Another remarkable theorem which depends on the Baire category theorem is the open mapping theorem.
Unlike Theorem 13.1.12 it requires both X and Y to be Banach spaces.

Theorem 13.1.13Let X and Y be Banach spaces, let L ∈ℒ(X,Y ), and suppose L is onto. ThenL maps open sets onto open sets.

To aid in the proof, here is a lemma.

Lemma 13.1.14Let a and b be positive constants and suppose

---------
B (0,a) ⊆ L(B(0,b)).

Then

---------
L(B(0,b)) ⊆ L(B(0,2b)).

Proof of Lemma 13.1.14: Let y ∈L(B(0,b)). There exists x_{1}∈ B(0,b) such that ||y −Lx_{1}|| <

a
2

. Now
this implies

2y− 2Lx ∈ B(0,a) ⊆ L(B-(0,b)).
1

Thus 2y − 2Lx_{1}∈L(B(0,b)) just like y was. Therefore, there exists x_{2}∈ B(0,b) such that
||2y − 2Lx_{1}− Lx_{2}|| < a∕2. Hence

||4y− 4Lx1 − 2Lx2||

< a, and there exists x_{3}∈ B

(0,b)

such that

||4y− 4Lx1 − 2Lx2 − Lx3||

< a∕2. Continuing in this way, there exist x_{1},x_{2},x_{3},x_{4},... in B(0,b) such
that

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 13.5 yields ||y − Lx|| = 0.Now

Proof of Theorem 13.1.13:Y = ∪_{n=1}^{∞}L(B(0,n)). By Corollary 13.1.7, the set, L(B(0,n_{0})) has
nonempty interior for some n_{0}. Thus B(y,r) ⊆L(B(0,n_{0})) for some y and some r > 0. Since L is linear
B(−y,r) ⊆L(B(0,n_{0})) also. Here is why. If z ∈ B(−y,r), then −z ∈ B(y,r) and so there exists
x_{n}∈ B

(0,n0)

such that Lx_{n}→−z. Therefore, L

(− xn)

→ z and −x_{n}∈ B

(0,n0)

also. Therefore
z ∈L(B(0,n_{0})). Then it follows that

B (0,r) ⊆ B (y,r) + B(− y,r)
≡ {y1-+-y2-: y1 ∈ B (y,r) and y2 ∈ B (− y,r)}
⊆ L (B (0,2n0))

The reason for the last inclusion is that from the above, if y_{1}∈ B

(y,r)

and y_{2}∈ B

(− y,r)

, there exists
x_{n},z_{n}∈ B

(0,n0)

such that

Lxn → y1,Lzn → y2.

Therefore,

||xn + zn || ≤ 2n0

and so

(y +y )
1 2

∈L(B(0,2n_{0})).

By Lemma 13.1.14, L(B(0,2n_{0}))⊆ L(B(0,4n_{0})) which shows

B (0,r) ⊆ L(B(0,4n0)).

Letting a = r(4n_{0})^{−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). (13.6)

(13.6)

Now let U be open in X and let x + B(0,r) = B(x,r) ⊆ U. Using 13.6,

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 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 :

(X,||⋅||)

→

(X,|⋅|)

is continuous and onto. Hence id is an open
map which implies id^{−1} is continuous. Theorem 13.1.8 gives the existence of the constant
k.