MATH 311-504
Topics in Applied Mathematics
Lecture 3-3:
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 |).
1/p
, p ≥ 1.
1/2
= |x|.
• kxkp = |x1 |p + |x2 |p + · · · + |xn |p
In particular,
• kxk1 = |x1 | + |x2 | + · · · + |xn |,
• kxk2 = |x1 |2 + |x2 |2 + · · · + |xn |2
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 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 =
Z
(uniform norm).
b
|f (x)| dx.
a
• kf kp =
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, yi ≥ 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 = x1 y1 + x2 y2 + · · · + xn yn .
• hx, yi = d1 x1 y1 + d2 x2 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 ).
Examples. V = C [a, b].
Z b
f (x)g (x) dx.
• hf , g i =
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.
Counterexamples. V = R2 .
• hx, yi = x1 y1 − x2 y2 .
Let v = (1, 2), then hv, vi = 12 − 22 = −3.
hx, yi is symmetric and bilinear, but not positive.
• hx, yi = 2x1 y1 + x1 x2 + 2x2 y2 + y1 y2 .
v = (1, 1), w = (1, 0) =⇒ hv, wi = 3, h2v, wi = 8.
hx, yi is positive and symmetric, but not bilinear.
• hx, yi = x1 y1 + x1 y2 − x2 y1 + x2 y2 .
v = (1, 1), w = (1, 0) =⇒ hv, wi = 0, hw, vi = 2.
hx, yi is positive and bilinear, but not symmetric.
Problem. Find an inner product on R2 such that
he1 , e1 i = 2, he2 , e2 i = 3, and he1 , e2 i = −1,
where e1 = (1, 0), e2 = (0, 1).
Let x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 .
Then x = x1 e1 + x2 e2 , y = y1 e1 + y2 e2 .
It follows that
hx, yi = hx1 e1 + x2 e2 , y1 e1 + y2 e2 i
= x1 he1 , y1 e1 + y2 e2 i + x2 he2 , y1 e1 + y2 e2 i
= x1 y1 he1 , e1 i + x1 y2 he1 , e2 i + x2 y1 he2 , e1 i + x2 y2 he2 , e2 i
= 2x1 y1 − x1 y2 − x2 y1 + 3x2 y2 .
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, xi + 2thx, yi + t 2 hy, yi.
The right-hand side is a quadratic polynomial in t
(provided that y 6= 0). Since hvt , vt i ≥ 0 for all t,
the discriminant D is nonpositive. But
D = 4hx, yi2 − 4hx, xihy, yi.
Cauchy-Schwarz Inequality:
p
p
|hx, yi| ≤ hx, xi hy, yi.
Cauchy-Schwarz Inequality:
p
p
|hx, yi| ≤ hx, xi hy, yi.
Corollary 1 |x · y| ≤ |x| |y| for all x, y ∈ Rn .
Equivalently, for all xi , yi ∈ R,
(x1 y1 + · · · + xn yn )2 ≤ (x12 + · · · + xn2 )(y12 + · · · + yn2 ).
Corollary 2 For any f , g ∈ C [a, b],
Z b
2 Z b
Z
f (x)g (x) dx ≤
|f (x)|2 dx ·
a
a
a
b
|g (x)|2 dx.