RieszFredhölm Theory


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1 RieszFredhölm Theory T. Muthukumar Contents 1 Introduction 1 2 Integral Operators 1 3 Compact Operators 7 4 Fredhölm Alternative 14 Appendices 18 A AscoliArzelá Result 18 B Normed Spaces and Bounded Operators 20 1 Introduction The aim of this lecture note is to show the existence and uniqueness of Fredhölm integral 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 be a compact subset of R n and C() denote the space of complex valued continuous functions on endowed with the uniform norm f = sup x f(x). Recall that C() is a Banach space with the uniform norm. 1
2 con Definition 2.1. Any continuous function K : C is called a tinuous kernel. Since K is continuous on a compact set, K is both bounded, i.e., there is a κ such that K(x, y) κ x, y and uniformly continuous. In particular, for each ε > 0 there is a δ > 0 such that K(x 1, y) K(x 2, y) ε y whenever x 1 x 2 < δ. xample 2.1 (Fredhölm integral operator). For any f C(), we define T (f)(x) = K(x, y)f(y) dy where x and K : R is a continuous function. For each ε > 0, T f(x 1 ) T f(x 2 ) K(x 1, y) K(x 2, y) f(y) dy ε f whenever x 1 x 2 < δ. Thus, T f C() and T defines a map from C() to C(). xample 2.2 (Volterra integral operator). For any f C[a, b], we define T (f)(x) = x a K(x, y)f(y) dy where x [a, b] and K : [a, b] [a, b] R is a continuous function. Note that, for x 1, x 2 [a, b], T f(x 1 ) T f(x 2 ) = = x1 a x1 + a x1 K(x 1, y)f(y) dy x2 a [K(x 1, y) K(x 2, y)]f(y) dy x 2 K(x 2, y)f(y) dy. K(x 2, y)f(y) dy 2
3 Therefore, for each ε > 0, T f(x 1 ) T f(x 2 ) x1 + a x1 x 2 K(x 1, y) K(x 2, y) f(y) dy K(x 2, y) f(y) dy ε f (x 1 a) + κ(x 1 x 2 ) f ε f (b a) + κδ f [(b a) + κ]ε f whenever x 1 x 2 < δ and, without loss of generality, we have assumed δ ε. Thus, T f 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 Fredhölm 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 Fredhölm operator 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 R n be a compact subset. A function K : C is said to be weakly singular kernel if it is continuous for all x, y such that x y and there exist positive constants M and α (0, n) such that K(x, y) M x y α n x, y ; x y. Theorem 2.3. Let R n be a compact subset with nonempty interior. Let K : C be a continuous function. Then the operator T : C() C(), defined as, (T f)(x) := K(x, y)f(y) dy, for each x is a bounded linear operator, i.e., is in B(C()) and T = sup K(x, y) dy. x 3
4 Proof. The fact that T is linear is obvious. For each f C() with f 1 and x, we have (T f)x K(x, y) dy. Thus, T = sup f C() f 1 T f sup K(x, y) dy. x It now only remains to show the other inequality. Since K is continuous, there is a x 0 such that K(x 0, y) dy = max K(x, y) dy. x For each ε > 0, choose g ε C() as g ε (y) := K(x 0, y), for y. K(x 0, y) + ε Then g ε 1 and K(x 0, y) 2 T g ε (T g ε )(x 0 ) = K(x 0, y) + ε dy K(x 0, y) 2 ε 2 K(x 0, y) + ε dy = K(x 0, y) dy ε. Hence T = sup T f T g ε K(x 0, y) dy ε, f C() f 1 and, since ε is arbitrary, we have T K(x 0, y) dy = max x K(x, y) dy. For any operator T : X X, one can define the iterated operators T k : X X as the ktimes composition of T with itself. 4
5 Theorem 2.4 (Neumann Series). Let X be a Banach space, T B(X) with T < 1 (contraction) and I : X X be the identity operator. Then the map I T : X X is invertible given by (I T ) 1 = T k and the inverse is bounded k=0 (I T ) 1 k=0 1 1 T. Proof. Since X is a Banach space, by Theorem B.12, B(X) is a Banach space. Thus, owing to Theorem B.4, to show the convergence of the series k=0 T k it is enough to show that it is absolutely convergent. Note that m m T k = lim T k lim T k = T k. m m k=0 Since T < 1, the geometric series converges to (1 T ) 1 and hence the series k=0 T k is absolutely convergent and, thus, convergence in B(X). Let S : X X be the limit of the series, i.e., S := T k. k=0 Note that S (1 T ) 1, hence, is a bounded linear operator on X, i.e., S B(X). It only remains to show that S is the inverse of I T. Note that m (I T )S = (I T ) lim T k = lim (I T )(I + T T m ) m m k=0 = lim m (I T m+1 ) = I. The last equality is due to the fact that the sequence T m+1 converges to 0 in B(X) because T m+1 T m+1 and, since T < 1, lim m T m+1 0. Similarly, S(I T ) = lim m m k=0 = lim m (I T m+1 ) = I. k=0 k=0 T k (I T ) = lim m (I + T T m )(I T ) 5
