Riesz-Fredh¨olmTheory T. Muthukumar [email protected] Contents 1 Introduction1 2 Integral Operators1 3 Compact Operators7 4 Fredh¨olmAlternative 14 Appendices 18 A Ascoli-Arzel´aResult 18 B Normed Spaces and Bounded Operators 20 1 Introduction The aim of this lecture note is to show the existence and uniqueness of Fredh¨olmintegral operators of second kind, i.e., show the existence of a solution x of x − T x = y for any given y, in appropriate function spaces. 2 Integral Operators Let E be a compact subset of Rn and C(E) denote the space of complex valued continuous functions on E endowed with the uniform norm kfk1 = supx2E jf(x)j. Recall that C(E) is a Banach space with the uniform norm. 1 Definition 2.1. Any continuous function K : E × E ! C is called a con- tinuous kernel. Since K is continuous on a compact set, K is both bounded, i.e., there is a κ such that jK(x; y)j ≤ κ 8x; y 2 E and uniformly continuous. In particular, for each " > 0 there is a δ > 0 such that jK(x1; y) − K(x2; y)j ≤ " 8y 2 E whenever jx1 − x2j < δ. Example 2.1 (Fredh¨olmintegral operator). For any f 2 C(E), we define Z T (f)(x) = K(x; y)f(y) dy E where x 2 E and K : E × E ! R is a continuous function. For each " > 0, Z jT f(x1) − T f(x2)j ≤ jK(x1; y) − K(x2; y)jjf(y)j dy ≤ "kfk1jEj E whenever jx1 − x2j < δ. Thus, T f 2 C(E) and T defines a map from C(E) to C(E). Example 2.2 (Volterra integral operator). For any f 2 C[a; b], we define Z x T (f)(x) = K(x; y)f(y) dy a where x 2 [a; b] and K :[a; b] × [a; b] ! R is a continuous function. Note that, for x1; x2 2 [a; b], Z x1 Z x2 T f(x1) − T f(x2) = K(x1; y)f(y) dy − K(x2; y)f(y) dy a a Z x1 = [K(x1; y) − K(x2; y)]f(y) dy a Z x1 + K(x2; y)f(y) dy: x2 2 Therefore, for each " > 0, Z x1 jT f(x1) − T f(x2)j ≤ jK(x1; y) − K(x2; y)jjf(y)j dy a Z x1 + jK(x2; y)jjf(y)j dy x2 ≤ "kfk1(x1 − a) + κ(x1 − x2)kfk1 ≤ "kfk1(b − a) + κδkfk1 ≤ [(b − a) + κ]"kfk1 whenever jx1 − x2j < δ and, without loss of generality, we have assumed δ ≤ ". Thus, T f 2 C[a; b] and T defines a map from C[a; b] to C[a; b]. One can think of Volterra integral operator as a special case of Fredh¨olm integral operators by considering a K :[a; b] × [a; b] such that K(x; y) = 0 for y > x. Geometrically this means, in the square [a; b] × [a; b], K takes the value zero in the region above y = x line. Thus, Volterra integral operator is a Fredh¨olmoperator for a K which may be discontinuous on the line y = x in the square. In fact, this particular K is a special case of weakly singular kernel. Definition 2.2. Let E ⊂ Rn be a compact subset. A function K : E×E ! C is said to be weakly singular kernel if it is continuous for all x; y 2 E such that x 6= y and there exist positive constants M and α 2 (0; n) such that jK(x; y)j ≤ Mjx − yjα−n 8x; y 2 E; x 6= y: Theorem 2.3. Let E ⊂ Rn be a compact subset with non-empty interior. Let K : E ×E ! C be a continuous function. Then the operator T : C(E) ! C(E), defined as, Z (T f)(x) := K(x; y)f(y) dy; for each x 2 E E is a bounded linear operator, i.e., is in B(C(E)) and Z kT k = sup jK(x; y)j dy: x2E E 3 Proof. The fact that T is linear is obvious. For each f 2 C(E) with kfk1 ≤ 1 and x 2 E, we have Z j(T f)xj ≤ jK(x; y)j dy: E Thus, Z kT k = sup kT fk1 ≤ sup jK(x; y)j dy: f2C(E) x2E E kfk1≤1 It now only remains to show the other inequality. Since K is continuous, there is a x0 2 E such that