Exercises for Section 1.1: Norm and Inner Product
1. Define the ℓ1 -norm on Rn by
kxk1 =
n
X
i=1
|xi |,
and define the sup-norm on Rn by
kxk∞ = sup |xi | .
Show that these satisfy Theorem ??.
Proof. (a) It should be clear that kxk1 ≥ 0, and kxk∞ ≥ 0, and that
eqality holds in each case if and only if x = 0.
(b)
kx + yk1 =
n
X
i=1
i
i
|x + y | ≤
n
X
i=1
|xi | + |y i| = kxk1 + kyk1.
This relies on the triangle inequality on R: |a + b| ≤ |a| + |b|.
kx + yk∞ = sup |xi + y i| .
i
But note that by the triangle inequality on R, |xi +y i| ≤ |xi |+|y i|,
and therefore,
sup |xi + y i| ≤ sup |xi | + |y i|
i
i
≤ sup |xi | + sup |y i|
i
i
= kxk∞ + kyk∞ .
(c)
Note that in general, sup {ai + bi } ≤ sup {ai } + sup {bi }.
kaxk1 =
X
|axi | =
X
|a| · |xi | = |a|
X
|xi | = |a| · kxk1 .
kaxk∞ = sup |axi | = sup |a| · |xi | = |a| sup |xi | = |a|·kxk∞ .
i
i
i
1
2. Prove that kxk ≤
than the ℓ1 -norm.
n
X
i=1
|xi |. In other words, the usual norm is no greater
Proof. First note that for a, b ≥ 0,
square both sides). Then
√
a+b ≤
√
a+
√
b (to see this,
v
u n
n
n q
X
X
uX
2
2
i
t
i
(x ) =
|xi | = kxk1 .
kxk =
(x ) ≤
i=1
i=1
i=1
3. Prove that kx − yk ≤ kxk + kyk. (Compare this with part (2) of
Theorem ??.) When does equality hold?
Proof. kx − yk = kx + (−y)k ≤ kxk + k − yk = kxk + kyk. Equality
holds if and only if y = −λx for some λ > 0. The “if” part is clear, let’s
prove the “only if.” Suppose kx − yk = kxk + kyk. Then kx − yk2 =
kxk2 + 2kxk · kyk + kyk2 But
kx − yk2 = hx − y, x − yi = hx, xi − 2 hx, yi + hy, yi ,
and
kxk2 + 2kxk · kyk + ky 2k2 = hx, xi + 2kxk · kyk + hy, yi .
Comparing these, we see that |hx, yi| = kxk · kyk. But Cauchy-Schwarz
says this can only happen if y = tx for some t ∈ R. We will now show
that t is negative.
− hx, yi = kxk · kyk,
but − hx, yi = −t hx, xi, and kxk · kyk = |t| · kxk2 . Thus |t| = −t, so t
is negative.
4. Prove that kxk − kyk ≤ kx − yk.
2
Proof. In proving this, we will see an alternative approach to Problem 3, above. Look at kx − yk2 .
kx − yk2 = hx − y, x − yi
= hx, xi − 2 hx, yi + hy, yi
= kxk2 − 2 hx, yi + kyk2.
But, by Cauchy Schwarz, hx, yi ≤ kxk · kyk. Therefore, we may
conclude that
kxk2 −2kxk·kyk+kyk2 ≤ kxk2 −2 hx, yi+kyk2 ≤ kxk2 +2kxk·kyk+kyk2.
The left-hand side is (kxk − kyk)2 , the center is kx − yk2, and the
right-hand side is (kxk + kyk)2 . So we have
(kxk − kyk)2 ≤ kx − yk2 ≤ (kxk + kyk)2 .
Take the square root throughout:
kxk − kyk ≤ kx − yk ≤ kxk + kyk.
5. The quantity ky − xk is called the distance between x and y. Prove
and interpret the “triangle inequality”:
kz − xk ≤ kz − yk + ky − xk.
Proof.
kz − xk = kz − y + y − xk = k(z − y) + (y − x)k ≤ kz − yk + ky − xk.
6. Let f and g be integrable on [a, b].
(a) Prove the integral version of the Cauchy-Schwarz inequality:
Z b Z b 1/2 Z b 1/2
≤
f
g
f2
g2
.
a
a
Hint: Consider separately the cases 0 =
Rb
and 0 < a (f − tg)2 for all t ∈ R.
3
a
Rb
a
(f −tg)2 for some t ∈ R,
Rb
Proof. Consider the integral a (f − tg)2 , where t is any real number. This is integrating a non-negative function, so the integral
itself must be non-negative. Furthermore, we can expand it:
Z
b
2
a
(f − tg) =
Z
b
2
f − 2t
a
Z
b
2
fg + t
a
Z
b
g2.
a
This, is a quadratic polynomial in t which is non-negative. Therefore it has either no real roots, or exactly one real root. Ruling
out the possiblity of two distinct real roots means that its discriminant must be non-positive. Computing the discriminant for this
polynomial, we get:
4
Z
b
fg
a
2
−4
Z
b
f
2
a
Z
b
g
2
a
≤ 0.