6 If x X is the solution of (I T )x = y, for any given y X, then x can be computed as x = (I T ) 1 y = k=0 Note that x m+1 = T x m + y, for m 0. T k y = lim m m k=0 T k y = lim m x m. Theorem 2.5. Let X be a Banach space, T B(X) with T < 1 (contraction) and I : X X be the identity operator. For any given y X and arbitrary x 0 X the sequence x m+1 := T x m + y for m = 0, 1, 2,... converges to a unique solution x X of (I T )x = y. Proof. Note that x 1 = T x 0 + y, x 2 = T x 1 + y = T 2 x 0 + (I + T )y. Thus, by induction, for m = 1, 2,..., Hence, lim x m = m x m = T m x 0 + m 1 k=0 T k y. T k y = (I T ) 1 y. k=0 Corollary 2.6. Let K be a continuous kernel satisfying K(x, y) dy < 1. max x Then, for each g C(), the integral equation of the second kind f(x) K(x, y)f(y) dy = g(x) x has a unique solution f C(). Further, for any f 0 C(), the sequence f m+1 (x) := K(x, y)f m (y) dy + g(x), m = 0, 1, 2,... converges uniformly to f C(). 6
7 3 Compact Operators Theorem 3.1. A subset of a normed space is compact iff it is sequentially compact. Definition 3.2. A subset of a normed space X is said to be relatively compact if, the closure of, 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 () is relatively compact in Y, for every bounded subset X. Let K(X, Y ) be the space of all compact linear maps from X to Y. To verify compactness of T, it is enough to check the relative compactness of T (B) in Y for the closed unit ball B X. Theorem 3.5. A linear operator T : X Y is compact iff for any bounded sequence {x n } X, the sequence {T x n } Y admits a convergent subsequence. Proposition 3.6. The set K(X, Y ) is a subspace of B(X, Y ). Proof. Any compact linear operator is bounded because any relatively compact set is bounded. Thus, K(X, Y ) B(X, Y ). Let S, T K(X, Y ). It is easy to check that, for every bounded subset of X and α, β C, (αs + βt )() is relatively compact in Y, since S() and T () are both relatively compact in Y. Theorem 3.7. Let X, Y and Z be normed spaces, S B(X, Y ) and T B(Y, Z). If either one of S or T is compact, then the composition T S : X Z is compact. Proof. Let {x n } be a bounded sequence in X. Suppose S is compact, then there is a subsequence {x nk } of {x n } such that Sx nk y in Y, as k. Since T is continuous, T (Sx nk ) T y in Z, as k. Thus, T S is compact. On the other hand, since S is bounded, Sx n is bounded sequence in Y. If T is compact, then there is a subsequence {x nk } of {x n } such that T (Sx nk ) z in Z, as k. Thus, T S is compact. 7
8 Theorem 3.8. If T B(X, Y ) such that T (X) is finite dimensional then T is compact. Proof. Let X be bounded. Since T is bounded, T () is bounded in Y. But T () is a bounded subset of the finite dimensional space T (X) and, hence, T () is relatively compact in Y. Theorem 3.9. Let R n be a compact subset with nonempty interior and K : C is a continuous kernel. Then the bounded linear operator T : C() C(), defined as, (T f)(x) := K(x, y)f(y) dy, for each x is compact. Proof. Let B C() be a bounded subset, i.e., there is a M > 0 such that f M for all f B. Thus, for all f B, we have T f(x) κm for all x and f B. Thus, T (B) is bounded in C(). By the uniform continuity of K, for each ε > 0, T f(x 1 ) T f(x 2 ) K(x 1, y) K(x 2, y) f(y) dy ε f εm for all x 1, x 2 such that x 1 x 2 < δ and for all f B. Therefore, T (B) is equicontinuous. By ArzeláAscoli theorem (cf. Appendix A) T (B) closure is compact and, hence, T is compact. Theorem Let R n be a compact subset with nonempty interior and K : C be a weakly singular kernel. Then the bounded linear operator T : C() C(), defined as, (T f)(x) := K(x, y)f(y) dy, for each x is compact. 8
9 Definition Let V and W be vector spaces over the same field. For any T L(V, W ), the kernel of T, denoted as N(T ), is defined as N(T ) = {x V T x = 0}. The kernel of T is also referred to as null space of T and the dimension of kernel of T is called nullity of T. Also, the range of T, denoted as R(T ), is defined as R(T ) = {T x x V }. The dimension of range of T is called the rank of T. xercise 3.1. If T L(V, W ), then N(T ) is a subspace of V and R(T ) is a subspace of W. Theorem Let X be a normed space and T : X X be a compact linear operator. Then N(I T ) is finite dimensional and R(I T ) is closed in X. Proof. Let B 1 be the closed unit ball in N(I T ), i.e., B 1 := {x N(I T ) x 1}. If x B 1, then x = T x and T x 1. Thus, B 1 T (B), where B is the closed unit ball of X. Since T is a compact