MTL 411: Functional Analysis Lecture C: Open mapping theorem and Closed-graph theorem Let (X; d) and (Y; ρ) be metric spaces. Definition 0.1. (continuous) We say that a function f : X ! Y is continuous on X if for every open set U in Y , the inverse image f −1(U) is open in X, i.e., the inverse image of an open set is open. Equivalently, the inverse image of a closed is closed (why?). Definition 0.2. (Open map) We say that a function f : X ! Y is open if for every open set G in X, the image f(G) is open in Y , i.e., the image of an open set is open. Definition 0.3. (Closed map) We say that a function f : X ! Y is closed if for every closed set F in X, the image f(F ) is closed in Y , i.e., the image of an closed set is closed. Remarks. 1. A continuous map need not be open/closed. For example, consider the map x 7! sin x from (0; π) to R: [Hint. See the image of the interval (0; π)] 2. An open, continuous map need not be closed, even if it is onto. For example, consider 2 2 the projection map P (x; y) = x from R to R: [Hint. Use f(x; y) 2 R : xy = 1g] 3. An closed, continuous map need not be open, even if it is onto. For example, consider the map x 7! cos x from [0; 2π] to [−1; 1]: 4. Similarly, we can construct functions which are open/closed need not be continuous. We will see that if a linear operator from a Banach space to a Banach space is continuous and surjective, then it is open map. This result play a important role to explain the sufficient condition to get a bounded inverse of an operator. Notations: 1. BX (x0; r) = fx 2 X : jjx − x0jj < rg is an open ball centered at x0 with radius r in X. 2. BX (x0; r) + z := fx + z : x 2 BX (x0; r)g where z 2 X. It is easy to verify that BX (0; r) + x0 = BX (x0; r): 3. c BX (x0; r) := fcx : x 2 BX (x0; r)g where c is scalar. It is easy to verify that BX (0; r) = rBX (0; 1): 1 1 Open Mapping Theorem Lemma 1.1 (Open unit ball). Suppose T is a bounded linear operator from a Banach space X onto a Banach space Y . Then BY (0; r) ⊂ T (BX (0; 1)) for some r > 0. 1 Proof. Claim 1. BY (y0; δ) ⊂ T (BX (0; 2 )) for some δ > 0: We can write 1 1 [ k [ 1 k X = BX (0; 2 ) = k BX (0; 2 ): (* x 2 X; jjxjj ≤ 2 ; for some k:) k=1 k=1 Thus, 1 1 [ 1 [ 1 T (X) = T (k BX (0; 2 )) = k T (BX (0; 2 )) (* T is linear) k=1 k=1 1 1 [ 1 [ 1 =) Y = k T (BX (0; 2 )) = k T (BX (0; 2 )): (* T is onto) k=1 k=1 Since Y is a Banach space and using Baire's category theorem, we get the interior of 1 1 k T (BX (0; 2 )) is non-empty for some k. Therefore, the interior of T (BX (0; 2 )) is non-empty, that is, there exists a y0 2 Y and δ > 0 such that 1 BY (y0; δ) ⊂ T (BX (0; 2 )): δ 1 Claim 2. BY (0; 2n ) ⊂ T (BX (0; 2n )) for all n ≥ 0: It is enough to show that for n = 0, BY (0; δ) ⊂ T (BX (0; 1) (why ?). 1 Let y 2 BY (0; δ): Then y + y0 2 BY (y0; δ) ⊂ T (BX (0; 2 )): By definition of closure of a set, 1 1 9 un 2 T (BX (0; 2 )); wn 2 BX (0; 2 ) such that T (wn) = un ! y + y0; 1 1 and 9 vn 2 T (BX (0; 2 )); zn 2 BX (0; 2 ) such that T (zn) = vn ! y0: From this we get un − vn = T (wn − zn) ! y. Notice that jjwn − znjj < 1, so we get y 2 T (BX (0; 1). Therefore, we get BY (0; δ) ⊂ T (BX (0; 1): δ Claim 3. BY (0; 2 ) ⊂ T (BX (0; 1): δ 1 Let y 2 BY (0; 2 ): Then by the above claim 2, y 2 T (BX (0; 2 )): So there exists x1 2 1 BX (0; 2 ) such that δ jjy − T x1jj < 4 : δ 1 Now y − T x1 2 BY (0; 4 ): Again by the above claim 2, y − T x1 2 T (BX (0; 4 )): So there exists 1 x2 2 BX (0; 4 ) such that δ jjy − T x1 − T x2jj < 23 : 1 By repeating this procedure and using induction, we get a sequence xn 