Lecture Notes in Differential Equations - Bruce E. Shapiro

130 LESSON 15. LINEAR OPERATORS AND VECTOR SPACES
3. The triangle follows from the properties of real numbers,
‖v + w‖ ∞ = ‖(x + p, y + q, z + r)‖ (15.23)
because |x + p| ≤ |x| + |p|, etc.Hence
Thus ‖v‖ ∞ is a norm.
= max(|x + p|, |y + q|, |z + r|) (15.24)
≤ max(|x| + |p|, |y| + |q|, |z| + |r|) (15.25)
‖v + w‖ ∞ ≤ max(|x|, |y|, |z|) + max(|p|, |q|, |r|) (15.26)
= ‖v‖ ∞ + ‖w‖ ∞ (15.27)
Example 15.6. Let V be the set of all functions y(t) defined on the interval
0 ≤ t ≤ 1. Then
(∫ 1
1/2
‖y‖ = |y(t)| dt) 2 (15.28)
is a norm on V.
0
The first property follows because the integral of an absolute value is a
positive number (the area under the curve) unless y(t) = 0 for all t, in
which case the area is zero.
The second property follows because
(∫ 1
‖cy‖ =
0
1/2 (∫ 1
1/2
|cy(t)| dt) 2 = |c| |y(t)| dt) 2 = |c|‖y‖ (15.29)
0
The third property follows because
‖y + z‖ 2 =
=
≤
∫ 1
0
∫ 1
0
∫ 1
0
|y(t) + z(t)| 2 dt (15.30)
|y(t) 2 + 2y(t)z(t) + z(t) 2 |dt (15.31)
|y(t)| 2 dt + 2
∫ 1
By the Cauchy Schwarz Inequality 3
∣
3 The Cauchy-Schwarz inequality is a property of integrals that says exactly this formula.
∫ 1
0
y(t)z(t)dt
∣
2
(∫ 1
≤
0
0
|y(t)||z(t)|dt +
∫ 1
0
|z(t)| 2 dt (15.32)
) (∫ 1
)
|y(t)| 2 dt |z(t)| 2 dt = ‖y‖ 2 ‖z‖ 2 (15.33)
0