- #1
in vacuo
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Homework Statement
Show that a closed symmetric operator has a matrix representation.
Homework Equations
There are lots. I'm hoping somebody familiar with linear operators in Hilbert spaces is reading this!
The Attempt at a Solution
Hi,
I'm trying to prove that a closed symmetric operator A with a dense domain
D(A) in a separable, infinite-dimensional Hilbert space H has a matrix
representation given by
a_{kl} = (Ae_l, e_k)
where {e_n} is a suitable basis in D(A) (the exercise asks to construct this
basis too). I constructed a proof, but I did not use the fact that A is
closed, so I'm worried that I made a mistake somewhere. I would appreciate
if someone could help point out where the closed property of A is used.
My proof follows. (Note that in all cases "inf" is the symbol for infinity.)
It consists of four parts:
STEP 1: There exists an everywhere dense sequence (g_n) \subset D(A) with the
following property: for every f in D(A) there exists a subsequence (g_n_k) of
(g_n) such that
f = lim_{k -> inf} g_n_k,
Af = lim_{k -> inf} Ag_n_k.
PROOF OF STEP 1: Let f be an element of D(A). Since H is separable, there
exists a sequence (h_n) in H such that {h_n} is dense in H. For each
ordered triple of natural numbers (l,m,n), let G_lmn be a set consisting of
all g in D(A) such that
|| h_m - g || < 1/l, || h_n - Ag || < 1/l
Next, let G be a set containing one member of each G_lmn, where (l,m,n)
is an ordered triple of natural numbers, which is non-empty. It is clear
that G is countable; let us label its elements thus:
G = {g_n}
We claim that G is dense in D(A). For let f be in D(A) and eps > 0; then
there exits an ordered triple of natural numbers (l,m,n) s.t. 2/l < eps
and
|| f - h_m || < 1/l, || Af - h_n || < 1/l
Therefore f is in G_lmn, so G_lmn is not empty. Accordingly, there exists
g_k in G such that
(eq1) || g_k - h_m || < 1/l, || Ag_k - h_n || < 1/l
Therefore
(eq2) || f - g_k || <= || f - h_m || + || h_m - g_k ||
< 1/l + 1/l < eps.
Hence G is dense in D(A). Since D(A) is dense in H, so is G. We also have
|| Af - Ag_k || <= || Af - h_n || + || h_n - Ag_k || < 1/l + 1/l < eps.
We now prove the main result. Let f be in D(A); if f is also in G, we don't really
need a subsequence. Otherwise, if f is not in G, for each natural number
k, Equations (eq1) and (eq2) imply that there exists a natural number
n_k such that g_n_k is in G and
|| f - g_n_k || < 1/k, || Af - Ag_n_k || < 1/k
Furthermore, there are an infinite number of elements in G which satisfy these
relations, so we can always choose a g_n_k such that n_k > n_{k-1}, etc.
In this way we construct the subsequence (g_n_k), which has the properties
f = lim_{k -> inf} g_n_k, Af = lim_{k -> inf} Ag_n_k
QED
STEP 2: there exits a basis {e_n} in D(A) for H with the following property:
let W be the linear manifold of finite complex linear combinations of basis
vectors e_k (k a natural number); for every f in D(A) there exists a
sequence {f_n} in A such that
f = lim_{n -> inf} f_n, Af = lim_{n -> inf} Af_n
PROOF OF STEP 2: We construct a linearly independent set from {g_n} in
step 1 as follows. Let n_1 be the smallest natural number such that
g_n_1 is not zero. If n_1 < n_2 < ... < n_k have already been chosen
so that g_n_1, ..., g_n_k are linearly independent and g_n is a finite
linear combination of g_n_1, ..., g_n_k for 1 <= n <= n_k, let n_{k+1} > n_k
be the smallest integer such that g_n_1, ..., g_n_{k+1} are linearly
independent. In this way, we define the linearly independent set
U = {g_n_k: k a natural number}
(This set is countably infinite since H is infinite-dimensional). We
apply the Gram-Schmidt orthogonalisation process to the members of U to
construct a new family of vectors:
U' = (g'_n: n a natural number}
We then define the following family of orthonormal vectors:
B = {e_n: n a natural number}
where e_n is defined by g'_n / || g'_n || for all natural numbers n. Now
each original g_n is a finite linear combination of members of U, and the
members of U are finite linear combinations of the members of U', and by
implication those of B. Hence each g_n is a finite linear combination of
vectors in B. Now, let f be in D(A). From step 1, there exists
a subsequence (g_n_k) of (g_n) such that
f = lim_{k -> inf} g_n_k, Af = lim_{k -> inf} Ag_n_k
But by the foregoing, each g_n is actually a member of W. We can relabel
the g_n_k's as f_n's to obtain the desired result.
Finally, we show that {e_n} is a basis for H. By construction, {e_n}
is an orthonormal set; all that remains is to show that it is maximal.
Suppose there exists u in H such that (e_n, u) = 0 for all natural numbers
n. Let f be in D(A); then by the earlier result, there exists a sequence
(f_n) in W such that
f = lim_{n -> inf} f_n, Af = lim_{n -> inf} Af_n
Since each f_n is a finite linear combination of members of {e_n}, we have
(f_n, u) = 0 for all natural numbers n. Thus,
(f,u) = lim_{n -> inf} (f_n, u) = 0.