operator, B 1 is compact in X and, hence, N(I T ) is finite dimensional (cf. Theorem B.8). Let {y n } R(I T ) and suppose that y n y in X. Thus, there is a sequence {x n } X such that y n = x n T x n. Since N(I T ) is finite dimensional, by Theorem B.6, for each n, there is a z n N(I T ) such that x n z n = inf z N(I T ) x n z. Thus, y n = (x n z n ) T (x n z n ). If the sequence {x n z n } is bounded in X, and since T is compact, there is a subsequence T (x nk z nk ) u in X. Then, x nk z nk y + u. Hence, T (y+u) = u. Set s := y+u. Now consider, (I T )s = s T s = y+u u = y. Therefore, y R(I T ) and, hence, R(I T ) is closed. It only remains to prove that {x n z n } is a bounded sequence in X. Suppose not, then, for a subsequence, x nk z nk as k. Set w nk := 1 x nk z nk (x n k z nk ) 9
10 so that w nk = 1. Further (I T )w nk = 1 x nk z nk y n k and hence the LHS converges to the zero vector, since the denominator in RHS blows up. Since T is compact, there is a subsequence {w nkl } of {w nk } such that T w nkl v in X. But, since (I T )w nkl 0 in X, we should have w nkl v in X. Thus, T w nkl T v and, hence, (I T )v = 0, i.e., v N(I T ). On the other hand, d(w nkl, N(I T )) = d(x n kl, N(I T ) x nkl z nkl = 1. Thus, d(v, N(I T )) = 1 and v N(I T ), which is impossible. Hence, the sequence {x n z n } is bounded in X. For any normed space X and compact operator T : X X, one can define the operators (I T ) n : X X, for n 1. Let us denote L := I T, then L n = (I T ) n = I T n where T n := n k=1 ( 1) k 1 n! k!(n k)! T k. Note that T n is compact and, hence N(L n ) is finite dimensional and R(L n ) is closed in X, for all n 1. Theorem Let X be a normed space, T K(X) and L := I T. Then there is a unique nonnegative integer r 0, called the Riesz number of T such that {0} N(L) N(L 2 )... N(L r ) = N(L r+1 ) =... and X R(L) R(L 2 )... R(L r ) = R(L r+1 ) =.... Proof. If x N(L n ), i.e., L n x = 0, then L n+1 x = L(L n x) = L0 = 0. Therefore, {0} N(L) N(L 2 ).... Suppose that all the inclusions are proper. Since N(L n ) is finite dimensional, it is closed proper subspace of N(L n+1 ). Therefore, by Riesz lemma, there is a x n N(L n+1 ) such that 10
11 x n = 1 and d(x n, N(L n )) 1/2. Thus, we have a bounded sequence {x n } X with d(x n, N(L n )) 1/2. Consider T (x n x m ) = (I L)(x n x m ) = x n (x m + Lx n Lx m ). Note that, for n > m, we have L n (x m + Lx n Lx m ) = L n m 1 L m+1 x m + L n+1 x n L n m L m+1 x m = 0. Therefore, for n > m, T x n T x m 1/2, which contradicts that T, since there can be no convergent subsequence of {T x n }. Thus, the sequence of inclusions cannot be proper for all. There exists two consecutive null spaces that are equal. Set r := min{k : N(L k ) = N(L k+1 )}. We claim that N(L r ) = N(L r+1 ) =.... Note that, for some k r, we have shown that N(L k ) = N(L k+1 ). Now, consider x N(L k+2 ), then 0 = L k+2 x = L k+1 Lx. Thus, Lx N(L k+1 ) = N(L k ), 0 = L k Lx = L k+1 x and x N(L k+1 ). Hence, N(L k+1 ) = N(L k+2 ) and {0} N(L) N(L 2 )... N(L r ) = N(L r+1 ) =.... Let y R(L n+1 ), then there is a x X such that L n+1 x = y. Thus, L n (Lx) = y and y R(L n ). We assume the inclusions are all proper. Since R(L k ) is closed subspace, by Riesz lemma, there is a y n R(L n ) such that y n = 1 and d(y n, R(L n+1 )) 1/2. Thus, we have a bounded sequence {y n } X with d(y n, R(L n+1 )) 1/2. Consider T (y n y m ) = (I L)(y n y m ) = y n (y m + Ly n Ly m ). Note that, for m > n, we have y m + Ly n Ly m = L n+1 (L m n 1 x m + x n L m n x m ) and y m +Ly n Ly m R(L n+1 ). Thus, for n > m, T y n T y m 1/2, which contradicts that T, since there can be no convergent subsequence of {T y n }. Thus, the sequence of inclusions cannot be proper for all. There exists two consecutive range spaces that are equal. Set s := min{k : R(L k ) = R(L k+1 )}. 11
