won't even know what completeness means.

An example space which is hilbert but not eucledian space can be set of all continuous function in a given interval..correct ?

You need to specify an inner product.

 

[R136a1]

If hilbert space can be finite or infinite dimension. what about eucledian space can it be infinite dimensional?

Underwhat special condition a hilber space becomes eucledian space?actually first i thought eucledian space as finite dimension but now I'm confused what it is.

Yes, Euclidean space is the same as a finite-dimensional Hilbert space. So the Euclidean space does not have infinite dimension by definition. A Hilbert space can be finite or infinite dimensional.

 

[R136a1]

I don't know who would describe a hilbert space as infinite dimensional only as one of the most important hilbert spaces in QM is two-dimensional. It's hard to talk about Hilbert spaces properly when a very good portion of the students won't even know what completeness means.

Sakurai:

Later, in section 1.6, we consider the case of continuous spectra - for example, the position (coordinate) or momentum of a particle - where the number of alternatives is nondenumerably infinite, in which case the vector space in question is known as a Hilbert space after D. Hilbert, who studied vector spaces in infinite dimensions.

I got to say that this is one of the absolute worst descriptions of Hilbert space that I've ever seen. I'm the last one to say that QM books should do rigorous math, but if they do mention things like Hilbert spaces, it should at the very least be approximately correct. /rant

 

[economicsnerd]

Sorry for the OT, but...

Some even define Hilbert space as $L^2(\mathbb{R})$ or $\ell^2$. I don't know why they don't teach it the right way though.

I think there's some out-of-date terminology that agrees with this. While you and I know a Hilbert space as an inner-product space whose induced metric space structure is complete, I've been told they used to refer to (the) Hilbert space, which is what we would call a separable, infinite-dimensional Hilbert space. Of course the word "the" is validated by there being only one such space (up to unitary equivalence).

 

[R136a1]

Sorry for the OT, but...

I think there's some out-of-date terminology that agrees with this. While you and I know a Hilbert space as an inner-product space whose induced metric space structure is complete, I've been told they used to refer to (the) Hilbert space, which is what we would call a separable, infinite-dimensional Hilbert space. Of course the word "the" is validated by there being only one such space (up to unitary equivalence).

Yeah. Originally, Hilbert only worked with the $\ell^2$ or $L^2$ space. He didn't consider more general spaces.

Probably apocryphal, but: once, a mathematician gave a lecture in front of other mathematicians. He said "consider a Hilbert space H and...". He was interrupted as somebody in the audience asked "What exactly is a Hilbert space?" That somebody was David Hilbert.

 

[Jorriss]

I got to say that this is one of the absolute worst descriptions of Hilbert space that I've ever seen. I'm the last one to say that QM books should do rigorous math, but if they do mention things like Hilbert spaces, it should at the very least be approximately correct. /rant

That's a very surprising description.

 

[economicsnerd]

Probably apocryphal, but: once, a mathematician gave a lecture in front of other mathematicians. He said "consider a Hilbert space H and...". He was interrupted as somebody in the audience asked "What exactly is a Hilbert space?" That somebody was David Hilbert.

A similar story, about Shizuo Kakutani. He proved a useful generalization of Brouwer's fixed point theorem (to set-valued functions), which is used all over the place in economic theory. Once, he presented at a conference and had a bunch of economists in the audience, so he asked a game theorist buddy why. When the game theorist told him that it was likely because of the Kakutani fixed point theorem, Kakutani replied, "What is the Kakutani fixed point theorem?"

 

[mathman]

It appears that this question has degenerated into a quibble about definition. The original question was about the difference between (finite dimensional) Euclidean space and (infinite dimensional) Hilbert space. I suggest that replies should try to address that question and not get hung up on definitions.

 

[pwsnafu]

A similar story, about Shizuo Kakutani. He proved a useful generalization of Brouwer's fixed point theorem (to set-valued functions), which is used all over the place in economic theory. Once, he presented at a conference and had a bunch of economists in the audience, so he asked a game theorist buddy why. When the game theorist told him that it was likely because of the Kakutani fixed point theorem, Kakutani replied, "What is the Kakutani fixed point theorem?"

And again, with Banach, when he was reading a paper asked what a Banach space was (he called them B-spaces IIRC).

 

[johnqwertyful]

It appears that this question has degenerated into a quibble about definition. The original question was about the difference between (finite dimensional) Euclidean space and (infinite dimensional) Hilbert space. I suggest that replies should try to address that question and not get hung up on definitions.

Uhh, the question is a question about definitions. I don't know what else he's asking for if not for clarification of the definitions.

 

[mathwonk]

i agree with mathman.