2 BX (0; 2n ) such that n X δ jjy − T xkjj < 2n+1 : (1.1) k=1 Pn Pn 1 Define zn = k=1 xk: Then jjzn − zmjj ≤ k=m+1 2k is Cauchy sequence in X. Since X is a Banach space, fzng converges to a element x 2 X and jjxjj < 1 (why ?). From the equation (1.1), we have T zn ! y. Since T is continuous, we get T x = y. Therefore, y 2 T (BX (0; 1): 2 Theorem 1.2 (Open mapping theorem). Suppose T is a bounded linear operator from a Banach space X onto a Banach space Y . Then T is an open map. Proof. Let G be a open subset of X. We have to show that T (G) is open in Y . Let y 2 T (G). Then we have a x 2 G such that T x = y: Since G is open, there exists a > 0 such that BX (x; ) ⊂ G: Thus BX (0; ) ⊂ G − x: By the above open map lemma, there exists a δ > 0 such that BY (0; δ) ⊂ T (BX (0; 1)): =) BY (0; δ) ⊂ T (BX (0; 1)) = T (BX (0; )) ⊂ T (G − x) = T (G) − T x = T (G) − y: [* T is linear] So we get BY (0; δ) + y ⊂ T (G): Therefore, y is an interior point of T (G). Hence T (G) is open in Y . 2 Bounded Inverse theorem Let X and Y be normed linear spaces, and T : X ! Y a linear operator. Then we have N (T ) = f0g () T is injective () T −1 : R(T ) ! X exists: Definition 2.1. We say that a linear operator T : X ! Y is bounded below if there exists a constant α > 0 such that jjT xjj ≥ αjjxjj; 8x 2 X: Proposition 2.2. Let T be a linear operator from a normed linear space X to a normed linear space Y . Then the following are equivalent: (a) T is bounded below. (b) T −1 : R(T ) ! X exists and bounded. Further, if T 2 B[X; Y ] and if X is a Banach space, then each of the above equivalent assertions implies that R(T ) = R(T ). Proof. (a) =) (b) Since T is bounded below, we have N (T ) = f0g (why?). So T −1 : R(T ) ! X exists. To show that T −1 is bounded, for y 2 R(T ), there is a vector x 2 X such that T x = y. Thus jjT −1yjj = jjT −1T xjj = jjxjj 1 1 ≤ α jjT xjj = α jjyjj; 8y 2 Y: (b) =) (a) Exercise. In addition to equivalent assertions, if T 2 B[X; Y ] and X is a Banach space, then we show that the range of T is closed. Let z 2 R(T ): Then there exists a sequence yn 2 R(T ) such that yn ! z in Y . We have to show that z 2 R(T ): For each n 2 N, there exists xn 2 X such that T xn = yn: Consider 1 jjxn − xmjj ≤ α jjT xn − T xmjj = jjyn − ymjj (n; m 2 N): Since fyng is a Cauchy, we get fxng is Cauchy in X. As X is a Banach space, fxng converges to some vector x 2 X. This implies that T xn = yn converges to T x. Hence T x = z: 3 Remark. If X is not complete then the above conclusion may not be true. For example, 1 consider the identity operator I(x) = x from (c00; jj · jj1) to ` space. Then I is isometry (in particular, I is bounded below) but the range of I is not closed in `1: Theorem 2.3 (Bounded inverse theorem). If X and Y are Banach spaces and T 2 B[X; Y ] is injective and surjective, then T −1 2 B[Y; X]: Proof. We know that the linear operator T −1 : Y ! X exists since that T is bijective and linear. Now we have to show that T −1 is continuous. Equivalently, the inverse image of an open set is open, i.e., for each open set G in X, the inverse image (T −1)−1(G) = T (G) is open in Y which is same as proving T is open map. Thus the result follows from the open mapping theorem. Corollary 2.4. If X and Y are Banach spaces and T 2 B[X; Y ] is bounded below, then T −1 2 B[R(T );X]: 3 Closed graph theorem In this section, we introduce closed linear operators which appears more frequently in the ap- plication. In particular, most of the practical applications we encounter unbounded operators which are closed linear operators.