Quotient Space

本文详细介绍了商空间的概念,包括其元素、运算和范数的定义,并通过几个关键定理探讨了商空间成为巴拿赫空间的条件。此外,还讨论了在巴拿赫空间中构造商空间的方法。

Quotient Space

Suppose that XX is a normed space, and Y is a closed subspace of XX, we can define an equivalent relation on X by x1x2x1∼x2 if and only if x1x2Yx1−x2∈Y.

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

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

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

The elements

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

The Operators

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

(x1+Y)+(x2+Y)=(x1+x2)+Yα(X+Y)=(αx)+Y.(x1+Y)+(x2+Y)=(x1+x2)+Yα(X+Y)=(αx)+Y.

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

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

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

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

THEOREM 1

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

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

Let x1+Y,x2+YX/Yx1+Y,x2+Y∈X/Y. Want to show ||(x1+Y)+(x2+Y)||||x1+Y||+||x2+Y||||(x1+Y)+(x2+Y)||≤||x1+Y||+||x2+Y||.

Let ϵ>0ϵ>0 be given, the by the definition we can find y1,y2Yy1,y2∈Y such that

||x1+y1||||x1+Y||+ϵ,||x2+y2||||x2+Y||+ϵ.||x1+y1||≤||x1+Y||+ϵ,||x2+y2||≤||x2+Y||+ϵ.

Then ||(x1+Y)+(x2+Y)||||x1+y1||+||x2+y2||||(x1+Y)+(x2+Y)||+2ϵ||(x1+Y)+(x2+Y)||≤||x1+y1||+||x2+y2||≤||(x1+Y)+(x2+Y)||+2ϵ. Finally, we can let ϵ0ϵ→0.

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

THEOREM 2

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

we have n=1xn∑n=1∞xn converges in XX. Which means, limkn=1kxnX exists.

Proof: If XX is complete and n=1||xn||, define yk=kn=1xnXyk=∑n=1kxn∈X. Then for k<jk<j

||ykyj||n=k+1j||xn||0,ask.||yk−yj||≤∑n=k+1j||xn||→0,ask→∞.

So we can see that ykyk is a Cauchy sequence in XX. Hence converges.

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

||yknykn+1||12n,n1.||ykn−ykn+1||≤12n,∀n≥1.

Then it is easy to see that n=1||yknykn+1||<∑n=1∞||ykn−ykn+1||<∞. By the assumption, we can conclude that limjjn=1(yknykn+1)limj→∞∑n=1j(ykn−ykn+1) exists in XX. Say the limit is y=limjn=1j(yknykn+1), then
limj(yk1ykj+1)=y.limj→∞(yk1−ykj+1)=y.

So we get
limjykj+1=yk1y.limj→∞ykj+1=yk1−y.

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

THEOREM 3

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

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

n=1||xn+Y||<.∑n=1∞||xn+Y||<∞.

We want to show that limk(xn+Y)limk→∞(xn+Y) exists in X/YX/Y.

By the definition of the quotient norm, for any nn we can find ynY such that

||xn+yn||||xn+Y||+12n.||xn+yn||≤||xn+Y||+12n.

So we have
n=1||xn+yn||n=1||xn+Y||+1<.∑n=1∞||xn+yn||≤∑n=1∞||xn+Y||+1<∞.

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

Now, since

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

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

Implies that
||(n=1k(xnz)+Y||||(n=1k(xnz)+n=1kyn||0.||(∑n=1k(xn−z)+Y||≤||(∑n=1k(xn−z)+∑n=1kyn||→0.

So we are done as kk→∞.
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