1. ## Schauder Basis

Show that if a Normed Space X has a Schauder Basis => X is Seperable.

Definition of a Schauder Basis:

If a Normed Space X contains a sequence (e_n) with the property that for every x element of X there exists:

unique sequence of scalars !(alpha_n) such that: ||x-(alpha_1.e_1 + alpha_2.e_2+....+alpha_ne_n|| ---> 0
as n-->infinity

then (e_n) is called a Schauder Basis for X.

2. Hint: Look at the set of finite linear combinations of the basis vectors with rational coefficients: $\Big\{\sum_{n=1}^N\alpha_ne_n\Bigr\}$, where the coefficients $\alpha_n$ are rational (if the scalar field is the real numbers), or have rational real and imaginary parts (if it is the complex numbers).

3. Hi Opalg. Thank you for you hint.
I know that if I take this set of finite linear combinations you proposed I can use it as a set, say called M. Then I must show that this set M is countable and this set is dense in X.

So it is easy to argue that this set is countable since there are finitre number of operations to multiply rational coefficients to basis vectors and then sum them.

But I must now take this set M and show it is dense in X by using the fact that rational numbers Q are dense in R (reals), but I do not know how to show this with norm estimates. So not sure how to write this in "analysis language".

4. Originally Posted by frater_cp
Hi Opalg. Thank you for you hint.
I know that if I take this set of finite linear combinations you proposed I can use it as a set, say called M. Then I must show that this set M is countable and this set is dense in X.

So it is easy to argue that this set is countable since there are finite number of operations to multiply rational coefficients to basis vectors and then sum them.

But I must now take this set M and show it is dense in X by using the fact that rational numbers Q are dense in R (reals), but I do not know how to show this with norm estimates. So not sure how to write this in "analysis language".

We can start by assuming that all the vectors in the Schauder basis have norm 1 (if not, replace them by suitable scalar multiples).

Given a vector x, and ε>0, we want to find an element of the countable set M that is within distance ε of x. We can find N, and scalars $\alpha_1,\ldots,\alpha_N$, so that $\Bigl\|x- \sum_{n=1}^N \alpha_ne_n\Bigr\| < \varepsilon/2$. Then, for n=1,2,...,N, we can find rational numbers $\beta_n$ such that $|\beta_n - \alpha_n|<2^{-(n+1)}$. Now (using the triangle inequality) check that $\Bigl\|\sum_{n=1}^N (\beta_n - \alpha_n)e_n\Bigr\| < \varepsilon/2$, and hence $\Bigl\|x- \sum_{n=1}^N \beta_ne_n\Bigr\| < \varepsilon$.

5. Hi Opalg.

Thank you again it is helping me a great deal!
I'm still confused not sure how to use the triangle inequality.

I know that ||x-y||<=||x-z|| + ||z-y||

I must use this but how to choose z?

Also why is difference between rational and real number < 2^-(n+1) ?

Im getting confused also between x which is infinite sum of alpha_n e_n

and y which is finite sum of beta_n e_n

6. Originally Posted by frater_cp
Hi Opalg.

Thank you again it is helping me a great deal!
I'm still confused not sure how to use the triangle inequality.

I know that ||x-y||<=||x-z|| + ||z-y||

I must use this but how to choose z?

Also why is difference between rational and real number < 2^-(n+1) ?

Im getting confused also between x which is infinite sum of alpha_n e_n

and y which is finite sum of beta_n e_n
Okay, so let $x = \sum_{n=1}^\infty \alpha_ne_n$, $y = \sum_{n=1}^N \alpha_ne_n$ and $z = \sum_{n=1}^N \beta_ne_n$. By choosing N large enough, we can ensure that $\|x-y\|<\varepsilon/2$. The series for y is finite, but its coefficients need not be rational. So we want to show that $\|y-z\|<\varepsilon/2$. Then the triangle inequality will show that $\|x-z\|<\varepsilon$. That will complete the proof, because the series for z has rational coefficients.

To show that $\|y-z\|<\varepsilon/2$ we use the triangle inequality again, but for N vectors instead of two:

\begin{aligned}\|y-z\| = \Bigl\|\sum_{n=1}^N (\alpha_n-\beta_n)e_n\Bigr\| &\leqslant \sum_{n=1}^N \|(\alpha_n-\beta_n)e_n\| \\ &= \sum_{n=1}^N |\alpha_n-\beta_n|\|e_n\| \leqslant \sum_{n=1}^N 2^{-(n+1)}\varepsilon < \varepsilon/2\end{aligned}
(the last inequality comes from summing a geometric series).

7. ## Tibi Gratulor

Has anyone recently told you that you are a G E N I U S!

Thank you very much have a wonderful day Opalg