HW1_p71-3- 7-14 (Deadline_30-1 Erwin Kreyszig)

MATH 4010 (2014-15) F unctional Analysis CUHK Suggested Solution to Homework 1 Yu Mei † P70, 3. In ∞ , let Y be the subset of all sequences with only finite ly many nonze ro terms. Sho w that Y is a subspace of ∞ but not a closed subspace. Proof. (1) Let x = {ξ j }, y = {η j } be any two elements in Y ⊂ ∞ . Then there exist N ∈ N + such that ξ j = η j = 0, ∀ j ≥ N, otherwise x, y has infinitely many nonzer o terms . Moreo ver , for any j , |ξ j | ≤ C x and |η j | ≤ C y for some nonnegetive constants C x ,C y since x, y ∈ ∞ . Hence, for any α, β ∈ R, αξ j + βη j = 0 , ∀ j ≥ N ; |αξ j + βη j | ≤ |α|C x + |β |C y , ∀ j ∈ N + , which implies that αx + βy ∈ Y . So, Y is a subspace of ∞ . (2) Y is not a close d subspace. F or example , let x n be a sequence such that x (n) j = 1/j, j ≤ n, 0, j > n. i.e. x n == {1, ···, 1 n , 0, ···}. It is clear that x n ∈ Y . Set x be a sequence in ∞ such that x j = 1 j . Thus, x n − x ∞ = 1 n + 1 → 0, as n → +∞. But x / ∈ Y since it has infinitely many nonzero terms. P71, 7. Show that convergence of y 1 + y 2 + y 3 + ··· may not imply convergence of y 1 + y 2 + y 3 + ···. Proof. Consider Y in the above problem. Set y n = {η (n) j } ∈ Y to be a sequence with η (n) n = 1/n 2 , η (n) j = 0, for all j = n. Then, for any n ∈ N + , y n = 1/n 2 which implies that ∞ n=1 y n = ∞ n=1 1/n 2 < +∞. Set y = {1, 1/2 2 , ···, 1/n 2 , ···}. Then, n j =1 y j − y ∞ = 1 (n + 1) 2 → 0, as n → +∞. But y has infinitely many nonzero terms, i.e. ∞ n=1 y n / ∈ Y . So, ∞ n=1 y n does not converge in Y . P71, 14, Let Y be a closed subspace of a normed space (X, · ). Show that a norm · 0 on X/ Y is defined by ˆ x 0 = inf x∈ˆ x x where ˆ x ∈ X/ Y , that is, ˆ x is any coset of Y . Proof. Recall that X/ Y = { ˆ x| ˆ x = x + Y , x ∈ X } and algebraic operations in X/Y are defined as: αˆ x = αx + Y ; ˆ x + ˆ z = x + z + Y. † Email address: [email protected]. (Any questions are welcome!) 1

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Transcript of HW1_p71-3- 7-14 (Deadline_30-1 Erwin Kreyszig)

MATH 4010 (2014-15) Functional Analysis CUHK

Suggested Solution to Homework 1

Yu Mei

P70, 3. In `, let Y be the subset of all sequences with only finitely many nonzero terms. Show that Y is asubspace of ` but not a closed subspace.

Proof.

(1) Let x = {j}, y = {j} be any two elements in Y `. Then there exist N N+ such thatj = j = 0, j N,

otherwise x, y has infinitely many nonzero terms. Moreover, for any j, |j | Cx and |j | Cy for somenonnegetive constants Cx, Cy since x, y `. Hence, for any , R,

j + j = 0, j N ; |j + j | ||Cx + ||Cy, j N+,which implies that x+ y Y. So, Y is a subspace of `.

(2) Y is not a closed subspace. For example, let xn be a sequence such that

x(n)j =

{1/j, j n,0, j > n.

i.e. xn == {1, , 1n , 0, }. It is clear that xn Y . Set x be a sequence in ` such that xj = 1j . Thus,

xn x` = 1n+ 1

0, as n +.

But x / Y since it has infinitely many nonzero terms.

P71, 7. Show that convergence of y1+ y2+ y3+ may not imply convergence of y1 + y2 + y3 + .Proof. Consider Y in the above problem. Set yn = {(n)j } Y to be a sequence with

(n)n = 1/n2,

(n)j = 0, for all j 6= n.

Then, for any n N+, yn = 1/n2 which implies that

n=1 yn =

n=1 1/n2 < +.

Set y = {1, 1/22, , 1/n2, }. Then,

nj=1

yj y` = 1(n+ 1)2

0, as n +.

But y has infinitely many nonzero terms, i.e.

n=1 yn / Y. So,

n=1 yn does not converge in Y .

P71, 14, Let Y be a closed subspace of a normed space (X, ). Show that a norm 0 on X/Y is definedby

x0 = infxxx

where x X/Y , that is, x is any coset of Y .Proof. Recall that X/Y = {x|x = x+ Y, x X} and algebraic operations in X/Y are defined as:

x = x+ Y ; x+ z = x+ z + Y.

Email address: [email protected]. (Any questions are welcome!)

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MATH 4010 (2014-15) Functional Analysis CUHK

(1) x0 = infxxx = inf

yYx+ y 0, since is a norm on X.

(2) On the one hand, 0 = Y yields that

00 = infyYy = 0 = 0.

On the other hand, if x0 = 0, then

infyYx+ y = inf

yYx y = 0

which implies x Y . Since Y is a closed subspace, then Y = Y . Thus, x Y so that x = 0.Hence x0 = 0 if and only if x = 0

(3) If = 0, then x0 = infyY0x+ y = 0. For any R {0}, it holds that

x0 = infyYx+ y = inf

yY(x+ y

) = || inf

yYx+ y

= || inf

yYx+ y = ||x0,

since Y is a subspace which yields that infyYx+ y = infyY x+ y.

(4) For any x, z X/Y , by the definition of infimum, for any > 0, there exist y1, y2 Y such that

x+ y1 x0 + , z + y2 z0 + .

Thus,x+ z + y1 + y2 x+ y1+ z + y2 x0 + z0 + 2.

which implies thatx+ z0 = inf

yYx+ z + y x0 + z0 + 2.

Since is arbitrary, it holds thatx+ z0 x0 + z0.

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