Quotient Space
Suppose that XX is a normed space, and is a closed subspace of XX, we can define an equivalent relation on by x1∼x2x1∼x2 if and only if x1−x2∈Yx1−x2∈Y.
Notation: for any x∈Xx∈X, the set x+Yx+Y is denoted by x+Y={x+y:y∈Y}x+Y={x+y:y∈Y}.
Note that in this notation, for any x∈Xx∈X the equivalence class under ∼∼ of xx is .
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:x∈X}X/Y={x+Y:x∈X}.
The Operators
Define addition and scalar multiplication on X/YX/Y by
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
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+Y∈X/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,y2∈Yy1,y2∈Y such that
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 ϵ→0ϵ→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 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, exists.
Proof: If XX is complete and , define yk=∑kn=1xn∈Xyk=∑n=1kxn∈X. Then for k<jk<j
So we can see that ykyk is a Cauchy sequence in XX. Hence converges.
On the other hand, let be a Cauchy sequence in XX and we choose a subsequence such that
Then it is easy to see that ∑∞n=1||ykn−ykn+1||<∞∑n=1∞||ykn−ykn+1||<∞. By the assumption, we can conclude that limj→∞∑jn=1(ykn−ykn+1)limj→∞∑n=1j(ykn−ykn+1) exists in XX. Say the limit is , then
So we get
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 is complete hence a Banach space.
THEOREM 3
Suppose XX is a Banach space and is a closed subspace of XX, the the quotient space 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
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 such that
So we have
Since XX is a Banach space which is complete by assumption, by Theorem 2 we obtain that converges to some z∈Xz∈X. Next, we only have to show that limk→∞∑kn=1(xn+Y)=z+Ylimk→∞∑n=1k(xn+Y)=z+Y. Then we can complete the proof.
Now, since
But
Implies that
So we are done as k→∞k→∞.