Measure, Integration & Real Analysis, 2021a

164 Chapter 6 Banach Spaces
Sometimes examples that do not satisfy a definition help you gain understanding.
6.35 Example not norms
• Let L 1 (R) denote the vector space of Borel (or Lebesgue) measurable functions
f : R → F such that ∫ | f | dλ < ∞, where λ is Lebesgue measure on R. Define
‖·‖ 1 on L 1 (R) by
∫
‖ f ‖ 1 = | f | dλ.
Then ‖·‖ 1 satisfies the homogeneity condition and the triangle inequality on
L 1 (R), as you should verify. However, ‖·‖ 1 is not a norm on L 1 (R) because
the positive definite condition is not satisfied. Specifically, if E is a nonempty
Borel subset of R with Lebesgue measure 0 (for example, E might consist of a
single element of R), then ‖χ E
‖ 1 = 0 but χ E ̸= 0. In the next chapter, we will
discuss a modification of L 1 (R) that removes this problem.
• If n ∈ Z + and ‖·‖ is defined on F n by
‖(a 1 ,...,a n )‖ = |a 1 | 1/2 + ···+ |a n | 1/2 ,
then ‖·‖ satisfies the positive definite condition and the triangle inequality (as
you should verify). However, ‖·‖ as defined above is not a norm because it does
not satisfy the homogeneity condition.
• If ‖·‖ 1/2 is defined on F n by
‖(a 1 ,...,a n )‖ 1/2 = ( |a 1 | 1/2 + ···+ |a n | 1/2) 2 ,
then ‖·‖ 1/2 satisfies the positive definite condition and the homogeneity condition.
However, if n > 1 then ‖·‖ 1/2 is not a norm on F n because the triangle
inequality is not satisfied (as you should verify).
The next result shows that every normed vector space is also a metric space in a
natural fashion.
6.36 normed vector spaces are metric spaces
Suppose (V, ‖·‖) is a normed vector space. Define d : V × V → [0, ∞) by
Then d is a metric on V.
Proof
Suppose f , g, h ∈ V. Then
d( f , g) =‖ f − g‖.
d( f , h) =‖ f − h‖ = ‖( f − g)+(g − h)‖
≤‖f − g‖ + ‖g − h‖
= d( f , g)+d(g, h).
Thus the triangle inequality requirement for a metric is satisfied. The verification of
the other required properties for a metric are left to the reader.
Measure, Integration & Real Analysis, by Sheldon Axler