Open Mapping Theorem and Closed-Graph Theorem

Home , Bounded inverse theorem, Closed graph, Closed graph theorem

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 {(x, y) ∈ R : xy = 1}] 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) = {x ∈ X : ||x − x0|| < r} is an open ball centered at x0 with radius r in X.

2. BX (x0, r) + z := {x + z : x ∈ BX (x0, r)} where z ∈ X. It is easy to verify that BX (0, r) + x0 = BX (x0, r).

3. c BX (x0, r) := {cx : x ∈ BX (x0, r)} 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 ∞ ∞ [ k [ 1 k X = BX (0, 2 ) = k BX (0, 2 ). (∵ x ∈ X, ||x|| ≤ 2 , for some k.) k=1 k=1 Thus, ∞ ∞ [ 1 [ 1 T (X) = T (k BX (0, 2 )) = k T (BX (0, 2 )) (∵ T is linear) k=1 k=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 ∈ 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 ∈ BY (0, δ). Then y + y0 ∈ BY (y0, δ) ⊂ T (BX (0, 2 )). By deﬁnition of closure of a set,

1 1 ∃ un ∈ T (BX (0, 2 )), wn ∈ BX (0, 2 ) such that T (wn) = un → y + y0, 1 1 and ∃ vn ∈ T (BX (0, 2 )), zn ∈ BX (0, 2 ) such that T (zn) = vn → y0.

From this we get un − vn = T (wn − zn) → y. Notice that ||wn − zn|| < 1, so we get y ∈ 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 ∈ BY (0, 2 ). Then by the above claim 2, y ∈ T (BX (0, 2 )). So there exists x1 ∈ 1 BX (0, 2 ) such that δ ||y − T x1|| < 4 . δ 1 Now y − T x1 ∈ BY (0, 4 ). Again by the above claim 2, y − T x1 ∈ T (BX (0, 4 )). So there exists 1 x2 ∈ BX (0, 4 ) such that δ ||y − T x1 − T x2|| < 23 . 1 By repeating this procedure and using induction, we get a sequence xn ∈ BX (0, 2n ) such that n X δ ||y − T xk|| < 2n+1 . (1.1) k=1 Pn Pn 1 Deﬁne zn = k=1 xk. Then ||zn − zm|| ≤ k=m+1 2k is Cauchy sequence in X. Since X is a Banach space, {zn} converges to a element x ∈ X and ||x|| < 1 (why ?). From the equation (1.1), we have T zn → y. Since T is continuous, we get T x = y. Therefore, y ∈ 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 ∈ T (G). Then we have a x ∈ 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 ) = {0} ⇐⇒ 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 ||T x|| ≥ α||x||, ∀x ∈ 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 ∈ 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 ) = {0} (why?). So T −1 : R(T ) → X exists. To show that T −1 is bounded, for y ∈ R(T ), there is a vector x ∈ X such that T x = y. Thus ||T −1y|| = ||T −1T x|| = ||x|| 1 1 ≤ α ||T x|| = α ||y||, ∀y ∈ Y. (b) =⇒ (a) Exercise. In addition to equivalent assertions, if T ∈ B[X,Y ] and X is a Banach space, then we show that the range of T is closed.

Let z ∈ R(T ). Then there exists a sequence yn ∈ R(T ) such that yn → z in Y . We have to show that z ∈ R(T ).

For each n ∈ N, there exists xn ∈ X such that T xn = yn. Consider 1 ||xn − xm|| ≤ α ||T xn − T xm|| = ||yn − ym|| (n, m ∈ N).

Since {yn} is a Cauchy, we get {xn} is Cauchy in X. As X is a Banach space, {xn} converges to some vector x ∈ 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, ∞ consider the identity operator I(x) = x from (c00, || · ||∞) to ` space. Then I is isometry (in particular, I is bounded below) but the range of I is not closed in `∞.

Theorem 2.3 (Bounded inverse theorem). If X and Y are Banach spaces and T ∈ B[X,Y ] is injective and surjective, then T −1 ∈ 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 ∈ B[X,Y ] is bounded below, then T −1 ∈ 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.

