13.1.3 Open Mapping Theorem
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 13.1.13 Let X and Y be Banach spaces, let L ∈ℒ(X,Y ), and suppose L is onto. Then
L maps open sets onto open sets.
To aid in the proof, here is a lemma.
Lemma 13.1.14 Let a and b be positive constants and suppose
Proof of Lemma 13.1.14: Let y ∈L(B(0,b)). There exists x1 ∈ B(0,b) such that ||y −Lx1|| <
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
, and there exists x3 ∈ B
2. Continuing in this way, there exist x1,x2,x3,x4,.
.. in B
Now consider the partial sums of the series, ∑
Therefore, these partial sums form a Cauchy sequence and so since X is complete, there exists
x = ∑
i=1∞2−(i−1)xi. Letting n →∞ in 13.5 yields ||y − Lx|| = 0.Now
This proves the lemma.
Proof of Theorem 13.1.13: Y = ∪n=1∞L(B(0,n)). By Corollary 13.1.7, 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
xn ∈ B
. Therefore, L
and −xn ∈ B
. Then it follows that
The reason for the last inclusion is that from the above, if y1 ∈ B
y2 ∈ B
, there exists
xn,zn ∈ B
By Lemma 13.1.14, L(B(0,2n0)) ⊆ L(B(0,4n0)) which shows
Letting a = r(4n0)−1, it follows, since L is linear, that B(0,a) ⊆ L(B(0,1)). It follows since L is
Now let U be open in X and let x + B(0,r) = B(x,r) ⊆ U. Using 13.6,
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
are two norms with respect to which a
is a Banach space such that
, then there exists a constant
This can be useful because sometimes it is not clear how to compute k
that is needed is its existence. To see the open mapping theorem implies this, consider the
identity map idx
. Then id
is continuous and onto. Hence
is an open
map which implies id−1
is continuous. Theorem 13.1.8
gives the existence of the constant