Page 235 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
We shall show that G = G .
T
T
Since G G always, it suffices to show that G T G .
T
T T
Let (x, y) G . Then there exists a sequence (x , T (x )) in G such that
T
n n T
(x , T (x )) (x, y)
n n
x x and T (x ) y.
n n
But T is continuous T (x ) T (x) and so y = T (x)
n
(x, y) = (x, T (x)) G
T
G T G
T
Hence G = G i.e. G is closed.
T
T T
Sufficient Part:
Let G is closed. Then we claim that T is continuous. Let B be the given linear space B renormed
T 1
by given by
1
x = x + T (x) for x B.
1
Now T (x) x + T (x) = x .
1
T is bounded (continuous) as a mapping from B to B .
1
So if B and B have the same topology then T will be continuous from B to B . To this end, we have
1
to show that B and B are homeomorphic.
1
Consider the identity mapping
I : B B defined by
1
I (x) = x for every x B .
1
Then I is always one-one and onto.

Further I (x) = x x + T (x) = x
1
I is bounded (continuous) as a mapping from B onto B.
1
Therefore if we show that B is complete with respect to , then B is a Banach space so by
1 1 1
theorem.
“Let B and B be Banach spaces and let T be one-one continuous linear transformation of B onto
–1
B . Then T is a homeomorphism. In particular, T is automatically continuous.”
I is homeomorphism. Therefore to complete the proof, we have to show that B is complete
1
under the norm .
1
Let (x ) be a Cauchy sequence in B . Then
n 1
x – x = x – x + T (x – x ) 0 as m, n
n m 1 n m n m
(x ) and (T (x )) are Cauchy sequences in B and B respectively.
n n
Since B and B are complete, we have
x x in B and T (x ) T (x) in B … (1)
n n

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