12 We claim that R(L s ) = R(L s+1 ) =.... Note that, for some k s, we have shown that R(L k ) = R(L k+1 ). Now, consider y R(L k+1 ), then y = L k+1 x = L(L k x). Thus, for some x 0 X, L k x = L k+1 x 0 and y = L(L k x) = L(L k+1 x 0 ) = L k+2 x 0. Hence R(L k+1 ) = R(L k+2 ) and X R(L) R(L 2 )... R(L r ) = R(L r+1 ) =.... It only remains to prove that r = s. Suppose r > s and let x N(L r ). Then L r 1 x R(L r 1 ) = R(L r ) and, hence, there is a y X such that L r y = L r 1 x. Therefore, L r+1 y = L r x = 0 and y N(L r+1 ) = N(L r ). This means that L r 1 x = 0 and x N(L r 1 ) which contradicts the minimality of r. On the other hand, if r < s. Let y R(L s 1 ). Then, for some x X, L s 1 x = y and Ly = L s x. Consequently, Ly R(L s ) = R(L s+1 ). Hence, there is a x 0 X such that L s+1 x 0 = Ly. Therefore, 0 = L s+1 x 0 Ly = L s (Lx 0 x), i.e., Lx 0 x N(L s ) = N(L s 1 ) and L s x 0 = L s 1 x = y. Thus, y R(L s ) which contradicts the minimality of s. Theorem Let X be a normed space, T K(X) and L := I T. Then, for each x X, there exists unique y N(L r ) and z R(L r ) such that x = y + z, i.e., X = N(L r ) R(L r ). Proof. Let x N(L r ) R(L r ). Then x = L r y for some y X and L r x = 0. Thus, L 2r y = 0 and y N(L 2r ) = N(L r ). Therefore, 0 = L r y = x. Let x X be an arbitrary element. Then L r x R(L r ) = R(L 2r ). Thus, there is a x 0 X such that L r x = L 2r x 0 and L r (x L r x 0 ). Define z := L r x 0 R(L r ) and y := x z. Since L r y = L r x L r z = L r x L 2r x 0 = 0, y N(L r ). Theorem Let X be a normed space, T K(X) and L := I T. Then L is injective iff L is surjective. If L is injective (and hence bijective), then its inverse L 1 B(X). Proof. The injectivity of L is equivalent to saying that the Riesz number is r = 0, which means L is surjective. The argument is also true viceversa. If L is injective and suppose L 1 is not bounded. Then there is a sequence {x n } X with x n = 1 such that L 1 x n n, for all n N. Define, for each n N, y n := 1 L 1 x n x n; z n := 12 1 L 1 x n L 1 x n.
13 Then Lz n = y n 0 in X, as n, and z n = 1 for all n. By the compactness of T, there is a subsequence {z nk } of z n such that, for some z X, T z nk z as k. Since Lz n = y n 0, we have z nk z, as k. Also, Lz = 0 and z N(L). By the injectivity of L, z = 0 which contradicts the fact that z n = 1. Thus, L 1 must be bounded. Corollary Let T : X X be a compact linear operator on a normed space X. If the homogeneous equation x T x = 0 has only the trivial solution x = 0, then for each f X the inhomogeneous equation x T x = f has a unique solution x X which depends continuously on f. If the homogeneous equation x T x = 0 has nontrivial solution, then it has m N linearly independent solutions x 1, x 2,..., x m and the inhomogeneous equation x T x = f is either unsolvable or its general solution is of the form m x = x 0 + α i x i where α i C for each i and x 0 is a particular solution of the inhomogeneous equation. The decomposition X = N(L r ) R(L r ) induces a projection operator P : X N(L r ) that maps P x := y, where x = y + z. Proposition The projection operator P : X N(L r ) is compact. Proof. We first show that P is a bounded linear operator. Suppose not, then there is a sequence {x n } X with x n = 1 such that P x n n for all n N. Define, for each n N, y n := 1 x P x n n. Then y n 0, as n, and P y n = 1 for all n N. Since N(L r ) is finitedimensional and {P y n } is bounded, by Theorem 3.3, there is a subsequence {y nk } such that P y nk z in N(L r ), as k. Also, since y nk 0, we have P y nk y nk z, as k. Note that P y nk y nk R(L r ), by direct decomposition, thus z R(L r ) because R(L r ) is closed. Since z N(L r ) R(L r ), z = 0 and P y nk 0 which contradicts P y nk = 1. Thus, P must be bounded. Moreover, since P (X) = N(L r ) is finite dimensional, by Theorem 3.8, P is compact. i=1 13
14 4 Fredhölm Alternative Definition 4.1. Let V and W be complex vector spaces. A mapping, : V W C is called a bilinear form if α 1 x 1 +α 2 x 2, y = α 1 x 1, y +α 2 x 2, y, x, β 1 y 1 +β 2 y 2 = β 1 y, x 1 +β 2 x, y 2 for all x 1, x 2, x X and y 1, y 2, y Y and α 1, α 2, β 1, β 2 C. Further, a bilinear form is called nondegenerate if for every nonzero x X there exists a y Y such that x, y 0 and, for every nonzero y Y there is a x X such that x, y 0. Definition 4.2. If two normed spaces X and Y are equipped with a nondegenerate bilinear form, then we call it a dual system denoted by X, Y. xample 4.1. Let R n be a nonempty compact subset. We define the bilinear form in C(), C() as f, g := f(x)g(x) dx. which makes the pair a dual system. Definition 4.3. Let X 1, Y 1 and X 2, Y 2 be two dual systems. The operators S : X 1 X 2 and T : Y 2 Y 1 are called adjoint if Sx, y = x, T y for all x X 1 and y Y 2. Theorem 4.4. Let R n be a nonempty compact subset and K be a continuous kernel on. Then the compact integral operators Sf(x) := K(x, y)f(y) dy x and T g(x) := K(y, x)g(y) dy are adjoint in the dual system C(), C(). Proof. Note that Sf, g = = x Sf(x)g(x) dx = K(x, y)f(y) dyg(x) dx f(y) K(x, y)g(x) dx dy = f(y)t g(y) dy = f, T g. 14
