Board.

Integral operators with Hilbert-Schmidt kernel are compact
Theorem 1. Let
Z
Au(x) =
k(x, y)u(y) dy
Ω
with kernel k ∈ L2 (Ω × Ω). Then A ∈ L(L2 (Ω), L2 (Ω)) is compact.
Proof. The space L2 (Ω) has a countable orthonormal basis (ONB). (A Hilbert space has a countable
orthonormal basis if and only if it is separable.) Note that if φi is an ONB of L2 (Ω) then φi (x)φj (y) is
an ONB of L2 (Ω × Ω). We write
∞
X
k(x, y) =
ki,j φi (x)φj (y),
i,j=1
with
Z Z
ki,j =
k(x, y)φi (x)φj (y) dx dy.
Ω
Ω
We now define
kn (x, y) =
n X
∞
X
ki,j φi (x)φj (y),
i=1 j=1
and
Z
An u(x) =
kn (x, y)u(y) dy.
Ω
Obviously, An maps from L2 (Ω) into a finite-dimensional subspace of L2 (Ω). The range of An is finitedimensional and hence An is compact. We find
2
Z
(k(x,
y)
−
k
(x,
y))u(y)
dy.
k(A − An )uk2 = n
Ω
2
Z Z
(k(x, y) − kn (x, y))u(y) dy
dx
Z
2
2
≤
(k(x, y) − kn (x, y)) dy
u (y) dy dx
Ω
Ω
Z
ΩZ
Z
2
2
=
(k(x, y) − kn (x, y)) dy dx
u (y) dy
Ω
 Ω Ω

∞ X
∞
X
=
|ki,j |2  kuk2
=
Ω
Ω
Z Z
i=n+1 j=1
P∞
Since kkk2L2 (Ω×Ω) = i,j=1 |ki,j |2 < ∞, the above sum has to go to zero as n → ∞. We conclude that
An → A in the operator norm which yields the assertion with the following Theorem:
Theorem 2. Let A : X → Y be a linear operator and An ∈ L(X, Y ) a sequence of compact operators. If
n→∞
An → A in the operator norm, then A is compact.
The proof of this theorem can be found on page 408, theorem 8.1-5, of Introductory Functional Analysis
with Applications by Erwin Kreyszig, Wiley 1989.
Proof. Let {xn } be a bounded sequence in X. Since A1 is compact, there exists a subsequence x1n such
that A1 x1n converges. Iteratively/inductively, we now construct a subsequence as follows: If xkn is a
subsequence of xn such that Ak xkn converges, xkn is still bounded and we can find a subsequence xk+1
of
n
xkn such that Ak+1 xk+1
converges.
n
1
Consider the sequence (xnn )n∈N . There exist natural numbers rn such that xrn = xnn . Clearly, for all
k ∈ N, Ak xrn converges. Now
kAxrn − Axrm k =kAxrn − Ak xrn + Ak xrn − Ak xrm + Ak xrm − Axrm k
≤kAxrn − Ak xrn k + kAk xrn − Ak xrm k + kAk xrm − Axrm k
≤kA − Ak kkxrn k + kAk xrn − Ak xrm k + kAk − Akkxrm k
Now the first and third term can become arbitrary small since xn is bounded and Ak converges to A and
the middle term can become arbitrary small by the construction of the subsequence. Thus, we found a
subsequence of xn for which Axrn is a Cauchy sequence and therefore convergent.
2

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