Quotient Space

Quotient Space

Suppose that X X is a normed space, and Y Y is a closed subspace of X X , we can define an equivalent relation on X X by x1∼x2 x 1 ∼ x 2 if and only if x1−x2∈Y x 1 − x 2 ∈ Y .

Notation: for any x∈X x ∈ X , the set x+Y x + Y is denoted by x+Y={x+y:y∈Y} x + Y = { x + y : y ∈ Y } .

Note that in this notation, for any x∈X x ∈ X the equivalence class under ∼ ∼ of x x is x+Y x + Y .

A very natural question is that what is the elements, operator and norm in X/Y X / Y ?

The elements

Let X/Y X / Y be the set of equivalence classes, which is X/Y={x+Y:x∈X} X / Y = { x + Y : x ∈ X } .

The Operators

Define addition and scalar multiplication on X/Y X / Y by

(x1+Y)+(x2+Y)=(x1+x2)+Yα(X+Y)=(αx)+Y. ( x 1 + Y ) + ( x 2 + Y ) = ( x 1 + x 2 ) + Y α ( X + Y ) = ( α x ) + Y .

This operators are well defined and make X/Y X / Y a vector space.

The Norm

Also, we define the norm of X/Y X / Y be

||x+Y||X/Y=inf{||x+y||X:y∈Y}. | | x + Y | | X / Y = inf { | | x + y | | X : y ∈ Y } .

Then next result claims that ||⋅|| | | ⋅ | | is well defined and is a norm on X/Y X / Y . It is called the quotient norm.

THEOREM 1

||⋅|| | | ⋅ | | defined above is a norm on X/Y X / Y .

Proof: We only chech the trianglar inequality for the quotient norm.

Let x1+Y,x2+Y∈X/Y x 1 + Y , x 2 + Y ∈ X / Y . Want to show ||(x1+Y)+(x2+Y)||≤||x1+Y||+||x2+Y|| | | ( x 1 + Y ) + ( x 2 + Y ) | | ≤ | | x 1 + Y | | + | | x 2 + Y | | .

Let ϵ>0 ϵ > 0 be given, the by the definition we can find y1,y2∈Y y 1 , y 2 ∈ Y such that

||x1+y1||≤||x1+Y||+ϵ,||x2+y2||≤||x2+Y||+ϵ. | | x 1 + y 1 | | ≤ | | x 1 + Y | | + ϵ , | | x 2 + y 2 | | ≤ | | x 2 + Y | | + ϵ .

Then ||(x1+Y)+(x2+Y)||≤||x1+y1||+||x2+y2||≤||(x1+Y)+(x2+Y)||+2ϵ | | ( x 1 + Y ) + ( x 2 + Y ) | | ≤ | | x 1 + y 1 | | + | | x 2 + y 2 | | ≤ | | ( x 1 + Y ) + ( x 2 + Y ) | | + 2 ϵ . Finally, we can let ϵ→0 ϵ → 0 .

The next theorem gives a sufficient and necessary condition for a normed vector space to become a Banach space.

THEOREM 2

X X is a normed space, then X X is a Banach space if and only if ∀(xn)⊂X ∀ ( x n ) ⊂ X , such that ∑∞n=1||xn||≤∞ ∑ n = 1 ∞ | | x n | | ≤ ∞

we have ∑∞n=1xn ∑ n = 1 ∞ x n converges in X X . Which means, limk→∞∑kn=1xn∈X lim k → ∞ ∑ n = 1 k x n ∈ X exists.

Proof: If X X is complete and ∑∞n=1||xn||≤∞ ∑ n = 1 ∞ | | x n | | ≤ ∞ , define yk=∑kn=1xn∈X y k = ∑ n = 1 k x n ∈ X . Then for k<j k < j

||yk−yj||≤∑n=k+1j||xn||→0,ask→∞. | | y k − y j | | ≤ ∑ n = k + 1 j | | x n | | → 0 , a s k → ∞ .

So we can see that yk y k is a Cauchy sequence in X X . Hence converges.

On the other hand, let (yk) ( y k ) be a Cauchy sequence in X X and we choose a subsequence (ykn) ( y k n ) such that

||ykn−ykn+1||≤12n,∀n≥1. | | y k n − y k n + 1 | | ≤ 1 2 n , ∀ n ≥ 1.

Then it is easy to see that ∑∞n=1||ykn−ykn+1||<∞ ∑ n = 1 ∞ | | y k n − y k n + 1 | | < ∞ . By the assumption, we can conclude that limj→∞∑jn=1(ykn−ykn+1) lim j → ∞ ∑ n = 1 j ( y k n − y k n + 1 ) exists in X X . Say the limit is y=limj→∞∑jn=1(ykn−ykn+1) y = lim j → ∞ ∑ n = 1 j ( y k n − y k n + 1 ) , then

limj→∞(yk1−ykj+1)=y. lim j → ∞ ( y k 1 − y k j + 1 ) = y .

So we get

limj→∞ykj+1=yk1−y. lim j → ∞ y k j + 1 = y k 1 − y .

Since yk y k is Cauchy and it has a subsequence which converges to a element in X X , then the whole sequence will converges to that element. So X X is complete hence a Banach space.

THEOREM 3

Suppose X X is a Banach space and Y Y is a closed subspace of X X , the the quotient space X/Y X / Y is also a Banach space.

Proof: We will use the previous theorem to show this result. Let (xn+Y)⊂X/Y ( x n + Y ) ⊂ X / Y be such that

∑n=1∞||xn+Y||<∞. ∑ n = 1 ∞ | | x n + Y | | < ∞ .

We want to show that limk→∞(xn+Y) lim k → ∞ ( x n + Y ) exists in X/Y X / Y .

By the definition of the quotient norm, for any n n we can find yn∈Y y n ∈ Y such that

||xn+yn||≤||xn+Y||+12n. | | x n + y n | | ≤ | | x n + Y | | + 1 2 n .

So we have

∑n=1∞||xn+yn||≤∑n=1∞||xn+Y||+1<∞. ∑ n = 1 ∞ | | x n + y n | | ≤ ∑ n = 1 ∞ | | x n + Y | | + 1 < ∞ .

Since X X is a Banach space which is complete by assumption, by Theorem 2 we obtain that ∑∞n=1(xn+yn) ∑ n = 1 ∞ ( x n + y n ) converges to some z∈X z ∈ X . Next, we only have to show that limk→∞∑kn=1(xn+Y)=z+Y lim k → ∞ ∑ n = 1 k ( x n + Y ) = z + Y . Then we can complete the proof.

Now, since

||∑n=1k(xn+Y)−(z+Y)||=||(∑n=1k(xn−z)+Y||. | | ∑ n = 1 k ( x n + Y ) − ( z + Y ) | | = | | ( ∑ n = 1 k ( x n − z ) + Y | | .

But

||∑n=1k(xn+yn)−z||→0, | | ∑ n = 1 k ( x n + y n ) − z | | → 0 ,

Implies that

||(∑n=1k(xn−z)+Y||≤||(∑n=1k(xn−z)+∑n=1kyn||→0. | | ( ∑ n = 1 k ( x n − z ) + Y | | ≤ | | ( ∑ n = 1 k ( x n − z ) + ∑ n = 1 k y n | | → 0.

So we are done as k→∞ k → ∞ .