15 The above result is also true for a weakly singular kernel K whose proof involves approximating K by continuous kernels. Lemma 4.5. Let X, Y be a dual system. Then to every set of linearly independent elements {x 1,..., x n } X, then there exists a set {y 1,..., y n } Y such that x i, y j = δ ij for all i, j. The result also holds true with the roles of X and Y interchanged. Proof. The result is true for n = 1, by the nondegeneracy of the bilinear form. We shall prove the result by induction. Let us assume the result for n 1 and consider the n + 1 linearly independent {x 1,..., x n+1 }. By induction hypothesis, for each m = 1, 2,..., n + 1, the linearly independent set {x 1,..., x m 1, x m+1,..., x n+1 } of n elements in X has a set of n elements {y m 1,..., y m m 1, y m m+1,..., y m n+1} in Y such that x i, y m j = δ ij for all i, j except i, j m. Since {x 1,..., x n+1 } is linear independent, we have n+1 x m x m, yj m x j 0. j=1 j m Thus, by nondegeneracy of bilinear form there is a z m Y such that n+1 x m x m, yj m x j, z m 0. j=1 j m The LHS is same as n+1 α m := x m, z m yj m x j, z m. Define y m := 1 α m x i, y m = 1 α m z m j=1 j m n+1 yj m x j, z m. j=1 j m Then x m, y m = 1, and for i m, we have n+1 x i, z m 15 j=1 j m x i, yj m x j, z m = 0
16 because x i, y m j = δ ij. Thus, we obtained {y 1,..., y n+1 } such that x i, y j = δ ij for all i, j. Theorem 4.6. Let X, Y be a dual system and S : X X, T : Y Y be compact adjoint operators. Then Proof. By Theorem 3.12, dim(n(i S)) = dim(n(i T )) <. dim(n(i S)) = m; dim(n(i T )) = n. We need to show that m = n. Suppose that m < n. If m > 0, we choose a basis {x 1,..., x m } N(I S) and a basis {y 1,..., y n } N(I T ). By Lemma 4.5, there exists elements {a 1, a 2,... a m } Y and {b 1, b 2,..., b n } X such that x i, a j = δ ij, for i, j = 1, 2,..., m, and b i, y j = δ ij for i, j = 1, 2,..., n. Define a linear operator F : X X by F x := m x, a i b i i=1 for m > 0. If m = 0 then F 0 is the zero operator. Note that F : N[(I S) r ] X is bounded by Theorem B.9 and P : X N[(I S) r ] is a compact projection operator by Proposition Then, by Theorem 3.7, F P : X X is compact. Since linear combination of compact operators are compact, S F P is compact. Consider Then x Sx + F P x, y j = x, (I T )y j + F P x, y j = F P x, y j. x Sx + F P x, y j = { P x, a j j = 1, 2,..., m 0 j = m + 1,..., n. If x N(I S + F P ), then by above equation P x, a j = 0 for all j = 1,..., m. Therefore, F P x = 0 and, hence, x N(I S). Consequently, x = m i=1 α ix i, i.e., α i = x, a i. But P x = x for x N(I S), therefore α i = P x, a i = 0 for all i = 1,..., m which implies that x = 0. Thus, I S + F P is injective. Hence the inhomogeneous equation x Sx + F P x = b n 16
17 has a unique solution x. Note that 0 = x Sx + F P x, y n = b n, y n = 1 is a contradiction. Therefore, m n. Arguing similarly by interchanging the roles of S and T, we get n m implying that m = n. Theorem 4.7. Let X, Y be a dual system and S : X X, T : Y Y be compact adjoint operators. Then R(I S) = {x X x, y = 0, y N(I T )} and R(I T ) = {y Y x, y = 0, x N(I S)}. Proof. The case of dim(n(i T )) = 0 is trivial because, in that case, dim(n(i S)) = 0 and R(I S) = X (by Theorem 3.15 and Theorem 4.6). Hence, the result is trivially true. Suppose that the dim(n(i T )) = m > 0. Let x R(I S), i.e., x = (I S)x 0 for some x 0 X. Then, for all y N(I T ), x, y = x 0 Sx 0, y = x 0, y T y = 0. Conversely, assume that x X satisfies x, y = 0 for all y N(I T ). From the proof of previous theorem, there is a unique solution x 0 X of (I S + F P )x 0 = x. Then P x 0, a j = (I S + F P )x 0, y j = x, y j = 0 j = 1, 2,..., m. Then F P x 0 = 0 and thus (I S)x 0 = x and x R(I S). The argument for R(I T ) is similar. The above two theorems together is called the Fredhölm alternative. Corollary 4.8. Let R n be a nonempty compact subset with nonempty interior and K be a continuous or weakly singular kernel on. Then either the homogeneous integral equations u(x) K(x, y)u(y) dy = 0 x (4.1) 17