 

[economicsnerd]

Okay, so rephrasing the question: What makes finite-dimensional Hilbert spaces (a.k.a. Euclidean spaces) special within the class of all Hilbert spaces?
- Of course, there is a finite spanning set. i.e. For some $n\in \mathbb Z_+$, there is some $\{e_i\}_{i=1}^n$ (which can be taken to be pairwise orthogonal) such that every vector is a linear combination of $\{e_i\}_{i=1}^n$.
- Every closed, bounded subset is compact.
- Every bounded sequence converges to something.
- Every linear subspace is itself a Hilbert space.
- Every linear map (into any other linear space) is continuous.
- Carathéodory's theorem: there is a finite number $k\in \mathbb N$ ($k$ can be taken to be $n+1$ with $n$ as above) such that: for every subset $P$ and every $x$ that can be written as a convex combination of points in $P$, there is a way to write $x$ as a convex combination of $k$ or fewer points in $P$.
- A lot of other stuff...

Basically, finite-dimensional spaces enjoy a lot of special properties that general Hilbert spaces don't.

 

[R136a1]

- Every bounded sequence converges to something.

I think you made a typo here. Of course it should be that any bounded sequence has a convergent subsequence.

 

[economicsnerd]

I think you made a typo here. Of course it should be that any bounded sequence has a convergent subsequence.

Hah! Thank you. Yes.

Every bounded sequence has a subsequence that converges to something.

 

[WWGD]

Hah! Thank you. Yes.

Every bounded sequence has a subsequence that converges to something.

Isn't this true in any metric space? Consider the bounded sequence as a metric (sub)space S. Then S is a compact (sub)space so that every sequence has a convergent subsequence.

 

[WWGD]

Now, another question:

Does every Hilbert space have a Schauder basis ?

 

[R136a1]

Isn't this true in any metric space? Consider the bounded sequence as a metric (sub)space S. Then S is a compact (sub)space so that every sequence has a convergent subsequence.

No, it's not true in any metric space. One example is an infinite-dimensional Hilbert space. Or the discrete metric space.

The failure of your proof is saying that the boundd sequence is compact. Not every bounded and closed space is compact.

 

[R136a1]

Now, another question:

Does every Hilbert space have a Schauder basis ?

No, only separable spaces have Schauder bases (and not even all separable spaces have one). So the only Hilbert spaces with Schauder bases are $\mathbb{R}^n$ and $\ell^2$.

A Hilbert spaces does always have an orthonormal basis.

 

[WWGD]

No, it's not true in any metric space. One example is an infinite-dimensional Hilbert space. Or the discrete metric space.

The failure of your proof is saying that the boundd sequence is compact. Not every bounded and closed space is compact.

Right, my bad; I forgot the obvious fact that compact== closed+ bounded does not always apply, even tho I've seen this a million times. Embarrassing.

 

[economicsnerd]

So the only Hilbert spaces with Schauder bases are $\mathbb{R}^n$ and $\ell^2$

Said differently, a Hilbert space has a Schauder basis if and only if it's separable.

 

[ajayguhan]

So we can say that hilbert space is a a inner product vector space which can be of finite or infinite dimension.if it's finite we can call it as eucledian space...?

Another question

Every vector space where inner product is defined is a hilbert space...is it True ?

 

[economicsnerd]

Every vector space where inner product is defined is a hilbert space...is it True ?

This isn't true. Do you know what a complete metric space is? Every inner-product space has a metric (i.e. distance) on the space induced by the inner product, via $$d(x,y)=\langle x-y, x-y\rangle^{\frac{1}{2}},$$ and an inner-product space is a Hilbert space if this metric is complete (i.e. has the property that every Cauchy sequence converges).

An example of an inner-product space is $l^2_0 = \{ x\in l^2:\enspace \exists n \text{ such that } x_i = 0 \text{ for all } i\geq n\}$, with the inner-product inherited from $l^2$.

As it turns out, every finite-dimensional inner-product space is a Hilbert space.

 

[WWGD]

So we can say that hilbert space is a a inner product vector space which can be of finite or infinite dimension.if it's finite we can call it as eucledian space...?

Another question

Every vector space where inner product is defined is a hilbert space...is it True ?

No; you need to have a specific relationship between the inner-product and the metric:

the inner-product needs to generate the norm ( and so generate the metric which is itself

generated by the norm.). The space has to be complete under this norm, altho a metric space is,

in a sense, as good as a complete metric space, since it has a completion (tho it is a nice exercise to show that the completion preserves the fact that the metric is generated by the inner-product.)

This happens iff the inner-product satisfies the parallelogram law;

out of all $$L^p$$ and $$l^p$$ spaces, only p=2 gives you a Hilbert space.

See, e.g:

http://math.stackexchange.com/questions/294544/parallelogram-law-valid-in-banach-spaces