Math 361: Homework 1 Solutions

January 31, 2014

Math 361: Homework 1 Solutions

1. We say that two norms | ? | and | ? | on a vector space V are equivalent or comparable if the topology they define on V are the same, i.e., for any sequence of vectors {xk} and x in V ,

lim |xk - x| = 0 if and only if lim |xk - x| = 0.

k

k

(a) Show that | ? | and | ? | are equivalent if and only if there exist C1, C2 > 0 such that

C1|x| |x| C2|x|

for any x V .

"=" We are given that | ? | and | ? | are equivalent, and we want to show that this implies that C1, C2 > 0 such that C1|x| |x| C2|x| .

Note that we can switch the definitions of the norms and get that C3|x| |x| C4|x|, which

means that if we prove there exists C1|x| |x|, we will also have proven that there exists C3|x|

|x| .

This implies (with C2 =

1 C3

)

that

there

exist

C1, C2

>

0

such

that

C1|x|

|x| C2|x| .

Therefore, we only need to show that one of the constants exists.

We do this by contradiction. Suppose that C > 0, x V such that C|x| > |x| . For each such C, |x| is a constant and hence we can say that C, x V such that:

1

1

x

C|x| > |x| = C|x| > |x| = C >

(1)

|x|

|x|

|x|

We

construct

a

sequence

of

vectors

{xk }

by

picking

the

appropriate

x

so that

for

C

=

1 k

,

C

>

x |x|

.

In

other

words,

for

arbitrary

k,

we

choose

C

=

1 k

and xk such that

1 k

>

|xk |

and

|xk |

=

1

(

x |x|

is

a unit vector for each k under | ? |).

Now,

since

limk

1 k

= 0 and

1 k

> |xk|

for each k, we deduce that limk |xk|

0.

Since

the norm is always positive, we know that limk |xk| = 0. As for the limit of the other norm,

because |xk| = 1 for each k, we can conclude that limk |xk| = 1.

Because the two norms are equivalent, we know that limk |xk -x| = 0 limk |xk -x| = 0 for any sequence of vectors {xk} in V . But for our sequence we just saw that limk |xk -0| = 0 and limk |xk - 0| = 1.

Therefore, we have a contradiction and we have shown that there must exist a C such that

C|x| < |x| , and by our earlier logic this is sufficient to prove the claim.

"=" We are given that C1, C2 > 0 such that C1|x| |x| C2|x| , and we want to show that this implies that | ? | and | ? | are equivalent.

We want to prove that limk |xk - x| = 0 = limk |xk - x| = 0 for any sequence of vectors {xk} in V . Note that this will prove the entire if and only if statement because we can just switch the definition of the norms to give us the other direction.

So, we begin by assuming that for any {xk}, limk |xk - x| = 0. This means that > 0, there exists an N such that if k > N , |xk - x| < . We want to show that > 0, there exists an N such that if k > N , |xk - x| < . Next, we observe that xk - x is just some vector in V for any k. We are given that for any x V , C1|x| < |x|. For each k, then, we get some C1,k so that C1,k|xk - x| |xk - x|.

1

Now we fix , and then we set k = C1,k for each k. We get N such that for k > N , |xk - x| < k = C1,k. Now we have for N = N :

C1,k|xk - x| |xk - x| <

C1,k

=

|xk - x|

<

C1,k C1,k

=

|xk - x|

<

(2)

Therefore, we have shown that for k large enough, the value |xk - x| will approach 0. This means that limk |xk - x| = 0 = limk |xk - x| = 0, and by our logic earlier, we have hence proven the if and only if statement.

(b) Show that any norm on Rn is equivalent to the Euclidean norm | ? |E. Thus all norms on a finite dimensional vector space are equivalent, which means topologically there is no difference. (Hint:

for any other norm |?|, first show |?| C|?|E for some C > 0, which implies the function f (x) = |x| is continuous with respect to the topology on Rn induced by the Euclidean norm. Recall that the unit ball in the Euclidean space is compact and verify the inequality in the other direction.)

