From: "Eduardo Chappa L." <chappa@math.washington.edu>
Subject: Re: QUESTION: adjoint operator in an inner product space
Date: Thu, 18 May 2000 01:41:26 -0700
Newsgroups: sci.math
Summary: When do linear maps on inner product spaces have adjoints?

*** bhalchin@hotmail.com wrote in the sci.math newsgroup today:

:)    In an inner product space X, if I have a linear operator A: X->X,
:) 
:) - does A's adjoint necessary exits??
:) 
:) - if so, how does one prove??

If X has finite dimension or A is continuous and X has any dimension you
have to use the Schwartz lemma which states that any linear functional
from  X to the field is given by inner product with a vector in the space
X, in this form the functional x -> <Ax,y> (for fixed y) must be the inner
product of <x,z> for some unique z in X. This element z is dependent only
on y and is the image under the adjoint of A of y.

-- 
Eduardo
http://www.math.washington.edu/~chappa/pine/
==============================================================================

From: spamless@Nil.nil
Subject: Re: QUESTION: adjoint operator in an inner product space
Date: 18 May 2000 08:59:00 -0400
Newsgroups: sci.math

bhalchin@hotmail.com wrote:

>    In an inner product space X, if I have a linear operator A: X->X,
> 
> - does A's adjoint necessary exits??

Consider <Ax,y> for a fixed y and let x vary. This does is a linear
operator on x (an element of the dual of X). Call it T_y(x) (some operator
on X). This gives a map from y to T_y (from X, where y lives, to the dual
of X). When you speak of an inner product space, I assume you mean one
where the dual space of X is X itself. So ... T_y is a map from X to
itself.
==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Subject: Re: QUESTION: adjoint operator in an inner product space
Date: 18 May 2000 19:24:38 GMT
Newsgroups: sci.math

In article <8fvv3g$311$1@nnrp1.deja.com>,  <bhalchin@hotmail.com> wrote:

>   In an inner product space X, if I have a linear operator A: X->X,
>
>- does A's adjoint necessary exits??

No (depending on what you mean), unless you assume X is a Hilbert 
space and A is bounded.  

The adjoint operator A* should go from X* (the bounded 
linear functionals on X) to itself, defined by (A*)(f)(x) = f(Ax).
Note that the identification of X* with X only works for a Hilbert
space.  For example, if X is a dense linear subspace of a Hilbert 
space H, X* = H* (i.e. every bounded linear functional on X extends
to a unique bounded linear functional on H with the same norm), and
H* can be identified with H.  The trouble is that if A is not a 
bounded linear operator, x -> f(Ax) need not be a bounded linear 
functional.  So A* in general will only be defined on a linear subspace 
of X*, and it is possible for that subspace to consist only of 0.
I don't know if you would count that as "A's adjoint exists".

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

From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: QUESTION: adjoint operator in an inner product space
Date: 18 May 2000 22:27:58 GMT
Newsgroups: sci.math

In article <8fvv3g$311$1@nnrp1.deja.com>, 
bhalchin@hotmail.com writes:
> Hello,
> 
>    In an inner product space X, if I have a linear operator A: X->X,
> 
> - does A's adjoint necessary exits??
> 
> - if so, how does one prove??


As several people have pointed out, there is trivially an "adjoint"
mapping on the dual spaces, so the result is true if each linear
function is given by inner product with something (or, when A is 
continuous, if each continuous linear function is given by such
an inner product).

But the answer is no in general. For example, take a space with
a (countably infinite) linear basis e1, e2, e3, e4, ...  
Define an inner product by making them orthonormal (that is,
<ei,ei> = 1 and <ei,ej> = 0 for i not equal to j). Define A
by A(e1) = e1,   A(ei) = e1 + ei  for  i > 1.  If there were
an adjoint A*, then A*(e1) would have to satisfy
   (A*(e1), ei> = <e1, A(ei)> = <e1, e1 + ei> =1 
for all i bigger than 1.  No linear combination of
the basis elements will do that.


William C. Waterhouse
Penn State