THE VOLTERRA OPERATOR

JOHN THICKSTUN

Let V be the indefinite integral operator defined by Z t V f(t) = f(s)ds. 0 This is a linear operator. It can be defined on any domain of integrable functions, but here we restrict ourselves to domains where it behaves nicely. For example, if f ∈ Lp[0, 1] for p ∈ (1, ∞] then by H¨older’sinequality Z x 1/q |V f(x) − V f(y)| ≤ |f(s)|ds ≤ kfkp|x − y| . y p So V f is (H¨older-)continuous on [0, 1]. If Bp is the unit ball in L [0, 1] then the image of Bp in V is equicontinuous. Furthermore, Z t |V f(t)| ≤ |f(s)|ds ≤ |t|1/q ≤ 1. 0 p Therefore the image of Bp is (uniformly) bounded. By Arzela-Ascoli, V : L [0, 1] → C[0, 1] is compact. The preceeding argument does not go through when V acts on L1[0, 1]. In this case equicontinuity fails, as is demonstrated by the following family {fn} ⊂ B1:

fn(s) = n1[0,1/n](s).

This suffices to preclude compactness of V ; in particular, V fn has no Cauchy subsequence. Suppose V f = λf for some λ 6= 0. By definition λf is (absolutely) continuous. We deduce that V f (and therefore f) is continuously differentiable and that f = λf 0. It follows that f(s) = ces/λ. But then 0 = λf − V f = c so V has no eigenvalues and by the spectral theorem for compact operators, σ(V ) = {0}. We now turn our attention to the operator norm of V . For now we restrict V to the square integrable domain L2. Note that C[0, 1] ⊂ L∞[0, 1] ⊂ L2[0, 1] and the identity mapping I : C[0, 1] → L2[0, 1] is a bounded linear operator. It follows that IV : L2[0, 1] → L2[0, 1] is compact and we will proceed to consider V : L2[0, 1] → L2[0, 1]. Recall that L2[0, 1] is a Hilbert space with an inner product defined by Z 1 hf, gi = f(t)g(t)dt. 0 If kfk ≤ 1 then by Cauchy-Schwartz kV fk2 = hV ∗V f, fi ≤ kV ∗V fkkfk ≤ kV ∗V k ≤ kV ∗kkV k. 1 2 JOHN THICKSTUN

Therefore kV k ≤ kV ∗k and by a symmetric argument, kV ∗k ≤ kV k. It follows that kV k2 = kV ∗V k and we can compute kV k in terms of V ∗V . The adjoint of V is Z 1 V ∗f(t) = f(s)ds. t This is easily verified by Fubini’s theorem: Z 1 Z t Z 1 Z 1 hf, V fi = f(t) f(s)dsdt = f(s) f(t)dtds = hV ∗f, fi. 0 0 0 s Because compact operators form an ideal, V ∗V is compact. Clearly V ∗V is self-adjoint. By the spectral theorem for compact self-adjoint operators, V ∗V is diagonalizable and therefore its operator norm is just the magnitude of its largest eigenvalue. Suppose V ∗V f = λf, f 6= 0. Then f is continuous and V f is continuously differentiable. It follows that f ∈ C2[0, 1] and 2 Z 1 Z t Z x 00 ∂ ∂ λf (x) = 2 f(s)dsdt = − f(s)ds = −f(x). ∂x x 0 ∂x 0 Let ω2 = 1/λ. We deduce that eigenfunctions of V ∗V must be of the form f(x) = aeiωx + be−iωx. Furthermore, routine integration shows that Z 1 Z t Z 1 Z t V ∗V f(x) = a eiωsds + b e−iωsds x 0 x 0 1 1 1 1 = f(x) + (a − b)x − aeiω + be−iω
− (a − b) ω2 iω ω2 iω From the second term, we must have a = b and f therefore has the form 2a cos(ωx). From the third term, since a 6= 0, cos(iω) = 0. It follows that f is an eigenfunction if and only 2n+1 ∗ if ω = 2 π. The eigenvalues of V V are therefore 4 λ = . n (2n + 1)2π2 Maximizing over n we see that the largest eigenvalue of V ∗V , and therefore its operator 4 2 norm, is λ0 = π2 . We conclude that kV k = π . The operator norm is crucially dependent upon the operator’s domain of definition. For example, consider instead V : L1[0, 1] → L1[0, 1]. Then Z 1 Z t Z 1 Z 1 kV fk ≤ |f(s)|dsdt ≤ |f(s)|dsdt = kfk. 0 0 0 0 Therefore kV k ≤ 1. Using the same example we used earlier to rule out compactness of V , as n → ∞, Z 1 Z t Z 1/n Z 1 1 kV fnk = n1[0,1/n](s)dsdt = ntdt + dt = 1 − → 1. 0 0 0 1/n 2n So in this case kV k = 1. THE VOLTERRA OPERATOR 3

1. hilbert-schmidt operators Let L(U) denote the space of linear operators on a linear space U over F . When U is finite dimensional we can identify U ⊗ U ∗ with L(U) by associating u ⊗ v∗ ∈ U ⊗ U ∗ with T ∈ L(U) such that T (w) = v∗(w)u. Let ev : U × U ∗ → F be the (bilinear) evaluation functional defined by ev(u, v∗) = v∗(u). By universality of the tensor product this induces a map tr : U ⊗ U ∗ → F . We identify this map with a map tr : L(U) → F , which we call the trace. In infinite dimensions, U ⊗ U ∗ identifies only with finite rank operators. From the pre- ceeding discussion, we can define a Hilbert-Schmidt inner product of finite rank operators S, T ∈ L(U), ∗ hS, T iHS = tr(S T ). And we define the space of Hilbert-Schmidt operators to be the completion of the finite ∗ rank operators with respect to this inner product. Let evu : U ⊗ U → U be the evaluation map on finite rank operators defined by evu(T ) = T u. This map is uniformly continuous and by universality of completion it induces a map on the Hilbert-Schmidt operators. Furthermore, if T is Hilbert-Schmidt then