.
Therefore,
Z
a
b
fg
2
≤
Z
b
f
a
2
Z
b
g
a
2
Taking the square root yields
Z b Z b 1/2 Z b 1/2
.
g2
f2
f g ≤
a
a
a
(b) If equality holds, must f = tg for some t ∈ R? What if f and g
are continuous?
Solution. If f and g are continuous functions then f must be a
multiple of g. But, if they are merely integrable, this is not so.
The best we can say in that f = tg “almost everywhere.”
(c) Show that the Cauchy-Schwarz inequality is a special case of (a).
Proof. Let x, y ∈ Rn . Define f, g : [0, n] −→ R via step functions
where the height of the ith step (where each step is width 1) of f
4
is the value of xi , and the height of the ith step of g is y i. Then
Z b = hx, yi ,
f
g
a
Z b 1/2
f2
= kxk,
a
Z
b
g
2
a
1/2
= kyk.
Therefore, the inequality in part (a) yields the Cauchy-Schwarz
inequality.
7. A linear transformation T : Rn −→ Rn is norm preserving if kT (x)k =
kxk, for all x ∈ Rn , and inner product preserving if hT x, T yi = hx, yi,
for all x, y ∈ Rn .
(a) Prove that T is norm preserving if and only if it is inner product
preserving.
Proof. Suppose T is norm preserving. Then kT xk = kxk. Now
use the Polarization Identity on hT x, T yi:
4 hT x, T yi =
=
=
=
kT x + T yk2 − kT x − T yk2
kT (x + y)k2 − kT (x − y)k2
kx + yk2 − kx − yk2
4 hx, yi .
The other direction is even easier: if T is inner product preserving,
then
kT xk2 = hT x, T xi = hx, xi = kxk2 .
(b) Prove that such a linear transformation is 1-1, and T −1 is norm
preserving (and inner product preserving).
Proof. Suppose T is norm preserving, and suppose T x = T y.
Then 0 = kT x − T yk = kT (x − y)k = kx − yk, so x = y, and T
5
is 1-1. Thus we can talk about its inverse. Use the fact that T is
norm preserving:
kxk = T T −1 x = T −1 x ,
to see that T −1 is norm preserving.
8. If T : Rm −→ Rn is a linear transformation, show that there is a
number M such that kT (h)k ≤ Mkhk for all h ∈ Rm . Hint: Estimate
kT (h)k in terms of khk and the entries in the matrix for T .
Proof. Let A = (aij ) be the matrix for T with respect to the standard
bases for Rm and Rn . Also write h = (h1 , . . . , hm ) with respect to the
standard basis. Let y = T (h) in Rn , and let ri denote the ith row of
A. Then y i = hri , hi. By Cauchy-Schwarz, |y i| ≤ kri k · khk.
Now look at kT (h)k:
kT (h)k =
≤
=
(y 1 )2 + · · · + (y n )2
1/2
(kr1 k · khk)2 + · · · + (krn k · khk)2
1/2
kr1 k2 + · · · + krn k2
khk
1/2
This completes the proof with
M = kr1 k2 + · · · + krn k
2 1/2
=
X
i,j
(aij )2
!1/2
.
m
9. If x, y ∈ Rn , and
pz, w ∈ R , show that h(x, z), (y, w)i = hx, yi + hz, wi,
and k(x, z)k = kxk2 + kzk2 . Note that (x, z) and (y, w) denote points
in Rn+m .
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Proof.
h(x, z), (y, w)i = (x1 , . . . , xn , z 1 , . . . , z m ), (y 1, . . . , y n , w 1, . . . , w m )
= x1 y 1 + · · · xn y n + z 1 w 1 + · · · + z m w m
= x1 y 1 + · · · xn y n + z 1 w 1 + · · · + z m w m
= hx, yi + hz, wi .
p
h(x, z), (x, z)i
p
=
hx, xi + hz, zi
p
kxk2 + kzk2 .
=
k(x, z)k =
10. If x, y ∈ Rn , then x and y are called perpendicular (or orthogonal),
and we write x ⊥ y, if hx, yi = 0. If x ⊥ y, prove that kx + yk2 =
kxk2 + kyk2.
Proof.
kx + yk2 =
=
=
=
=
=
hx + y, x + yi
hx, x + yi + hy, x + yi
hx, xi + hx, yi + hy, xi + hy, yi
hx, xi + 2 hx, yi + hy, yi
hx, xi + 2(0) + hy, yi
kxk2 + kyk2 .
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