# Normed space

• Mar 14th 2009, 11:01 AM
math8
Normed space
Let X be a normed space, F be a closed linear subspace of X,
Let z be in X, z is not in F.
Let S={x+az:a is in the field Phi}=Span of F and z
We show S is closed.

I would define a function f : S--> Phi by f(x+az)=a
and show that |f|< or equal to 1/d where d= distance(z,F)
hence f is in S* (dual of S).
So we let {x_n +a_n z} be a sequence converging to w (x_n is in F,a_n is in Phi). We show w is in S.
We note {f(x_n +a_n z)} is a cauchy sequence.

But I am not sure how to proceed.
• Mar 14th 2009, 01:58 PM
Opalg
Quote:

Originally Posted by math8
Let X be a normed space, F be a closed linear subspace of X,
Let z be in X, z is not in F.
Let S={x+az:a is in the field Phi}=Span of F and z
We show S is closed.

I would define a function f : S--> Phi by f(x+az)=a
and show that |f|< or equal to 1/d where d= distance(z,F)
hence f is in S* (dual of S).
So we let {x_n +a_n z} be a sequence converging to w (x_n is in F,a_n is in Phi). We show w is in S.
We note {f(x_n +a_n z)} is a cauchy sequence.

But I am not sure how to proceed.

That's a very good start. In fact, $(f(x_n +a_n z))$ is a Cauchy sequence of scalars, and since the scalar field (presumably $\mathbb{R}$ or $\mathbb{C}$) is complete, that sequence must converge to a scalar a. In other words, $a_n\to a$.

Then $x_n = w-a_nz\to w-az = y$ say, as n→∞. But F is closed, and therefore y∈F. Thus $w = y + az \in S$.
• Mar 14th 2009, 03:05 PM
math8
Thanks, this makes a lot of sense.

But how do I show that |f|< or equal to 1/d where d= distance(z,F)

I know it involves some corollary to the Hahn-Banach theorem:
" If Y is a linear subspace of the normed space X, x is in X and d=dist(x,Y)>0, then there exists x* in X* such that x*(x)=1 and |x*|=1/d and x*|Y=0 (x* restricted to Y is 0)".

I think maybe the reason we want this to be true is that in that case, f would be bounded hence continuous, thus f is in the dual S*.

But how does this help?
• Mar 14th 2009, 03:25 PM
Opalg
Quote:

Originally Posted by math8
Thanks, this makes a lot of sense.

But how do I show that |f|< or equal to 1/d where d= distance(z,F)

I know it involves some corollary to the Hahn-Banach theorem:
" If Y is a linear subspace of the normed space X, x is in X and d=dist(x,Y)>0, then there exists x* in X* such that x*(x)=1 and |x*|=1/d and x*|Y=0 (x* restricted to Y is 0)".

I think maybe the reason we want this to be true is that in that case, f would be bounded hence continuous, thus f is in the dual S*.

But how does this help?

Yes, that is exactly what you need. In fact, if you apply the H–B theorem as you have stated it, taking Y to be the subspace F and x to be the element z, then the functional x* satisfies x*|F=0 and x*(z)=1. So for each element v in F, x*(v+az) = a. In other words, x* is the functional f that you want (and you need it to be continuous because you want it to take a Cauchy sequence to a Cauchy sequence).
• Mar 14th 2009, 04:52 PM
math8
Thanks a lot that is very helpful :D