18 and v(x) K(y, x)v(y) dy = 0 x (4.2) only have the trivial solutions u = 0 and v = 0, and the inhomogeneous integral equations u(x) K(x, y)u(y) dy = f(x) x and v(x) K(y, x)v(y) dy = g(x) x have unique solution u, v C() for given f, g, C(), respectively, or both (4.1) and (4.2) have the same finite number m N of linearly independent solutions and the inhomogeneous integral equations are solvable iff f(x)v(x) dx = 0 for all v solving (4.2) and for all u solving (4.1), respectively. Appendices u(x)g(x) dx = 0 A AscoliArzelá Result Definition A.1. Let X be a topological space. A set X is said to be totally bounded if, for every given ε > 0, there exists a finite collection of points {x 1, x 2,, x n } X such that n i=1b ε (x i ). xercise A.1. If X is totally bounded then n X n is also totally bounded. Definition A.2. A subset A C(X) is said to be bounded if there exists a M N such that f M for all f A. 18
19 Definition A.3. A subset A C(X) is said to be equicontinuous at x 0 X if, for every given ε > 0, there is an open set U of x 0 such that f(x) f(x 0 ) < ε x U; f A. A is said to be equicontinuous if it is equicontinuous at every point of X. Theorem A.4. Let X be a compact topological space and Y be a totally bounded metric space. If a subset A C(X, Y ) is equicontinuous then A is totally bounded. Proof. Let A be equicontinuous and ε > 0. Then, for each x X, there is a open set U x containing x such that f(y) f(x) < ε 3 y U x ; f A. Since X is compact, there is a finite set of points {x i } n 1 X = n i=1u xi. Define the subset A of Y n as, X such that A := {(f(x 1 ), f(x 2 ),, f(x n )) f A} which is endowed with the product metric, i.e., d(y, z) = max 1 i n { y i z i } where y, z Y n are ntuples. Since Y is totally bounded, Y n is also totally bounded (cf. xercise A.1). Thus, A is totally bounded and there are m number of ntuples, y j := (f j (x 1 ), f j (x 2 ),, f j (x n )) Y n, for each 1 j m, such that A m j=1b ε/3 (k j ). For any f A, there is a j such that d(y j, z f ) < ε 3 where z f = (f(x 1 ), f(x 2 ),, f(x n )). In particular, given any f A, there is a j such that, for all 1 i n, f j (x i ) f(x i ) < ε 3. Given f A, fix the j as chosen above.now, for any given x X, there is a i such that x U xi. For this choice of i, j, we have f(x) f j (x) f(x) f(x i ) + f(x i ) f j (x i ) + f j (x i ) f j (x) The first and third term is smaller that ε/3 by the continuity of f and f j, respectively, and the second term is smaller than ε/3 by choice of f j. Hence A is totally bounded, i.e., A m j=1b ε (f j ), equivalently, for any f A there is a j such that f f j < ε. 19
20 Lemma A.5. Let X be compact topological space. If A C(X) is bounded then there is a compact subset K R such that f(x) K for all f A and x X. Proof. Choose an element g A. Since A is bounded in the uniform topology, there is a M such that f g < M for all f A. Since X is compact, g(x) is compact. Hence there is a N > 0 such that g(x) [ N, N]. Then f(x) [ M N, M + N] for all f A. Set K := [ M N, M + N] and we are done. Corollary A.6 (other part of AscoliArzela Theorem). Let X be a compact topological space. If a subset A C(X) is closed, bounded and equicontinuous then A is compact. Proof. Since A is bounded, by Lemma above, we have A C(X, K) C(X) for some compact subset K R. Then, by the Theorem above, A is totally bounded. Since A is a closed and totally bounded subset of the metric space C(X), A is compact. B Normed Spaces and Bounded Operators Definition B.1. Let V and W be real or complex vector spaces. A linear map from V to W is a function T : V W such that T (αx + βy) = αt (x) + βt (y) x, y V and α, β R or C. Observe that a linear map is defined between vector spaces over the same field of scalars. xercise B.1. Show that a linear map T satisfies T (0) = 0. Let L(V, W ) be the space of linear maps from V to W. Definition B.2. A normed space is a pair (X, ), where X is a vector space over C or R and : X [0, ) is a function such that (i) x = 0 iff x = 0, (ii) λx = λ x for all x X and λ F (absolute homogeneity), (iii) x + y x + y for all x, y X. inequality) (subadditivity or triangle 20
