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Isometrically Isomorphic

Let

X=\{n^{-1}\mid n\ge 1\}\cup\{0\}

.
Let

C(X)=

all continuous functions

f:X\to\mathbb{F}\mid\|f\|\equiv\sup{\{|f(x)|\mid x\in X\}}<\infty

.
Let

c=

the set of all sequences

\{\alpha_{n}\}_{1}^{\infty}

\alpha_{n}\in\mathbb{F}

, such that

\lim{\alpha_{n}}

exists.

Show that

C(X)

and the space of

c

are isometrically isomorphic.'

So I said that

C(X)

and

c

are isometrically isomorphic if there is a surjective linear isometry from

c

onto

C(X)

.

Let

x

be in

c

. Define the function

f

C(X)

as follows:

f(\frac{1}{n})=x_{n}

\forall n

.
The image of

c

by this function is a subspace and the norm is conserved.
Therefore...

Is this right?

Anyone?

Original post by RamocitoMorales

Anyone?

What is F? If I were marking your assignment and you wrote that as an answer I would give you at most 1 mark. You should explicitly show that the norm is preserved. A trivial point, but you should probably show that the map is well defined, i.e. that it actually does take a point x in c to a function in C(X).

(edited 10 years ago)

Reply 3

Shillington

Original post by RamocitoMorales

Anyone?

Assuming that

\mathbb{F}

is an arbitrary field, the function you give works. In order to answer the question, you need to show that it works. i.e. you need to show the following (mostly one line answers):

The function is well defined (i.e. it actually does map from c to C(X) and each element of c has a single image).
The function is bijective.
Write out details of the line "the image of c by this function is a subspace and the norm is conserved".

All pretty straightforward.