“Proof” of Hilbert-Schmidt Theorem
The goal of this note is to give a proof of the following general theorem in the special
case that is needed for our applications.
Theorem 0.1 (Hilbert-Schmidt). Let H 6= 0 be an infinite dimensional, separable Hilbert
space and let K ∈ L(H) be compact and self adjoint. Then there exists a countable orthonormal basis (O.N.B.) of H consisting of eigenvectors of K.
Recall that the eigenvlaues of a compact operator K ∈ L(H) form a discrete set that
is either finite or forms a sequence tending to zero. Furthermore, each non-zero eigenvalue
has a finite dimensional eigenspace. From the Hilbert-Schmidt theorem, it follows that if
K is also self adjoint, then σp (K) being a finite set implies that λ = 0 is an eigenvalue of K
with an infinite dimensional eigenspace. Indeed, the Hilbert-Schmidt theorem implies that
[
N (K − λI) = H,
λ∈σp (K)
which, since H is inifinte dimensional, is impossible unless either σp (K) is inifinite or at least
one of the eigenspaces is infinite dimensional (or both!). For our intended application, it is
sufficient to consider the case that 0 ∈
/ σp (K), in which case the Hilbert-Schmidt theorem
implies that K has infinitely many (real) eigenvalues {λj }∞
j=1 such that |λj | → ∞ as j → ∞.
We now prove the above theorem in the case where 0 ∈
/ σp (K).
Proof. Assume that, in addition to the stated hypothesis, 0 ∈
/ σp (K). Recall, however, that
0 ∈ σ(K) due to the infinite dimensionality of H.
Claim: λ = 0 is not anSislodated point of σ(K).
Indeed, set M = λ∈σp (K) N (K − λI) and note that N (K − λI) is finite dimensional
for each λ ∈ σp (K). Using the fact that λ = 0 is the only possible accumulation point of
σp (K), it follows that if λ = 0 is an isolated point of σ(K) then σp (K) is a finite set, and
hence M is finite dimensional. Now, using a Gram-Schmidt process, for each λ ∈ σp (K) we
can find a (finite) O.N.B. of the eigenspace N (K − λI). Since eigenspaces corresponding
to distinct eigenvalues are orthogonal, this provides us with a (finite) set of orthonormal
dim(M )
vectors {vj }j=1
that provide an O.N.B. for M . Since M is closed, we can now define the
projection operator Π : H → M ⊥ onto the orthogonal complement of M given by
dim(M )
Π(u) = u −
X
hφj , ui φj .
j=1
The restriction of K to M ⊥ , given explicitly by K|M ⊥ := ΠKΠ : M ⊥ → M ⊥ , which, by
definition of M , is1 a self adjoint compact operator with σ(K|M ⊥ ) = {0}. Since K|M ⊥ is
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You should check these claims for yourself!
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self adjoint, it follows that
kK|M ⊥ k =
sup
|λ| = 0
λ∈σ(K|M ⊥ )
and hence K|M ⊥ is the zero operator. But then it follows that Kv = 0 for all v ∈ M ⊥ so
that 0 ∈ σp (K), a contradiction. Thus, the claim holds.
From the above claim, it follows that σp (K) consists of infinitely many eigenvalues
{λj }∞
j=1 with λj → 0. As above, using a Gram-Schmidt process we can use the corresponding eigenspaces to construct a countably infinte collections {φj }∞
j=1 of eigenvectors of K.
∞
Set X := span{φj }j=1 .
Claim X = H.
If not, then the operators K|X : X → X and K|X ⊥ : X ⊥ → X ⊥ (defined similarly as
K|M ⊥ above) are both compact, self adjoint operators. If X ⊥ 6= {0}, then either K|X ⊥ = 0,
in which case 0 ∈ σp (K) (again, a contradiction), or else kK|X ⊥ k > 0. In the latter case,
the self adjointness of K|X ⊥ implies at least one of the numbers ±kK|X ⊥ k is an eigenvalue
of K|X ⊥ . In particular, there would exist an eigenvector of K in the subspace X ⊥ , which
contradicts the definition of X. Thus, X ⊥ = {0} as desired.
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