Page 233 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Theorem 1: Let N and N be normed linear spaces. Then N × N is a normed linear space with
coordinate-wise linear operations and the norm.

1
p p p
(x, y) = x x , where x N, y N
and | p < . Moreover, this norm induces the product topology on N × N , and N × N is
complete iff both N and N are complete.

Proof:
(i) It needs to prove the triangle inequality since other conditions of a norm are immediate.
Let (x, y) and (x , y ) be two elements of N × N .
Then (x, y) + (x , y ) = (x + x , y + y )

1
p p p
x x y y

1
p p p
= x x y y
1 1
p p p p p p
= x y x y

(By Minkowski’s inequality)
= (x, y) + (x , y ) .

This establishes the triangular inequality and therefore N × N is a normed linear space.
Furthermore (x , y ) (x, y) x = x and y = y. Hence theorem on N × N induces the
n n n n
product topology.
(ii) Next we show that N × N is complete N, N are complete.
Let (x , y ) be a Cauchy sequence in N × N . Given > 0, we can find a n such that
n n o
(x , y ) – (x , y ) < m, n n . … (1)
n n m m o
(x – x ) < and y – y < m, n n
n m n m o
(x ) and (y ) are Cauchy sequences in N and N respectively.
n n
Since N, N are complete, let

x x N and y y N in their norms,
n o n o
i.e. (x – x ) < and y – y < m, n n . … (2)
n o n m o
since x N, y N , (x , y ) N × N .
o o o o
Further (x , y ) – (x , y ) < n n (using (2))
n n o o o
(x , y ) (x , y ) in the norm of N × N and (x , y ) N × N .
n n o o o o
N × N is complete.

The converse follows by reversing the above steps.
This completes the proof of the theorem.

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