We begin by constructing the standard basis ei for Rn, and letting | ? | be any norm on Rn. We can write any x = (x1, x2, . . . , xn) Rn as a linear combination of the standard basis elements:

n

x = xiei

(1)

i=1

Now we try to obtain an upper bound for |x|, the | ? | norm of x:

n

n

n

|x| = xiei |xiei| = |xi||ei|

(2)

i=1

i=1

i=1

Note that we used the triangle inequality of | ? | in (2). We make the expression in (2) bigger using the Cauchy-Schwartz inequality:

n

n

n

n

|x| |xi||ei|

|xi|2

|ei|2 = |x|E

|ei|2

(3)

i=1

i=1

i=1

i=1

Notice that the term multiplying |x|E in (3) is a constant, so we define C2 =

n i=1

|ei

|2.

Therefore, we shown that |x| C2|x|E. We claim that this implies that f (x) = |x| is continuous

with respect to the topology on Rn induced by the Euclidean norm.

To show this, we need to prove that for all > 0 there exists > 0 such that |x - y|E < = |f (x) - f (y)|E < . Fix and let = C2 .

|f (x) - f (y)|E = ||x| - |y||E |x - y| C2|x - y|E

C2 ? C2 =

(def. of f )

(4)

(reverse triangle inequality)

(5)

(|x| C2|x|E)

(6)

(def. of )

(7)

Therefore, f (x) = |x| is continuous with respect to the Euclidean norm on Rn. We also know that the unit ball in Rn is compact--that is, closed and bounded by the Heine-Borel theorem. We let B be the unit ball, i.e., B = {x Rn : |x|E = 1}.

2

Since | ? | is continuous on Rn and B is compact, it must be the case that for all y B there exists

p B such that |y| |p|. Now, we can scale any x Rn using the Euclidean norm so that they

appear within B:

x Rn

=

x |x|E

B

=

x |p|

|x|E

(8)

For a given x, |x|E is just a constant, so if we look at

x |x|E

, we can just move the constant outside

the norm:

x

1

|x|E

=

|x| |x|E

=

|x| |x|E ? |p|

(9)

Finally, taking C1 = |p|, we have that for any x Rn, |x| C1|x|E. Together with our earlier upper bound for |x|:

C1|x|E |x| C2|x|E

(10)

Based on our work in part (a) of the question, this implies that | ? | is equivalent to | ? |E, i.e., any norm in Rn is equivalent to the Euclidean norm.

(c) Consider the norms

1

|f |L1 = |f (t)|dt and |f |C0 = max {|f (t)|}

0

t[0,1]

on the space of C0([0, 1]) of continuous functions f : [0, 1] R. Show that the two norms are not equivalent. (Note that C0([0, 1]) is a vector space of infinite dimension.)

Consider the following sequence of functions {f (t)k} defined by f (t)k = tk. Consider the following two limits:

lim |f (t)k|L1 = lim

k

k

1

|tk|dt = lim

0

k

1

tkdt = lim

tk+1

1

=

lim

1 =0

0

k k + 1 0 k k + 1

(1)

lim |f (t)k|C0 = lim max {|tk|} = lim 1 = 1

(2)

k

k t[0,1]

k

However, these limits imply that for x = 0, limk |f (t)k - x|L0 = limk |f (t)k - x|C0 , which means by the definition that the two norms are not equivalent.

2. Solve Pugh's Chapter 5, Problem #9, 15, 16, 18.

9) Give an example of two 2 ? 2 matrices such that the norm of the products is less than the product of the norms. Consider the following two matrices:

A=

1 0

0 0

B=

0 0

0 1

(1)

First, let us calculate A . Let x R2.

A = sup |Ax| = sup

|x|1

|x|1|

1 0

0 0

x1 x2

= sup

|x|1

x1 0

= sup

|x|1

x12 = sup x1 = 1

|x|1

(2)

Note that if x = (1, 0), then |x| = 1 and we are restricted to vectors for which |x| 1.

3

Now, let us calculate B . Let x R2.

00

B = sup |Bx| = sup

|x|1

|x|1|

0

1

x1 x2

0

= sup

|x|1

x2

= sup

|x|1

x22 = sup x2 = 1

|x|1

(3)

Note that if x = (0, 1), then |x| = 1 and we are restricted to vectors for which |x| 1.

Now that we have A and B , we can calculate the product of the norms:

A ? B =1?1=1

(4)

Next, we find the norm of the product. First we must find what A ? B is:

A?B =

1 0

0 0

?

0 0

0 1

=

0 0

0 0

(5)

Now, we finally calculate A ? B . Let x R2.

