THE OPEN MAPPING THEOREM AND RELATED THEOREMS ANTON. R SCHEP We start with a lemma, whose proof contains the most ingenious part of Banach's open mapping theorem. Given a norm k · ki we denote by Bi(x; r) the open ball fy 2 X : ky − xki < rg. Lemma 1. Let X be a vector space with two norms k · k1, k · k2 such that (X; k · k1) is a Banach space and assume that the identity map I :(X; k · k1) ! (X; k · k2) is continuous. k·k2 If B2(0; 1) ⊂ B1(0; r) , then B2(0; 1) ⊂ B1(0; 2r) and the two norms are equivalent. 1 Proof. From the hypothesis we get B2(0; 1) ⊂ B1(0; r) + B2(0; 2 ), so by scaling we get that 1 r 1 B2(0; 2n ) ⊂ B1(0; 2n ) + B2(0; 2n+1 ) for all n ≥ 1. let now kyk2 < 1. Then we can write 1 1 y = x1 + y1, where kx1k1 < r and ky1k2 < 2 . Assume we have kynk2 < 2n we can write r 1 yn = xn+1 + yn+1, where kxn+1k1 < 2n and kyn+1k2 < 2n+1 . By completeness of (X; k · k1) P1 there exists x 2 X such that x = n=1 xn, where the series converges with respect to the norm k · k1. By continuity of the identity map I :(X; k · k1) ! (X; k · k2) it follows that the same series also converges to x with respect to k · k2. On the other hand the equation Pn+1 P1 y = k=1 xk + yn+1 shows that the series n=1 xn converges to y with respect to k · k2. P1 Hence y = x and thus kyk1 = kxk1 ≤ n=1 kxnk1 < 2r. It follows thatB2(0; 1) ⊂ B1(0; 2r) and thus kyk1 ≤ 2rkyk2 for all y 2 X. As the continuity of I gives that there exists C such that kyk2 ≤ Ckyk1 for all y 2 X, we get that the two norms are equivalent. Theorem 2. Let X be a vector space with two norms k · k1, k · k2 such that (X; k · k1) and (X; k · k2) are Banach spaces. Assume that the identity map I :(X; k · k1) ! (X; k · k2) is continuous. Then the norms k · k1 and k · k2 are equivalent. 1 Proof. Applying the Baire Category theorem in (X; k · k2) to X = [n=1B1(0; n) we can k·k2 find n0, x0 and r0 > 0 such that B2(x0; r0) ⊂ B1(0; n0) . Translating over −x0 we get k·k2 that B2(0; r0) ⊂ B1(−x0; n0) . Now by the triangle inequality we get that B2(0; r0) ⊂ k·k2 B1(0; n0 + kx0k1) . By the above lemma the two norms are equivalent. Theorem 3 (Bounded Inverse Theorem). Let X, Y be Banach spaces and assume T : X ! Y is an one-to-one, onto continuous linear operator. Then T −1 : Y ! X is continuous. −1 Proof. Define kykT = kT yk. Then k · kT is a norm on Y and kyk ≤ kT kkykT for P1 y 2 Y , so I :(Y; k · kT ) ! (Y; k · k) is continuous. Moreover, if n=1 kynkT < 1, then P1 −1 PN PN −1 x0 = n=1 T yn exists in X and kT x0 − n=1 ynkT = kx0 − n=1 T ynk ! 0 as N ! 1. Hence (Y; k · kT ) is also a Banach space. By the above Theorem the two norms 1 2 ANTON. R SCHEP on Y are equivalent, so there exists C such that kT −1(y)k ≤ Ckyk for all y 2 Y , i.e. T −1 is continuous. We recall now that a linear map T : X ! Y is called open if T (O) is open for all open O ⊂ X. It is easy to see that an open linear map is surjective. The Open Mapping theorem gives a converse to that statement. Before stating and proving that theorem, we recall a few basic facts about quotient maps. Let X be a Banach space and M ⊂ X a closed subspace. Then X=M is a Banach space with respect to the quotient norm k[x]k = inffkyk; y 2 [x]g. Denote by Q the quotient map Q(x) = [x]. Then Q is open. In fact, it is easy to see from the definition of the quotient norm that Q(fx : kxk < 1g) = f[x]: k[x]k < 1g. Theorem 4 (Open Mapping Theorem). Let X, Y be Banach spaces and assume T : X ! Y is an onto continuous linear operator. Then T is an open map. Proof. Let Q : X ! X=ker(T ) be the quotient map. Then by the above remarks Q is an open mapping. Let T^ : X=ker(T ) ! Y be the induced map such that T = T^ ◦ Q. Then T^ is one to one and onto, so by the above Theorem T^−1 is continuous, so T^ is open and thus T is open. Let now A : D(A) ! Y be a linear operator, where D(A) is a (not necessarily closed) linear subspace of the Banach space X. The subspace D(A) is called the domain of A. Given a linear operator A : D(A) ! Y we define the graph Γ(A) = (x; Ax): x 2 D(A); It is clear that Γ(A) is linear subspace of X × Y . We can equip X × Y with the product norm k(x; y)k = kxk+kyk. Then we say that A has a closed graph (or is a closed operator), if Γ(A) is a closed subspace of X × Y . Example 5. Let X = Y = C[0; 1] with the supremum norm. Let D(A) = C0[0; 1] the subspace of X consisting of continuously differentiable functions and define A : D(A) ! Y 0 n by Af = f . One can can see that A is not bounded, by taking fn(t) = t , and noting that kfnk = 1 and kAfnk = n. On the other hand A has a closed graph. To see that A 0 has a closed graph, let (fn; fn) ! (f; g) in X × Y . Then by the Fundamental Theorem of R t 0 R s R t Calculus fn(t) − fn(0) = 0 fn(s) ds ! 0 g(s) ds. It follows that f(t) = f(0) + 0 g(s) ds. Hence f 2 D(A) and f 0 = g, i.e., (f; g) 2 Γ(A). The following proposition is immediate from the definition. Proposition 6. Let X and Y be Banach spaces and assume A : D(A) ! Y is a linear operator, where D(A) is a subspace of X. Then the following are equivalent. (1) A has a closed graph. (2) If xn 2 D(A), xn ! x 2 X, and Axn ! y 2 Y , then x 2 D(A) and Ax = y. (3) D(A) is a Banach space with respect to the graph norm kxkA = kxk + kAxk. Theorem 7 (Closed Graph Theorem). et X and Y be Banach spaces and assume A : X ! Y is a closed linear operator. Then A is bounded. THE OPEN MAPPING THEOREM AND RELATED THEOREMS 3 Proof. Define P : Γ(A) ! X by P (x; Ax) = x. Then P is clearly a bounded, one-to-one, onto linear operator, so by the Bounded Inverse Theorem the inverse operator P −1 : X ! Γ(A) is bounded. Hence there exists a constant C such that kxk + kAxk ≤ Ckxk, i.e., kAxk ≤ (C − 1)kxk for all x 2 X. We now present a proof of the Uniform Boundedness Principle, based on the Closed Graph Theorem. Theorem 8 (Banach-Steinhaus). Let X and Y be Banach spaces and assume Aα 2 L(X; Y ) (α 2 F) is a pointwise bounded family of bounded operators, i.e., for all x 2 X there exists a constant Cx such that kAαxk ≤ Cx for all α 2 F. Then there exists a constant C such that kAαjj ≤ C for all α 2 F. Proof. Define the space ⊕αY = f(yα): yα 2 Y; supα kyαk < 1g with norm k(yα)k = supα kyαk. It is straightforward to verify that ⊕αY is also a Banach space. Now define T : X ! ⊕αY by T x = (Aαx). Note T x 2 ⊕αY , since the collection Aα is pointwise bounded. Clearly T is linear and we claim that T is closed. To see this, let xn ! 0 and T xn ! (yα). Then Aαxn ! yα for all α 2 F, but also Aαxn ! 0 for all α. Hence (yα) = (0) and T is closed. By the Closed Graph Theorem T is bounded, i.e., there exists a constant C such that for all kxk ≤ 1 we have supα kAαxk ≤ C. Hence kAαk ≤ C for all α 2 F. .