Page 229 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Proof: By the theorem,
T : B B is a homeomorphism. So that T and T are both continuous and hence bounded. Hence
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by theorem,
Let N and N be normed linear spaces. Then N and N are topologically isomorphic if and only
if there exists a linear transformation T of N onto N and positive constants m and M such that
m x T (x) M x , for every x N.

constants m and M such that
m x T (x) M x .
Theorem 2: If a one-to-one linear transformation T of a Banach space B onto itself is continuous,
then its inverse T is continuous.
–1
–1
Proof: T is a homeomorphism (using theorem of B onto B. Hence T is continuous.
This completes the proof of the theorem.

Note The following examples will show that the completeness assumption in the open
mapping theorem and theorem can neither be omitted in the domain of definition of T nor
in the range of T.

Example 1: Let C [0, 1] be the set of all continuous differentiable function on [0, 1]. We
know that C [0, 1] is an incomplete space with the norm

f = sup {|f (x)| : 0 x 1}
But it is complete with respect to the norm
f = f + f .
Now let us choose B = [C [0, 1], ] and N = [C [0, 1], ].
Consider the identity mapping I : B N. The identity mapping is one-to-one onto and continuous.
I is not continuous. For, if it were continuous, then it is a homeomorphism. Mapping of a
–1
complete space into an incomplete space which cannot be. Hence I does not map open sets into
open sets.
Thus the open mapping theorem fails if the range of T is not a Banach space.

Example 2: Let B be an infinite dimensional Banach space with a basis { : i I} with
i
= 1 for each i I. Let N be the set of all functions from I to C which vanish everywhere except
i
a finite member of points in I. Then N is a linear space under addition and scalar multiplication.
We can define the norm on N as
f = |f (i)|, i I.
Then N is an incomplete normed linear space. Now consider the transformation
T : N B defined as follows.
For each f N, let T (f) = f (i) .
i

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