00

A?B

= sup |(A ? B)x| = sup

|x|1

|x|1|

0

0

x1 x2

0

= sup

|x|1

0

= sup 0 = 0

|x|1

(6)

According to (4) and (6), we have found two 2 ? 2 matrices A and B such that A ? B < A ? B .

15) Show that the partial derivatives of the function

xy

f (x, y) = x2 + y2

if (x, y) = (0, 0)

0 if (x, y) = (0, 0)

exist at the origin, but the function is not differentiable there. First we show that the partial derivatives exist at (0, 0):

f (0, 0) =

lim

f (h, 0) - f (0, 0)

=

lim

h?0 h2 +0

-0

=

lim

0

=

lim 0 = 0

(1)

x

h0

h

h0

h

h0 h h0

f (0, 0) =

lim

f (0, h) - f (0, 0)

=

lim

0?h 0+h2

-0

=

lim

0

=

lim 0 = 0

(2)

y

h0

h

h0

h

h0 h h0

That is, the partial derivatives exist at the origin and both are equal to 0 there. Now we must show

that, despite this, f is not differentiable at the origin.

Assume that f is differentiable. Let h = (h1, h2). For some linear transformation A of R2 into R, since f is differentiable:

|f (x + h) - f (x - Ah)|

lim

=0

(3)

h0

|h|

Since we assumed f is differentiable at (0, 0), A must be equal to the matrix determined by its partial derivatives in (1) and (2).

A(0,0) =

f x

f y

=

(0,0)

0

0

(4)

Acting A(0,0) on h, we get:

A(0,0)h = 0

0

?

h1 h2

=0

(5)

4

Plugging in x = (0, 0) and (5) into (3):

lim |f (0 + h1, 0 + h2) - f (0, 0) - 0| = lim

h0

| h12 + h22|

h0 |

| | h1h2

h1 2 +h2 2

h12 + h22|

=

lim

h0

(h12

h1h2 + h22)3/2

(6)

According to (3), the limit in (6) has to equal 0 from any direction. We choose h = (t, t) (approaching along the line y = x) and get:

lim

h0

h1h2 (h12 + h22)3/2

t2 = lim

t0 23/2t3

1 = lim

t0 22/3t

=0

(7)

Therefore, our assumption of the limit in (3) existing and equaling zero was wrong, which means that despite the fact that the partial derivatives of f exist at the origin, f is not differentiable there.

16) Let f : R2 R3 and g : R3 R be defined by f = (x, y, z) and g = w where

w = w(x, y, z) = xy + yz + zx x = x(s, t) = st

y = y(s, t) = s cos t

z = z(s, t) = s sin t

(a) Find the matrices that represent the linear transformations (Df )p and (Dg)q where p = (s0, t0) = (0, 1) and q = f (p).

First, note that f (s, t) = (x, y, z) = (st, s cos t, s sin t). Now, let us find the partial derivatives Djfi of f :

D1f1

=

fx s

=

x s

=t

D1f2

=

fy s

=

y s

= cos t

D1f3

=

fz s

=

z s

= sin t

D2f1

=

fx t

=

x t

=

s

(1)

D2f2

=

fy t

=

y t

= -s sin t

(2)

D2f3

=

fz t

=

z t

=

s cos t

(3)

According to theorem 9.17 in Rudin, we construct Df and then evaluate at (0, 1):

D1f1 D2f1 t

s

1 0

(Df )(0,1) = D1f2 D2f2 cos t -s sin t = cos(1) 0

(4)

D1f3 D2f3 (0,1) sin t

s cos t

(0,1)

sin(1) 0

Next, we note that g(x, y, z) = w = xy + yz + zx. Now we proceed similarly to find the partial derivatives Djgi of g:

g

D1g = x = y + z

(5)

g

D2g = y = x + z

(6)

g

D3g = z = y + x

(7)

As before, we construct Dg and evaluate at q = f (p) = f (0, 1) = (0 ? 1, 0 ? cos(1), 0 ? sin(1)) = (0, 0, 0):

(Df )(0,0,0) =

g x

g y

g z

=

(0,0,0)

y+z

x+z

y + x (0,0,0) = 0

0

0

(8)

Therefore, we have constructed the two matrices (Df )(0,1) and (Dg)(0,0,0).

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