Hilbert and Euclidian space

  • Thread starter ajayguhan
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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

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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.
 
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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.
 
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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 ?
 
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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 [tex]d(x,y)=\langle x-y, x-y\rangle^{\frac{1}{2}},[/tex] 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 [itex]l^2_0 = \{ x\in l^2:\enspace \exists n \text{ such that } x_i = 0 \text{ for all } i\geq n\}[/itex], with the inner-product inherited from [itex]l^2[/itex].

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

WWGD

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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
 
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