Deﬁnition 3.1. Let X and Y be normed spaces. Then a linear operator T : X → Y is said to be closed operator if for every sequence {xn} in X such that

xn → x and T xn → y =⇒ T x = y.

Equivalent definition: Define a normed space X × Y , where the two algebraic operations are defined as,

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) α(x, y) = (αx, αy), and the norm on X × Y is deﬁned by

||(x, y)|| = ||x|| + ||y||.

Then a linear operator T : X → Y is closed operator if the graph of T , G(T ) = {(x, T x): x ∈ X} is closed in X × Y. Exercises.

1. With above deﬁned norm, show that X × Y is a Banach space whenever X and Y are Banach spaces.

2. Show that the above two deﬁnitions are equivalent.

3. Show that the graph G(T ) is a subspace of X × Y.

0 1 Example. Consider the diﬀerential operator T : f 7→ f from (C [a, b], || · ||∞) to (C[a, b], || · ||∞). We know that the operator is not continuous (why?). Now we show that the operator 0 is closed using uniform convergence property. Let {(fn, fn)} be a sequence in G(T ) such that

4 0 0 (fn) converges to f and fn converges to g in sup-norm. We have to show that g = f . Using fundamental theorem integral calculus, we write

x Z 0 fn(x) = fn(a) + fn(t) dt a x Z f(x) = f(a) + g(t) dt (as n → ∞; limit and integral can be interchanged ?) a The result follows by fundamental theorem of integral calculus. Remark. Continuous linear operator =⇒ Closed linear operator. The converse is not true (see the above example). Under certain conditions, the converse is true which is stated as

Theorem 3.2 (Closed graph theorem). If X and Y are Banach spaces and T : X → Y is linear operator, then T is continuous ⇐⇒ T is closed.

Proof. If T is continuous, then T is closed (verify!). Conversely, suppose T is a closed operator. Then the graph of T , G(T ) is closed in X ×Y . Moreover, it is a subspace and so it is a complete space. Deﬁne P : G(T ) → X by P (x, T x) = x. It is easy to verify that P is continuous, injective and surjective. By bounded inverse theorem, P −1 : X → G(T ) is continuous, that is,

||P −1(x)|| ≤ c||x||, ∀x ∈ X for some c > 0. Hence T is bounded because of

||T x|| ≤ ||T x|| + ||x|| = ||(x, T x)|| = ||P −1(x)|| ≤ c||x||, ∀x ∈ X.

Exercises.

1. Let X = c00 be the normed space with sup-norm. Let T : X → X be deﬁned by

x2 x3 T x = (x1, 2 , 3 ,...), x = (xn). Show that T is linear and bounded but T −1 is unbounded. Does this contradict bounded inverse theorem?

2. Show that the null space N (T ) of a closed linear operator T : X → Y is a closed subspace of X.

3. Suppose T is a bounded linear operator on a Banach Space X with ||I − T || < 1, where I denotes the identity operator on X. Then show that T −1 exists and it is bounded operator on X.

4. Let (X, || · ||X ) and (Y, || · ||Y ) be Banach spaces and T : X → Y be a surjective linear operator from X onto Y such that

∃ c > 0 ∀x ∈ X : ||T x||Y ≥ c ||x||X .

Then T is bounded.

5 5. Let (X, k · k) be a Banach space with Schauder basis {xn, n ≥ 1}. Let Y be the ∞ X vector space consisting of those sequences of scalars (cn) for which the series cnxn n=1 Pn is convergent in X. Deﬁne k(cn)kY = sup k k=1 ckxkk for (cn) ∈ Y. n≥1

(a) Show that the function k · kY defines a norm on Y . P∞ (b) Define the linear mapping T : Y → X by T ((cn)) = n=1 cnxn. Show that T is bijective and continuous, and further T −1 is continuous. P∞ (c) For n ∈ N, x = k=1 ckxk ∈ X, define the linear (coefficient) functional fn(x) = cn. Prove that the linear functional fn is continuous on X.