Z Z jK(x0; y)j dy = max jK(x; y)j dy: E x2E E For each " > 0, choose g" 2 C(E) as K(x0; y) g"(y) := ; for y 2 E: jK(x0; y)j + " Then kg"k1 ≤ 1 and Z 2 jK(x0; y)j kT g"k1 ≥ j(T g")(x0)j = dy E jK(x0; y)j + " Z 2 2 Z jK(x0; y)j − " ≥ dy = jK(x0; y)j dy − "jEj: E jK(x0; y)j + " E Hence Z kT k = sup kT fk1 ≥ kT g"k1 ≥ jK(x0; y)j dy − "jEj; f2C(E) E kfk1≤1 and, since " is arbitrary, we have Z Z kT k ≥ jK(x0; y)j dy = max jK(x; y)j dy: E x2E E For any operator T : X ! X, one can define the iterated operators T k : X ! X as the k-times composition of T with itself. 4 Theorem 2.4 (Neumann Series). Let X be a Banach space, T 2 B(X) with kT k < 1 (contraction) and I : X ! X be the identity operator. Then the map I − T : X ! X is invertible given by 1 X (I − T )−1 = T k k=0 and the inverse is bounded 1 k(I − T )−1k ≤ : 1 − kT k Proof. Since X is a Banach space, by Theorem B.12, B(X) is a Banach space. P1 k Thus, owing to Theorem B.4, to show the convergence of the series k=0 T it is enough to show that it is absolutely convergent. Note that 1 m m 1 X X X X kT kk = lim kT kk ≤ lim kT kk = kT kk: m!1 m!1 k=0 k=0 k=0 k=0 Since kT k < 1, the geometric series converges to (1 − kT k)−1 and hence the P1 k series k=0 T is absolutely convergent and, thus, convergence in B(X). Let S : X ! X be the limit of the series, i.e., 1 X S := T k: k=0 Note that kSk ≤ (1 − kT k)−1, hence, is a bounded linear operator on X, i.e., S 2 B(X). It only remains to show that S is the inverse of I − T . Note that m X (I − T )S = (I − T ) lim T k = lim (I − T )(I + T + ::: + T m) m!1 m!1 k=0 = lim (I − T m+1) = I: m!1 The last equality is due to the fact that the sequence T m+1 converges to 0 in m+1 m+1 m+1 B(X) because kT k ≤ kT k and, since kT k < 1, limm!1 kT k ! 0. Similarly, m X S(I − T ) = lim T k(I − T ) = lim (I + T + ::: + T m)(I − T ) m!1 m!1 k=0 = lim (I − T m+1) = I: m!1 5 If x 2 X is the solution of (I − T )x = y, for any given y 2 X, then x can be computed as 1 m −1 X k X k x = (I − T ) y = T y = lim T y = lim xm: m!1 m!1 k=0 k=0 Note that xm+1 = T xm + y, for m ≥ 0. Theorem 2.5. Let X be a Banach space, T 2 B(X) with kT k < 1 (con- traction) and I : X ! X be the identity operator. For any given y 2 X and arbitrary x0 2 X the sequence xm+1 := T xm + y for m = 0; 1; 2;::: converges to a unique solution x 2 X of (I − T )x = y. 2 Proof. Note that x1 = T x0 + y, x2 = T x1 + y = T x0 + (I + T )y. Thus, by induction, for m = 1; 2;:::, m−1 m X k xm = T x0 + T y: k=0 Hence, 1 X k −1 lim xm = T y = (I − T ) y: m!1 k=0 Corollary 2.6. Let K be a continuous kernel satisfying Z max jK(x; y)j dy < 1: x2E E Then, for each g 2 C(E), the integral equation of the second kind Z f(x) − K(x; y)f(y) dy = g(x) x 2 E E has a unique solution f 2 C(E). Further, for any f0 2 C(E), the sequence Z fm+1(x) := K(x; y)fm(y) dy + g(x); m = 0; 1; 2;::: E converges uniformly to f 2 C(E). 6 3 Compact Operators Theorem 3.1. A subset of a normed space is compact iff it is sequentially compact. Definition 3.2. A subset E of a normed space X is said to be relatively compact if E, the closure of E, is compact in X. Theorem 3.3. Any bounded and finite dimensional subset of a normed space is relatively compact. Definition 3.4. Let X and Y be normed spaces. A linear operator T : X ! Y is said to be compact if T (E) is relatively compact in Y , for every bounded subset E ⊂ X.