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Unit 20: The Closed Graph Theorem

Since G is closed, we have Notes
T
(x T (x)) G and if we take
1 T
y = T(x); then (x, y) G .
T
Now x – x = x – x + T (x – x)
n 1 n n
= x – x + T (x ) – T(x)
n n
= x – x + T (x – y) 0 as n . (Using (1))
n n
Hence, the sequence (x ) in B x B and consequently B is complete.
n 1 1 1
This completes the proof of the theorem.

Theorem 3: Let B and B be Banach spaces and let T : B B be linear. If G is closed in B × B and
T
if T is one-one and onto, then T is a homeomorphism from B onto B .
Proof: By closed graph theorem, T is continuous.

–1
Let T = T : B B. Then T is linear.
Further (x, y) G (y, x) G .
T T
G is closed in B × B.
T
T is continuous (By closed graph theorem)
T is a homeomorphism on B onto B .
This completes the proof of the theorem.
Theorem 4: Let a Banach space B be made into a Banach space B by a new norm. Then the
topologies generated by these two norms are the same if either is stronger than the other.
Proof: Let the new norm on B be . Let is stronger than . Then a constant k such that
x k x for every x B.
Consider the identity map
I : B B .
We claim that G is closed.
1
Let x x in B and x y in B .
n n
Then x k x x B, I (x ) = x y in also.
n n
Since a sequence cannot converge to two distinct points in , y = x. Consequently G is closed.
1
Hence closed graph theorem, I is continuous. Therefore a k s such that
x = I(x) k x for every x B. Hence is stronger than . Hence two topologies are
same.
20.2 Summary

 Let N and N be a normal linear space and let T : N N be a mapping with domain N and
range N . The graph of T is defined to be a subset of N × N which consist of all ordered
pairs (x, T (x)). It is generally denoted by G .
T

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