Page 230 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 19: The Open Mapping Theorem

Then T is linear and Notes
T (f) |f(i)| i
i I
= |f | = f for every f N.
i
i I
Hence T is bounded transformation from N to B . It is also one-to-one and onto. But T does not
map open subsets of N onto B . For, if it maps, it is a linear homeomorphism from N onto B
which cannot be since N is incomplete.
Theorem 3: Let B be a Banach space and N be a normed linear space. If T is a continuous linear
open map on B onto N, then N is a Banach space.

Proof: Let (y ) be a Cauchy sequence in N. Then we can find a sequence of positive integer (n )
n k
such that n < n and for each k
k k + 1
1
y y
n k 1 n k k
2
Hence by theorem: “Let N and N be normed linear spaces. A linear map T : N N is open and
onto if and only if there is a M > 0 such that for any y N , there is a x N such that Tx = y and
x M y .”
For y y N , there is a n B and a constant M such that
n k 1 n k
T (x ) = y y n and x y y n .
k n k 1 k n k 1

By on choice y y is convergent so that x is convergent. Since B is a Banach
n k 1 n k n k
k 1 k 1
space there is a x B such that
n
x = Lim x
n n k
k 1
n
Since T is continuous T(x ) T (x) as n
k
k 1
But T(x ) y n k 1 y n 1 so that
k
k 1
y y T(x) y y T(x)
n k 1 n 1 n k 1 n 1
Since (y ) is a Cauchy sequence such that every subsequences is convergent, (y ) itself converges
n n
and y y + T (x) in N.
n n 1
Hence N is complete. Consequently, N is a Banach space.

This completes the proof of the theorem.

Example: Let N be complete in two norms and respectively. If there is a number
1 2
a > 0 such that x a x for all x N, then show that the two norms are equivalent.
1 2
Solution: The identity map
i : (N, ) (N, ) is an one-one onto map.
2 1

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