The completeness of Hilbert Space

[Ed Quanta]

Can anyone guide me through or point me to a link of a proof that Hilbert space is complete? I am doing a paper on Hilbert space so I introduced some of its properties and now want to show it is complete.

 

[mathwonk]

what is your,definition ,of hilbert space? i.e. sometimes one defines "a hilbert space" as a complete inner product space, so it then complete by definition, and then task is to produce some exampels.

the standard oens are all spaces of square integrable functions, so completeness would be found in any real analysis book, or functional analysis book.

 

[Jimmy Snyder]

A Hilbert space is defined to be a complete inner product space. You cannot 'prove' that it is complete. Here is a link to a definition:

http://mathworld.wolfram.com/HilbertSpace.html

 

[fourier jr]

A Hilbert space is just a Banach space with an inner product. A Banach space is by definition a complete normed vector space with the metric d(x,y) = ||x-y||

The spaces $$L^2$$ (the space of square-integrable functions) & $$l^2$$ (square-summable sequences) are examples of Hilbert spaces, and the $$L^p$$ spaces are complete by the Riesz-Fischer theorem. Have a look at p.125 of Royden's Real Analysis (what else?!) for the proof.

 

[Ed Quanta]

Is Royden's real analysis on the web?

 

[dextercioby]

I doubt it.Does Rudin's book have it...?(The proof that $L^{2}\left(\mathbb{R}\right)$ and its complex counterpart are complete preHilbert spaces).

Daniel.

 

[fourier jr]

I don't know about that but I looked up Riesz-Fischer theorem & found that it was only mentioned twice (& not proved).

 

[saltydog]

A complete function space is a function set in which no Cauchy sequence of functions in the set converge to limits which are not in the set.

The Riesz-Fischer Theorem identifies the set of "square integrable functions" as a complete function (inner-product) space (a Hilbert Space):

Riesz-Fischer Theorem:

Let the functions $f_1(x),f_2(x),...$ be elements in a function space. If:

$$\mathop\lim\limits_{m,n\to\infty}||f_n-f_m||^2\equiv\mathop\lim\limits_{m,n\to\infty}\int_a^b|f_n(x)-f_m(x)|^2dx=0$$

then there exists a "square-integrable function" f(x) to which the sequence $f_n(x)$converges such that:

$$\mathop\lim\limits_{n\to\infty}\int_a^b|f(x)-f_n(x)|^2 dx=0$$

Edit: Thus all Cauchy sequences of square-integrable functions converge to functions which themselves are square-integrable. Makes sense right? If they converged to a function which was not square-integrable, then the set of square integrable functions would not be complete. You know . . . I'm pretty sure that's right. Correct me if it could be said differently.