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MATH 675 HOMEWORK 5SOLUTIONS

Exercise 1. Show that the operator norm satises all of the properties of a norm.

Let T L(X, Y ). Then T = supx =0T x Y x X

. We have to prove three things:

(a) T 0 and T = 0 if and only if T = 0 . Since for all x, T x Y and x X are nonnegative so is each term in the above supremum and hence so is T .T = 0 means that 0 supx =0 T x / x = 0 so that T x = 0 for all x = 0.This means that T x = 0 for all x = 0 and since T (0) = 0 anyway, T = 0. If T = 0 then T x = 0 for all x and hence sup x =0 T x / x = T = 0.(b)

T =

|

|T .

T = supx =0

(T )x / x = supx =0

(T x) / x = || supx =0 T x / x = || T .

(c) T 1 + T 2 T 1 + T 2 for all T 1, T 2 L(X, Y ). Given x X ,(T 1 + T 2)x Y = T 1x + T 2x Y

T 1x Y + T 2x Y T 1 x X + T 2 x X = ( T 1 + T 2 ) x X .

Since T 1 + T 2 is the smallest such constant T 1 + T 2 T 1 + T 2 .Exercise 2. Determine whether or not each of the given normed linear spaces is a Banachspace, that is, is complete with respect to the metric induced by the norm. If it is, prove it,and if it is not, prove that it is not by nding a Cauchy sequence in the space that does notconverge to an element of the space.

(a) = {{xn }n =1 C : supn

N |xn | < } equipped with the norm {xn } = sup n |xn |.

is a Banach space. To see this, let {s n }n =1 be a Cauchy sequence in l.Then for each n, sn = {xnk }k =1 and for every > 0 there is an N such thatif n, m N then s n sm = sup k |xmk x

nk | < . Now for each xed k,we have that for n,m > N , |xmk xnk | < . This means that for each k thesequence {x

nk }n =1 is a Cauchy sequence of numbers hence convergent. Denexk = lim n xnk , and let s = {xk }k =1 .

Now I will show that l is complete by showing that in fact sn s in thel norm, that is lim n s s n = 0. To see that this is true, let > 0

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and choose N so large that if n, m > N then sm s n < /2. For eachk, choose m = m(k) N so that |xk x

m (k )k | < /2. Then for each k,

|xk

xnk

| |xk

x

m (k )k

|+

|x

m (k )k

xnk

| |xk

x

m (k )k

|+ sm (k )

s n

< /2+ /2 = .

Since k was arbitrary and the only assumption made on n is that it wasgreater than N , we can take the supremum of the right side and concludethat if n > N then s s n < .It follows from this that in fact s l by choosing N so that s sN < 1and observing that s s sN + sN 1 + sN < infty .

(b) c0 = {{xn }n =1 C : limn xn = 0} equipped with the same norm as .c0 is a Banach space. To see this let {s n }n =1 be a Cauchy sequence in c0.Then the same argument as for part (a) shows that

{sn

}is in fact convergent

to s l. To show that c0 is complete it remains only to show that in thiscase the limit sequence s is in c0. To this end let > 0 and x N so largethat s sN < . Then for each xed k,

|xk | |xk xN k |+ |x

N k | s sN + |x

N k | < + |x

N k |.

Since sN c0, letting k on both sides of the inequality giveslim sup

k |xk | < + limk |x

N k | = .

Since > 0 was arbitrary, limsup k |xk | = 0 and hence lim k |xk | = 0.We conclude that s

c0.

(c) f = {{xn }n =1 C :N = N (x), xn = 0 n N } equipped with the same norm as .f is not a Banach space. To see this consider the sequence

sn = {1, 1/ 2, 1/ 3, . . . , 1/n, 0, 0, 0, . . .}.Then {s n } f and it is Cauchy since if n > m then

sn sm = {0, 0, . . . , 0, 1/m +1 , 1/m +2 , . . . , 1/n, 0, . . .} = 1m + 1

.

However {sn } does not converge to any element of f , for if we suppose thatit did converge to s = {xn } f then there is an N such that xn = 0 if n > N and s sn > 1/N for all n > N .Exercise 3. Show that for T : V 1 V 2 a linear operator on the normed linear spaces V 1and V 2,

T = supv 1 =1

T v 2 = supv 1 1

T v 2v 1

.

Here 1 denotes the norm in V 1 and 2 denotes the norm in V 2.

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Since T is dened as supv =0T v 2v 1

, and since the supremum of a quantity taken

over a small set is always less than the supremum of the same quantity takenover a larger set, we have that

supv 1 =1

T v 2 supv 1 1T v 2v 1 supv =0

T v 2v 1

= T .

Hence all that needs to be shown is that T sup v 1 =1 T v 2. To that end letv V 1 \ {0}. Then clearly x = v/ v 1 has norm 1 and by linearity of T ,T v 2v 1

= T vv 1 2

= T x 2 supv 1 =1 T v 2.

Taking the supremum of the left side over all such v gives the result.

Exercise 4. Let T : V 1 V 2 be an unbounded linear operator. Here 1 denotes thenorm in V 1 and 2 denotes the norm in V 2. Show the following.(a) There exists a sequence {xn } V 1 such that xn 1 = 1 for all n and T xn 2 .(b) There exists a sequence {xn } V 1 such that xn 1 0 and T xn 2 = 1 for all n.(c) There exists a sequence {xn } V 1 such that xn 1 0 and T xn 2 .

Note rst that the denition of a bounded linear operator asserts that T isbounded provided that there is a number C such that for all x V 1, T x 2 C x 1. The negation of this fact is the following: For all numbers C there is anx V 1 such that T x 2 > C x 1.Now taking C = n we can nd vn V 1 such that T vn 2 > n vn 1. Deningxn = vn / vn 1 we have that xn 1 = 1 and T xn 2 > n as n . Thisis (a).Taking the same sequence {xn } as above, dene yn =

x nT x n 2

. Then clearlyyn 1 < 1n 0 as n and T yn 2 = T (xn / T xn 2) 2 =

T x n 2T x n 2 = 1 for all

n. This is (b).

Taking again the same sequence as above, dene z n = 1 n xn . Then z n 1 =1 n 0 as n and T z n 2 = T (xn / n) 2 = 1 n T xn 2 > n asn . This is (c).