Page 205 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Again since T is continuous, it is bounded so that there exists a positive constant k such that
–1
T (y) k y x k T(x)
–1
1
m x T(x) where m = > 0.
k
This completes the proof of the theorem.
Theorem 7: Let T : N N be a linear transformation. Then T is bounded if and only if T maps
bounded sets in N onto bounded set in N .
Proof: Since T is a bounded linear transformation,
T (x) k x for all x N.
Let B be a bounded subset of N. Then

x k k x B.
1
We now show that T(B) is bounded subset of N .
From above we see that

T (x) k x B.
1
T (B) is bounded in N .
Conversely, let T map bounded sets in N into bounded sets in N . To prove that T is a bounded
linear transformation, let us take the closed unit sphere S [0, 1] in N as a bounded set. By
hypothesis, its image T (S[1, 0]) must be bounded set in N .

Therefore there is a constant k such that
1
T (x) k for all x S [0, 1]
1
x
Let x be any non-zero vector in N. Then S[0,1] and so we get
x

x
T k 1
x

T (x) k x .
1
Since this is true for x = 0 also, T is a bounded linear transformation.

This completes the proof of the theorem.

16.2 Summary

 Let N be a normed linear space. Then we know the set R of real numbers and the set C of
complex numbers are Banach spaces with the norm of any x R or x C be the absolute
value of X. (N, R) or (N, C) denote respectively the set of all continuous linear
transformations from N into R or C.

 A linear functional on a normed linear space N is said to be bounded, if there exists a
constant k such that

|f (x)| k x x N.

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