Thus u is perpendicular to all vectors in D(A). Next, let x be in H. Since
D(A) is dense in H, there exists a sequence (x_n) in D(A) such that
x = lim_{n -> inf} x_n
Now (x_n, u) = 0 for all natural numbers n, so
(x, u) = lim_{n -> inf} (x_n, u) = 0
In particular, letting x = u gives (u,u) =0, so u =0. Therefore the set
{e_n} is maximal in H, so it is a basis.
QED.
STEP 3: Suppose A is symmetric, let {e_k} be a basis as in step 2, and
define
a_{kl} = (Ae_l, e_k) for all natural numbers k and l.
Furthermore, define the operator A' on the linear manifold
D(A') = {g = sum{k=1 to inf} b_l e_l: sum_{k=1 to inf} | sum_{l=1 to inf} a_{kl} b_l|^2 < inf }
by
A'g = sum_{k = 1 to inf} ( sum_{l = 1 to inf} a_{kl} b_l ) e_k.
Then A' = A* (where A* is the adjoint of A).
PROOF OF STEP 3: We first show that A* \subset A'. Let g be in D(A*),
where D(A*) is the domain of A*. Since {e_k} is a basis for H, g has the
Fourier expansion
g = sum_{l = 1 to inf} b_l e_l
Now Parseval's identity gives
inf > || A*g ||^2 = sum_{k=1 to inf} |(A*g, e_k)|^2 = sum_{k=1 to inf} |(Ae_k, g)|^2
From the Fourier expansion of g and the continuity of the inner product,
we have
(Ae_k, g) = sum_{l = 1 to inf) (Ae_k, b_l e_l) = sum_{l = 1 to inf) conj(b_l) (Ae_k, el)
= sum_{l = 1 to inf} conj(b_l) a_{lk}.
Note that conj(x) denotes the complex conjugate of x. Since A is symmetric,
from the definition of a_{kl} we see that a_{lk} = conj(a_{kl}, so
(Ae_k, g) = sum_{l = 1 to inf} conj(a_{kl} b_l).
Therefore
inf > || A*g ||^2 = sum_{k = 1 to inf} | sum_{l = 1 to inf} a_{kl} b_l |^2,
hence g is in D(A'), the domain of A'. To show that A*g = A'g,
A*g = sum_{k = 1 to inf} (A*g, e_k) e_k = sum_{k = 1 to inf} conj( (e_k, A*g) )e_k
= sum{k = 1 to inf} conj( (Ae_k, g) ) e_k
= sum{k = 1 to inf} ( sum_{l = 1 to inf} a_{kl} b_l ) e_k
= A'g.
Therefore A* \subset A'.
We now show that A' \subset A*. Let g be in D(A'). Now g has the Fourier
expansion
g = sum_{l = 1 to inf} b_l e_l
Let l be a natural number; then
(Ae_l, g) = sum_{k = 1 to inf} conj(b_k) (Ae_l, e_k)
= sum{k = 1 to inf) conj(b_k) a_{kl}
But A'g = sum{k = 1 to inf} ( sum_{l = 1 to inf} a_{kl} b_l ) e_k, so
(e_l, A'g) = sum_{k = 1 to inf) conj( a_{lk} b_k ) = sum_{k = 1 to inf} conj(b_k) a_{kl}
= (Ae_l, g).
It follows then that if u is any vector which is a finite linear
combination of basis vectors, we have
(eq3) (Au, g) = (u, A'g).
We now let f be in D(A) and let eps > 0. By step 2, there exists a
sequence (f_n) in W such that
f = lim_{n -> inf} f_n, Af = lim_{n -> inf} Af_n
Now
|(Af, g) - (f, A'g)|
<= |(Af - Af_n, g)| + |(Af_n, g) - (f_n, A'g)| + |(f-f_n, A'g)|
But from (eq3), we know that |(Af_n, g) - (f_n, A'g)| = 0, so
|(Af, g) - (f, A'g)| <= || Af - Af_n || || g || + || f - f_n || || A'g ||
since || g || and || A'g || are bounded, there exits a natural number N
such that
|| Af - Af_n || || g || < eps/2, || f - f_n || || A'g || < eps/2
for all n >= N. Therefore
|(Af, g) - (f, A'g)| < eps.
Since eps is arbitrary, (Af, g) = (f, A'g). Thus g is in D(A*) and
A'g = A*g, so A' \subset A*.
From the above considerations, we see that A' = A*.
QED
STEP 4: The operator A has a matrix representation.
PROOF OF STEP 4: Since A is symmetric, we have for all f in D(A):
Af = A*f = A'f.
QED
As one can see, I have used the assumptions that A is symmetric, that H
is separable and infinite-dimensional, and that D(A) is dense in H. But
I have not used (at least explicitly) the fact that A is closed. I know the
above proof is quite long but again I would greatly appreciate it if someone
could point out to me either where I have made a mistake or I have implicitly
used the assumption that A is closed.
Thanks in advance.