MATH 311
Topics in Applied Mathematics I
Lecture 28:
Norms and inner products.
Norm
The notion of norm generalizes the notion of length
of a vector in Rn .
Definition. Let V be a vector space. A function
α : V → R is called a norm on V if it has the
following properties:
(i) α(x) ≥ 0, α(x) = 0 only for x = 0 (positivity)
(ii) α(r x) = |r | α(x) for all r ∈ R (homogeneity)
(iii) α(x + y) ≤ α(x) + α(y) (triangle inequality)
Notation. The norm of a vector x ∈ V is usually
denoted kxk. Different norms on V are
distinguished by subscripts, e.g., kxk1 and kxk2.
Examples. V = Rn , x = (x1, x2, . . . , xn ) ∈ Rn .
• kxk∞ = max(|x1|, |x2|, . . . , |xn |).
Positivity and homogeneity are obvious. Let
x = (x1, . . . , xn ) and y = (y1, . . . , yn ). Then
x + y = (x1 + y1, . . . , xn + yn ).
|xi + yi | ≤ |xi | + |yi | ≤ maxj |xj | + maxj |yj |
=⇒ maxj |xj + yj | ≤ maxj |xj | + maxj |yj |
=⇒ kx + yk∞ ≤ kxk∞ + kyk∞ .
• kxk1 = |x1| + |x2 | + · · · + |xn |.
Positivity and homogeneity are obvious.
The triangle inequality: |xi + yi | ≤ |xi | + |yi |
P
P
P
=⇒
|x
+
y
|
≤
|x
|
+
j
j
j
j
j
j |yj |
Examples. V = Rn , x = (x1, x2, . . . , xn ) ∈ Rn .
1/p
• kxkp = |x1|p + |x2|p + · · · + |xn |p
, p > 0.
Remark. kxk2 = Euclidean length of x.
Theorem kxkp is a norm on Rn for any p ≥ 1.
Positivity and homogeneity are still obvious (and
hold for any p > 0). The triangle inequality for
p ≥ 1 is known as the Minkowski inequality:
1/p
≤
|x1 + y1|p + |x2 + y2|p + · · · + |xn + yn |p
1/p
1/p
.
+ |y1 |p + · · · + |yn |p
≤ |x1|p + · · · + |xn |p
Normed vector space
Definition. A normed vector space is a vector
space endowed with a norm.
The norm defines a distance function on the normed
vector space: dist(x, y) = kx − yk.
Then we say that a vector x is a good
approximation of a vector x0 if dist(x, x0) is small.
Also, we say that a sequence x1, x2, . . . converges
to a vector x if dist(x, xn ) → 0 as n → ∞.
Unit circle: kxk = 1
kxk = (x12 + x22 )1/2
1/2
kxk = 21 x12 + x22
kxk = |x1 | + |x2|
kxk = max(|x1 |, |x2|)
black
green
blue
red
Examples. V = C [a, b], f : [a, b] → R.
• kf k∞ = max |f (x)|.
a≤x≤b
• kf k1 =
• kf kp =
Z
b
|f (x)| dx.
a
Z
a
b
p
|f (x)| dx
1/p
, p > 0.
Theorem kf kp is a norm on C [a, b] for any p ≥ 1.
Inner product
The notion of inner product generalizes the notion
of dot product of vectors in Rn .
Definition. Let V be a vector space. A function
β : V × V → R, usually denoted β(x, y) = hx, yi,
is called an inner product on V if it is positive,
symmetric, and bilinear. That is, if
(i) hx, xi ≥ 0, hx, xi = 0 only for x = 0 (positivity)
(ii) hx, yi = hy, xi
(symmetry)
(iii) hr x, yi = r hx, yi
(homogeneity)
(iv) hx + y, zi = hx, zi + hy, zi (distributive law)
An inner product space is a vector space endowed
with an inner product.
Examples. V = Rn .
• hx, yi = x · y = x1y1 + x2y2 + · · · + xn yn .
• hx, yi = d1 x1y1 + d2x2 y2 + · · · + dn xn yn ,
where d1 , d2, . . . , dn > 0.
• hx, yi = (Dx) · (Dy),
where D is an invertible n×n matrix.
Remarks. (a) Invertibility of D is necessary to show
that hx, xi = 0 =⇒ x = 0.
