Topics In Analysis

15.1.3 Open Mapping Theorem

Another remarkable theorem which depends on the Baire category theorem is the open mapping theorem. Unlike Theorem 15.1.8 it requires both X and Y to be Banach spaces.

To aid in the proof, here is a lemma.

Proof of Lemma 15.1.10: Let y ∈L(B(0,b)). There exists x₁ ∈ B(0,b) such that ||y − Lx₁|| <

. Now this implies

--------- 2y − 2Lx1 ∈ B(0,a) ⊆ L (B (0,b)).

Thus 2y − 2Lx₁ ∈L(B(0,b)) just like y was. Therefore, there exists x₂ ∈ B(0,b) such that ||2y − 2Lx₁ − Lx₂|| < a∕2. Hence

< a, and there exists x₃ ∈ B such that < a∕2. Continuing in this way, there exist x₁,x₂,x₃,x₄,... in B(0,b) such that

n ∑n n−(i−1) ||2 y− 2 L(xi)|| < a i=1

which implies

( ) ∑n ∑n ||y− 2−(i− 1)L(xi)|| = ||y − L 2−(i−1)(xi) || < 2−na (15.1.2) i=1 i=1

(15.1.2)

Now consider the partial sums of the series, ∑ _i=1^∞2⁻⁽ⁱ⁻¹⁾x_i.

∑n ∞∑ || 2−(i− 1)xi|| ≤ b 2−(i−1) = b2−m+2. i=m i=m

Therefore, these partial sums form a Cauchy sequence and so since X is complete, there exists x = ∑ _i=1^∞2⁻⁽ⁱ⁻¹⁾x_i. Letting n →∞ in 15.1.2 yields ||y − Lx|| = 0.Now

∑n ||x|| = nl→im∞ || 2−(i− 1)xi|| i=1

∑n ∑n ≤ lim 2−(i−1)||xi|| < lim 2−(i−1)b = 2b. n→∞ i=1 n→∞ i=1

This proves the lemma.

Proof of Theorem 15.1.9: Y = ∪_n=1^∞L(B(0,n)). By Corollary 15.1.3, the set, L(B(0,n₀)) has nonempty interior for some n₀. Thus B(y,r) ⊆L(B(0,n₀)) for some y and some r > 0. Since L is linear B(−y,r) ⊆L(B(0,n₀)) also. Here is why. If z ∈ B(−y,r), then −z ∈ B(y,r) and so there exists x_n ∈ B

such that Lx_n →−z. Therefore, L → z and −x_n ∈ B also. Therefore z ∈L(B(0,n₀)). Then it follows that

The reason for the last inclusion is that from the above, if y₁ ∈ B and y₂ ∈ B, there exists x_n,z_n ∈ B such that

Lxn → y1,Lzn → y2.

Therefore,

||xn + zn|| ≤ 2n0

and so

∈L(B(0,2n₀)).

By Lemma 15.1.10, L(B(0,2n₀)) ⊆ L(B(0,4n₀)) which shows

B(0,r) ⊆ L(B(0,4n0)).

Letting a = r(4n₀)⁻¹, it follows, since L is linear, that B(0,a) ⊆ L(B(0,1)). It follows since L is linear,

L(B (0,r)) ⊇ B (0,ar). (15.1.3)

(15.1.3)

Now let U be open in X and let x + B(0,r) = B(x,r) ⊆ U. Using 15.1.3,

L(U) ⊇ L(x+ B (0,r))

= Lx + L(B(0,r)) ⊇ Lx +B (0,ar) = B(Lx,ar).

Hence

Lx ∈ B(Lx,ar) ⊆ L (U).

which shows that every point, Lx ∈ LU, is an interior point of LU and so LU is open. This proves the theorem.

This theorem is surprising because it implies that if

and are two norms with respect to which a vector space X is a Banach space such that ≤ K, then there exists a constant k, such that ≤ k. This can be useful because sometimes it is not clear how to compute k when all that is needed is its existence. To see the open mapping theorem implies this, consider the identity map idx = x. Then id : → is continuous and onto. Hence id is an open map which implies id⁻¹ is continuous. Theorem 15.1.4 gives the existence of the constant k.