21 The function is called the norm of a vector from X. Norm is a generalisation of the notion length of a vector in a uclidean space. xercise B.2. Show that every normed space is a metric space with the metric d(x, y) = x y. xercise B.3. Show that the map : X [0, ) is uniformly continuous on X. Proof. Observe that x = x y + y x y + y. Thus, x y x y. Similarly, y y x + x. Thus, x y x y. xercise B.4. The operations addition (+) and scalar multiplication are continuous from X X and X C to X, respectively. xercise B.5. Show that for a Cauchy sequence {x n } in X, we have x m x n < 1 2 n m n. Definition B.3 (Infinite Series). An infinite series in a normed space X, say i=1 x i = x 1 + x , is said to be convergent if the sequence s n is convergent, where s n = n i=1 x i is the sequence of partial sums. An infinite series is said to be absolutely convergent if the series i=1 x i is convergent. Theorem B.4. A normed space X is a Banach space iff every absolutely convergent series in X is convergent. Proof. Let X be Banach space and let x = i=1 x i be an absolutely convergence series. Let y n = n i=1 x i be the partial sum. It is enough to show that {y n } is Cauchy in X. Given ε > 0, there exists a N 0 such that i=n 0 x i < ε. We choose m, n N 0 and, without loss of generality, fix N 0 m < n. Then y n y m = n i=m+1 x i n i=m+1 x i i=n 0 x i < ε. Thus, {y n } is a Cauchy sequence in X and hence converges. Hence, the given absolutely convergent series converges. Conversely, let every absolutely convergent series in X converge. We need to show that every Cauchy sequence in X converges. Let {x n } be a Cauchy sequence in X. Therefore, by xercise B.5, x m x n < 1 2 n m n. 21
22 Now, let us construct a series in X using the given Cauchy sequence. Set x 0 = 0 and define y k = x k x k 1 for all i 1. Then, observe that n k=1 y k = x n. Therefore, the n th partial sum of the series k=1 y k is x n. Observe that y k < 1/2 k 1. Thus, by comparison test, the series absolutely convergent and hence, by hypothesis, converges. Therefore its sequence of partial sums {x n } converges. Therefore X is Banach since {x n } was a arbitrary sequence in X. Theorem B.5. Let {x 1, x 2,..., x n } be a linearly independent set of vectors in a normed space X. Then there is a constant c > 0 such that for every choice of λ 1, λ 2,..., λ n we have ( n n ) λ i x i c λ i. (c independent of the scalars) (B.1) i=1 i=1 Proof. Set s = n i=1 λ i. If s = 0 then λ i = 0, for each i = 1, 2,..., n. Thus, (B.1) holds trivially, for any c > 0. Suppose that s > 0. Then, observe that proving (B.1) is equivalent to showing the existence of a constant c > 0 such that for all scalars α i of the satisfying n i=1 α i = 1, we have n α i x i c. i=1 The equivalence is obtained by dividing s on both sides of (B.1) and setting α i = λ i. Suppose our claim is false, then for every m N, there is a set of s scalars {αi m } n 1 such that n i=1 αm i = 1 and n y m = αi m x i < 1 m. i=1 Thus, y m 0 as m. Since n i=1 αm i = 1, for each i, αi m 1. Fixing i = 1, we observe that the sequence {αi m } m is bounded in R. By invoking BolzanoWeierstrass theorem, {α1 m } m has a convergent subsequence {γ1 m } that converges to α 1. Let ym 1 = γ1 m x 1 + n i=2 αm i x i which is a subsequence of y m. Repeating the argument for ym, 1 we get a subsequence ym 2 = 2 i=1 γm i x i + n i=3 αm i x i with α 2 being the limit of the subsequence of {α2 m }. Thus, repeating the procedure n times, we have a subsequence {ym} n m of y m which is given by n ym n = γi m x i i=1 22
23 where, for each i, γi m have α i and n i=1 γm i = 1. Thus, letting m, we y n m y = n α i x i and n i=1 α i = 1. Thus, α i 0 for some i. Since the set {x 1, x 2,..., x n } is linearly independent y 0. By xercise B.3, if ym n m y, then ym n m y. But y m 0, hence the subsequence ym n 0. Therefore y = 0 implies y = 0 which is a contradiction. Theorem B.6. Let Y be finite dimensional subspace of a normed space X. Then, for any x X, there is a y Y such that i=1 x y = inf x z. z Y Lemma B.7 (Riesz Lemma). Let Y be a proper closed subspace of a normed space X. Then, for every 0 < ε < 1, there is a point x ε X such that x ε = 1 and ε d(x ε, Y ) 1, where d(x, Y ) = inf y Y x y. Proof. Since Y X, choose x X such that x Y. Since Y is closed d(x, Y ) > 0. Now, for any 0 < ε < 1, there is a y 0 Y such that d(x, Y ) x y 0 d(x, Y ). ε The above inequality can be rewritten as ε d(x, Y ) x y 0 1. (B.2) Set x ε = x y 0 x y 0. Observe that x ε y = = 1 x y 0 (x y 0 x y 0 y) 1 x y 0 x y 1 ( where y 1 = y 0 x y 0 y Y ). Therefore, d(x ε, Y ) = 1 x y 0 d(x, Y ) and by (B.2), we have our claim. 23