(b) The second example is a particular case of the
1/2
1/2
1/2
third one when D = diag(d1 , d2 , . . . , dn ).
Problem. Find an inner product on R2 such that
he1 , e1 i = 2, he2 , e2 i = 3, and he1 , e2i = −1,
where e1 = (1, 0), e2 = (0, 1).
Let x = (x1, x2), y = (y1, y2) ∈ R2 .
Then x = x1 e1 + x2e2 , y = y1e1 + y2 e2.
Using bilinearity, we obtain
hx, yi = hx1 e1 + x2e2 , y1e1 + y2 e2i
= x1 he1 , y1e1 + y2 e2i + x2he2 , y1e1 + y2e2 i
= x1y1he1 , e1i + x1 y2he1 , e2i + x2y1he2 , e1i + x2y2he2 , e2i
= 2x1y1 − x1y2 − x2y1 + 3x2y2.
It remains to check that hx, xi > 0 for x 6= 0.
Indeed, hx, xi = 2x12 − 2x1 x2 + 3x22 = (x1 − x2 )2 + x12 + 2x22 .
Example. V = Mm,n (R), space of m×n matrices.
• hA, Bi = trace (AB T ).
If A = (aij ) and B = (bij ), then hA, Bi =
n
m P
P
i=1 j=1
Examples. V = C [a, b].
Z b
• hf , g i =
f (x)g (x) dx.
a
• hf , g i =
Z
b
f (x)g (x)w (x) dx,
a
where w is bounded, piecewise continuous, and
w > 0 everywhere on [a, b].
w is called the weight function.
aij bij .
Theorem Suppose hx, yi is an inner product on a
vector space V . Then
hx, yi2 ≤ hx, xihy, yi for all x, y ∈ V .
Proof: For any t ∈ R let vt = x + ty. Then
hvt , vt i = hx + ty, x + tyi = hx, x + tyi + thy, x + tyi
= hx, xi + thx, yi + thy, xi + t 2hy, yi.
hx, yi
Assume that y 6= 0 and let t = −
. Then
hy, yi
hx, yi2
hvt , vt i = hx, xi + thy, xi = hx, xi −
.
hy, yi
Since hvt , vt i ≥ 0, the desired inequality follows.
In the case y = 0, we have hx, yi = hy, yi = 0.
Cauchy-Schwarz Inequality:
p
p
|hx, yi| ≤ hx, xi hy, yi.
Corollary 1 |x · y| ≤ kxk kyk for all x, y ∈ Rn .
Equivalently, for all xi , yi ∈ R,
(x1y1 + · · · + xn yn )2 ≤ (x12 + · · · + xn2)(y12 + · · · + yn2).
Corollary 2 For any f , g ∈ C [a, b],
2 Z b
Z b
Z
2
|f (x)| dx ·
f (x)g (x) dx ≤
a
a
b
|g (x)|2 dx.
a
Norms induced by inner products
Theorem Suppose hx, yi is anpinner product on a
vector space V . Then kxk = hx, xi is a norm.
Proof: Positivity is obvious. Homogeneity:
p
p
p
kr xk = hr x, r xi = r 2hx, xi = |r | hx, xi.
Triangle inequality (follows from Cauchy-Schwarz’s):
kx + yk2 = hx + y, x + yi
= hx, xi + hx, yi + hy, xi + hy, yi
≤ hx, xi + |hx, yi| + |hy, xi| + hy, yi
≤ kxk2 + 2kxk kyk + kyk2 = (kxk + kyk)2.
Examples. • The length of a vector in Rn ,
p
|x| = x12 + x22 + · · · + xn2,
is the norm induced by the dot product
x · y = x1 y1 + x2 y2 + · · · + xn yn .
• The norm kf k2 =
Z
a
b
|f (x)|2 dx
1/2
on the
vector space C [a, b] is induced by the inner product
Z b
hf , g i =
f (x)g (x) dx.
a
Angle
Since |hx, yi| ≤ kxk kyk, we can define the angle
between nonzero vectors in any vector space with
an inner product (and induced norm):
hx, yi
.
∠(x, y) = arccos
kxk kyk
Then hx, yi = kxk kyk cos ∠(x, y).
In particular, vectors x and y are orthogonal
(denoted x ⊥ y) if hx, yi = 0.