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 2 Lp[0; 1] for p 2 (1; 1] then by H¨older'sinequality Z x 1=q jV f(x) − V f(y)j ≤ jf(s)jds ≤ kfkpjx − yj : 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 jV f(t)j ≤ jf(s)jds ≤ jtj1=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 ffng ⊂ 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 ) = f0g. 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] ⊂ L1[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 2 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 !2 = 1/λ. We deduce that eigenfunctions of V ∗V must be of the form f(x) = aei!x + be−i!x: Furthermore, routine integration shows that Z 1 Z t Z 1 Z t V ∗V f(x) = a ei!sds + b e−i!sds x 0 x 0 1 1 1 1 = f(x) + (a − b)x − aei! + be−i! − (a − b) !2 i! !2 i! From the second term, we must have a = b and f therefore has the form 2a cos(!x). From the third term, since a 6= 0, cos(i!) = 0. It follows that f is an eigenfunction if and only 2n+1 ∗ if ! = 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 ≤ jf(s)jdsdt ≤ jf(s)jdsdt = kfk: 0 0 0 0 Therefore kV k ≤ 1. Using the same example we used earlier to rule out compactness of V , as n ! 1, 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∗ 2 U ⊗ U ∗ with T 2 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 2 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!1 n!1 n!1 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)i2B and let u 2 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: i2B i2B ∗ 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): i2B i2B 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: i2B i2B i2B 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 : i2B i2B 2 Observe that kAkHS < 1 and therefore kAeik = 0 for all but countably many i 2 B. If 2 we reorder (ei) as a countable sequence then (Aei) as a sequence in ` . Furthermore, by 4 JOHN THICKSTUN Bessel's inequality 1 X 2 2 jhu; eiij ≤ kuk < 1: i=1 2 2 Therefore hu; eii is also a sequence in ` . By Cauchy-Schwarz in ` , 1 1 1 1 1 ! 2 1 ! 2 X X X 2 X 2 kAuk ≤ khu; eiiAeik = jhu; eiijkAeik ≤ jhu; eiij 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 2 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; ·) 2 L2(X) almost everywhere. If k(x; ·) 2 L2(X) and u 2 L2(X) then by Cauchy- Schwarz on L2(X), Z jk(x; y)u(y)jdy ≤ kk(x; ·)kkuk < 1: X Therefore k(x; y)u(y) 2 L1(X) and K is defined on L2(X) (modulo null sets). Another application of Fubini shows that Z Z Z jKu(x)j2dx ≤ kk(x; ·)k2kuk2dx = kuk2 jk(x; y)j2dy ⊗ dx = kuk2kkk2 < 1: Y X X×X It follows that Ku 2 L2(X) and we consider K as a linear operator K : L2(X) ! L2(X). Our work further implies that kKk ≤ kkk. Let i 2 B index an orthonormal basis (ei) of X. By Bessel's inequality 2 ∗ X 2 X X 2 kKkHS = tr(K K) = kKeik ≤ jhKei; ejij : i2B i2B j2B 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).