24 Theorem B.8. If a normed space X is such that the unit ball B(X) is compact, then X is finite dimensional. Proof. Let us suppose that X is infinite dimensional. Let x 1 X such that x 1 = 1. The X 1 = [x 1 ] is a one dimensional subspace of X. Since X is infinite dimensional, [x 1 ] is a proper subspace of X. By Riesz lemma, there is a x 2 X with x 2 = 1 such that x 2 x 1 1/2. Now, X 2 = [x 1, x 2 ] is a twodimensional proper subspace of X. Therefore, again by Riesz lemma, there is a x 3 X with x 3 = 1 such that x 3 x 1/2 for all x X 2. In particular, x 3 x 1 1/2 and x 3 x 2 1/2. Arguing further in a similar way, we obtain a sequence {x n } in B(X) such that x m x n 1/2 for all m n. Thus, we have obtained a bounded sequence in B(X) which cannot converge for any subsequence, which contradicts the hypothesis that B(X) is compact. Therefore dim(x) =. Theorem B.9. Let X and Y be normed spaces. If X is finite dimensional, then every linear map from X to Y is continuous. Proof. Let X be finite dimensional and and let T L(X, Y ). If T = 0, then the result is trivial, since L(X, Y ) = {0}. Let X 0 and {e 1, e 2,..., e m } be a basis for X. Let x n x in X. For some scalars, since x n = m i=1 λn i e i and x = m i=1 λ ie i, we have λ n i λ i for all 1 i m (by Theorem B.5). Consider, T x n = m λ m i T e i (by linearity of T ) i=1 m λ i T e i = T x. i=1 The convergence is valid by the continuity of addition and scalar multiplication (cf. xercise B.4). Thus T is continuous. Definition B.10. For any given normed spaces X and Y, a linear map T L(X, Y ) is said to be bounded if there is a constant c > 0 such that T x c x, x X. Basically, bounded operators map bounded sets in X to bounded sets in Y. Let B(X, Y ) be the space of bounded linear maps from X to Y. Any map in L(X, Y ) \ B(X, Y ) is said to be unbounded linear map. We shall now prove an interesting result which says that B(X, Y ) = C(X, Y ), the 24
25 space of continuous linear maps from X to Y. In other words, by looking at bounded linear maps we are, in fact, looking at continuous linear maps. Such a result is possible only because of the underlying linear structure of the space. The following theorem proves these remarks rigorously. Theorem B.11. Let X and Y be normed spaces and let T L(X, Y ). Then the following are equivalent: (i) T C(X, Y ). (ii) T is continuous at some point x 0 X. (iii) T B(X, Y ). Proof. The above equivalence are true if T = 0. Hence, henceforth, we assume T 0. (i) implies (ii) is trivial from the definition of continuity. Let us now assume (ii). Then, for any ε > 0, there is a δ > 0 such that T x T x 0 ε whenever x x 0 δ. Set B δ (x 0 ) = {x X x x 0 δ. For any nonzero x X, x 0 + δ x B x δ(x 0 ). Therefore δ T x ε and x hence T x ε x. Thus, T is bounded. δ We shall now assume T is bounded and prove (i). There is a c > 0 such that T x c x. Therefore, T x T y c x y for all x, y X. Thus, for any given ε > 0, T x T y ε whenever x y ε/n. We shall now introduce a norm in B(X, Y ) to make it a normed space. Observe that, in the definition of bounded linear map, we seek the existence of a constant c > 0. The question is what is the smallest such constant c > 0? Note that T x c x X and x 0. x Thus, the smallest such constant would be sup x X x 0 xercise B.6. For any T B(X, Y ) T x sup x X x 0 x = sup x X x =1 T x x c. T x = sup T x. x X x 1 25
26 Proof. Observe that for any x X, the vector z = z = 1. Thus, by the linearity of T, ( T x sup x X x = sup x T x X x x 0 x 0 x x ) = sup T z. z X z =1 X such that xercise B.7. Show that the function : B(X, Y ) [0, ) defined as T = sup x X x 0 T x, T B(X, Y ) x is a norm on B(X, Y ). Thus, B(X, Y ) is a normed space. xercise B.8. Show that if T B(X, Y ) and S B(Y, Z), then the composition S T = ST is in B(X, Z) and ST S T. In particular, show that B(X) is an algebra under composition of operators. Theorem B.12. If Y is complete then B(X, Y ) is a complete normed space. In particular, X is a Banach space. Proof. Let {T n } be a Cauchy sequence in B(X, Y ). Then {T n x} is a Cauchy sequence in Y, for all x X. Since Y is complete, there is a y Y such that T n x y. Set T x = y. It now remains to show that T B(X, Y ) and T n T in B(X, Y ). For any given x 1, x 2 X and scalars λ 1, λ 2 R, we have T (λ 1 x 1 + λ 2 x 2 ) = lim T n (λ 1 x 1 + λ 2 x 2 ) n = lim λ 1 T n x 1 + lim λ 2 T n x 2 n n = λ 1 T x 1 + λ 2 T x 2. Thus, T L(X, Y ). Since T n is Cauchy, for every ε > 0, there is a N 0 N such that T m T n < ε for all m, n N 0. Therefore, for all x X, we have, T n x T m x T n T m x < ε x m, n N 0. Now, by letting m, we have T n x T x ε x for all x X and n N 0. Thus, T n T 0. Also, since we have T B(X, Y ). T x = T x T n x + T n x (ε + T n ) x, 26
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