Page 228 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 19: The Open Mapping Theorem

19.1.2 Proof of the Open Mapping Theorem Notes

Statement: If T is a continuous linear transformation of a Banach space B onto a Banach space B ,
then T is an open mapping.
Proof: Let G be an open set in B. We are to show that T (G) is an open set in B

i.e. if y is any point of T (G), then there exists an open sphere centered at y and contained in T (G).
y T (G) y = T(x) for some x G.
x G, G open in B there exists an open sphere S (x; r) with centre x and radius r such that
S (x; r) G.
But as remarked earlier we can write S (x; r) = x + S , where S is open sphere of radius r centered
r r
at the origin in B.
Thus x + S G … (1)
r
By lemma (2) (prove it),

T (S ) contains some S . Therefore
r r 1
S (y; r ) = y + S
1 r 1
y + T (S )
r
= T (x) + T (S )
r
= T (x + S )
r
T (G), (Using (1))
since x + S = S (x; r) G.
r
Thus we have shown that to each y T (G), there exists an open sphere in B centered at y and
contained in T (G) and consequently T (G) is an open set.
This completes the proof of the theorem.

19.1.3 Theorems and Solved Examples

Theorem 1: Let B and B be Banach spaces and let T be an one-one continuous linear transformation
of B onto B . Then T is a homeomorphism.
–1
In particular, T is automatically continuous.
Proof: We know that a one-to-one continuous open map from B onto B is a homeomorphism.

By hypothesis T : B B is a continuous one-to-one onto mapping.
By the open mapping theorem, T is open. Hence T is a homeomorphism. Since T is
–
–
homeomorphism, T exists and continuous from B to B so that T is bounded and hence
–1
T (B , B).
This completes the proof of the theorem.
Cor. 1: Let B and B be Banach spaces and let T (B, B ). If T : B B is one-to-one and onto, there
are positive numbers m and M such that
m x T (x) M x .

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