T (αu + v) = lim Tn(αu + v) = α lim Tnu + lim Tnv = αT u + T v. n→∞ n→∞ n→∞ Limits here are taken in the sense of the topology induced by the Hilbert-Schmidt inner product. We therefore identify Hilbert-Schmidt operators with a subspace of L(U). Suppose H is a Hilbert space with (possibly uncountable) orthonormal basis (ei)i∈B and let u ∈ H. If A is a bounded linear operator on H then A is continuous and has an abstract fourier representation X X Au = A hu, eiiei = hu, eiiAei. i∈B i∈B ∗ The latter sum converges in the sense of the operator topology. Let ei = hei, ·iHS be the dual basis associated with the Hilbert-Schmidt inner product. Then ! X ∗ X ∗ Au = ei (u)Aei = ei ⊗ Aei (u). i∈B i∈B Here the latter sum converges in the Hilbert-Schmidt topology. It follows that

X ∗ X ∗ X tr(A) = tr ei ⊗ Aei = tr(ei ⊗ Aei) = hAei, eii. i∈B i∈B i∈B We can now compute the Hilbert-Schmidt norm of an operator A:

2 ∗ X ∗ X 2 kAkHS = hA, AiHS = tr(A A) = hA Aei, eii = kAeik . i∈B i∈B 2 Observe that kAkHS < ∞ and therefore kAeik = 0 for all but countably many i ∈ B. If 2 we reorder (ei) as a countable sequence then (Aei) as a sequence in ` . Furthermore, by 4 JOHN THICKSTUN

Bessel’s inequality ∞ X 2 2 |hu, eii| ≤ kuk < ∞. i=1 2 2 Therefore hu, eii is also a sequence in ` . By Cauchy-Schwarz in ` ,

1 1 ∞ ∞ ∞ ! 2 ∞ ! 2 X X X 2 X 2 kAuk ≤ khu, eiiAeik = |hu, eii|kAeik ≤ |hu, eii| kAeik . i=1 i=1 i=1 i=1 This sum is finite from preceeding calculations and moreover

2 kAuk ≤ kAkHSkuk.

It follows that kAkop ≤ kAkHS. Therefore Hilbert-Schmidt limits of finite rank operators are operator norm limits of finite rank operators, which are compact. We conclude that Hilbert-Schmidt operators are compact. Does L2(X × X, µ ⊗ µ) = L2(X, µ) ⊕ L2(X, µ)? Let X be a σ-finite measure space with k ∈ L2(X × X). We associate k (which we call a kernel) with an integral operator K by the map Z Ku(x) = k(x, y)u(y)dy. X We will show that K is a Hilbert-Schmidt operator on L2(X). By Fubini’s theorem, k(x, ·) ∈ L2(X) almost everywhere. If k(x, ·) ∈ L2(X) and u ∈ L2(X) then by Cauchy- Schwarz on L2(X), Z |k(x, y)u(y)|dy ≤ kk(x, ·)kkuk < ∞. X Therefore k(x, y)u(y) ∈ L1(X) and K is defined on L2(X) (modulo null sets). Another application of Fubini shows that Z Z Z |Ku(x)|2dx ≤ kk(x, ·)k2kuk2dx = kuk2 |k(x, y)|2dy ⊗ dx = kuk2kkk2 < ∞. Y X X×X It follows that Ku ∈ L2(X) and we consider K as a linear operator K : L2(X) → L2(X). Our work further implies that kKk ≤ kkk. Let i ∈ B index an orthonormal basis (ei) of X. By Bessel’s inequality

2 ∗ X 2 X X 2 kKkHS = tr(K K) = kKeik ≤ |hKei, eji| . i∈B i∈B j∈B And by Fubini’s theorem again, Z Z hKei, eji = Kei(x)¯ej(x)dx = k(x, y)ei(y)e ¯j(x)dy ⊗ dx. X X×X THE VOLTERRA OPERATOR 5

2 Let uij(x, y) = ei(y)e ¯j(x) and observe that (uij) forms an orthonormal basis for L (X ×X). Then by another application of Bessel’s inequality, 2 X X 2 X 2 2 kKkHS ≤ |hKei, eji| = |hk, uiji| ≤ kkk . i∈B j∈B i,j∈B×B When L2(X) is separable Bessel’s inequality is replaced by Parseval’s identity and equality holds. We deduce that K is a Hilbert-Schmidt operator and (for separable spaces) the map k 7→ K is (Hilbert-Schmidt)-isometric. Remarkably, any Hilbert-Schmidt operator can be characterized by a kernel. Suppose K is Hilbert-Schmidt. Then K is can be written as a Hilbert-Schmidt limit of finite rank 2 operators Kn. For ui, vi ∈ L (X), we can write n n n X X Z Z X Knw(x) = ui ⊗ vi(x) = ui(x)v ¯i(y)w(y)dy = ui(x)v ¯i(y)w(y)dy. i=1 i=1 X X i=1 The kernel here is given by n X kn(x, y) = ui(x)v ¯i(y). i=1 Because kn 7→ Kn is a Hilbert-Schmidt isometry and (Kn) converges,

kkn − kmkL2 = kKn − KmkHS → 0. 2 Therefore (kn) is Cauchy and by completness of L (X) we deduce that kn → k for some k ∈ L2(X). This permits us to write Z Kw(x) = k(x, y)w(y)dy. X