From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: inner product space
Date: 13 Mar 2000 22:47:24 GMT
Newsgroups: sci.math.research
Summary: [missing]

In article <080320001443299965%edgar@math.ohio-state.edu.nospam>, 
"G. A. Edgar" <edgar@math.ohio-state.edu.nospam> writes:
> This is a problem in Kolmogorov & Fomin, Introductory Real
> Analysis.
> 
> Give an example of an inner product space such that there is
> no orthonormal set whose closed span is the whole space.
> 
> [Certainly such an example must be incomplete and nonseparable.]
>...

There is an example in N. Bourbaki, Espaces Vectoriels Topologiques,
 Chap. V, Section 2, Exercise 2.

Here is the construction (hints for the proof can be looked up
in Bourbaki).  Take a Hilbert space E1 with countably infinite
orthonormal basis, and Hilbert spaces E2 and E3 with orthonormal
bases the cardinality of the continuum.  Let  a_i be such an 
orthonormal basis of E2. Choose continuum many linearly independent
elements a_i in E1 and let H be the (algebraic) span of all ai+bi
inside the orthogonal sum of E1 and E2. Show that no orthonormal
subset of H can have an element in E2.  Let G = H + E3 inside the
orthogonal sum of E1 and E2 and E3. Then any orthonormal subset T of G
will meet the sum of E2 and E3 only in E3, and hence the projection
of T to E2 will be countable; but the projection of G onto E2 has
dense image.  

William C. Waterhouse
Penn State

==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Subject: Re: inner product space
Date: Mon, 13 Mar 2000 14:53:22 -0800 (PST)
Newsgroups: sci.math.research

In article <080320001443299965%edgar@math.ohio-state.edu.nospam>,
 Gerald Edgar <edgar@math.ohio-state.edu> wrote:
> This is a problem in Kolmogorov & Fomin, Introductory Real
> Analysis.
 
> Give an example of an inner product space such that there is
> no orthonormal set whose closed span is the whole space.
 
> [Certainly such an example must be incomplete and nonseparable.]

See Dixmier, "Sur les bases orthonormales dans les espaces prehilbertiens",
Acta Sci. Math. Szeged 15 (1953) 29-30.

Let H_1 be an infinite-dimensional separable Hilbert space.  There is a 
linearly independent set {x_b: b in B} in H_1, indexed by a set B of cardinality
c (e.g. in L^2([0,1]) you could take x_b for 0 < b < 1 as the step function
x_b(t) = 0 for t < b, 1 for t >= b).  Let H_2 be the non-separable Hilbert
space with orthonormal basis {y_b: b in B} indexed by B.  Let K be the linear
span of the vectors {x_b + y_b: b in B} in the direct sum H_1  + H_2.
Then there is no orthonormal set in K whose span is dense in K.
In fact:

1) Any orthonormal set in K is countable. 

Proof: Suppose { z_a: a in A } is an orthonormal set in K.  We can write each
z_a as a finite sum: sum_{i=1}^n c_i (x_{b_i} + y_{b_i}), where 
sum_i c_i x_{b_i} <> 0 since the x_b are linearly independent.  
Let { w_j: j in N } be an orthonormal basis of H_1.  For each j, we can
have <w_j, z_a> <> 0 for at most countably many a's, so there are at only
countably many a's for which some <w_j, z_a> <> 0, and thus at most countably
many a's for which any <w, z_a> <> 0 with w in H_1.  But since 
<sum_{i=1}^n c_i x_{b_i}, z_a> = ||sum_{i=1}^n c_i x_{b_i}||^2 > 0,
this says that A is at most countable.

2) The span of any countable orthonormal set is separable, and therefore
is not dense in K. 

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2