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[SOLVED] Show Y is dense in X

Fermat

Active member
Nov 3, 2013
188
Let X be the subset of \(\displaystyle l_{1}\) consisting of those sequences \(\displaystyle (x_{n})\)such that

\(\displaystyle x_{1}+x_{3}+x_{5}+.....=x_{2}+x_{4}+x_{6}+....\)

Let Y be the linear span of \(\displaystyle {{e_{n}+e_{n+1}:n=1,2,....}}\) where \(\displaystyle e_{n}\) is the sequence with a 1 in the nth position, zeroes elsewhere.

You are given that X is a closed subspace, and you use the following result:
Let F be a subspace of E and x be in E. Suppose that for all maps f in the dual space of E satisfying f(y)=0 for all y in F, we have that f(x)=0. Then x is in the closure of Y.

Prove that the closure of Y is equal to X.

My work: I show Y is a subset of X: an aritrary member of Y is of the form:\(\displaystyle t_{1}(e_{1}+e_{2})+.....+t_{k}(e_{k}+e_{k+1})\) and its easy to show that the sum of the odd positioned terms is equal to the even positioned ones.

So since X is closed, the closure of Y is contained in X. Now I use result with X in place of E and Y in place of F. So let f in the dual of X satisfy f(y)=0 for all y in Y.
taking k=1, then k=2, and so in the above, we see this implies that \(\displaystyle f(e_{n}+e_{n+1})=0\) for all n. Can anyone help me finish to show f(x)=0 for any x in X?

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let X be the subset of \(\displaystyle l_{1}\) consisting of those sequences \(\displaystyle (x_{n})\)such that

\(\displaystyle x_{1}+x_{3}+x_{5}+.....=x_{2}+x_{4}+x_{6}+....\)

Let Y be the linear span of \(\displaystyle {{e_{n}+e_{n+1}:n=1,2,....}}\) where \(\displaystyle e_{n}\) is the sequence with a 1 in the nth position, zeroes elsewhere.

You are given that X is a closed subspace, and you use the following result:
Let F be a subspace of E and x be in E. Suppose that for all maps f in the dual space of E satisfying f(y)=0 for all y in F, we have that f(x)=0. Then x is in the closure of Y.

Prove that the closure of Y is equal to X.

My work: I show Y is a subset of X: an aritrary member of Y is of the form:\(\displaystyle t_{1}(e_{1}+e_{2})+.....+t_{k}(e_{k}+e_{k+1})\) and its easy to show that the sum of the odd positioned terms is equal to the even positioned ones.

So since X is closed, the closure of Y is contained in X. Now I use result with X in place of E and Y in place of F. So let f in the dual of X satisfy f(y)=0 for all y in Y.
taking k=1, then k=2, and so in the above, we see this implies that \(\displaystyle f(e_{n}+e_{n+1})=0\) for all n. Can anyone help me finish to show f(x)=0 for any x in X?

Thanks
The function $f$ vanishes on $x_1(e_1 + e_2)$. It also vanishes on $(x_2-x_1)(e_2+e_3)$, on $(x_3-x_2+x_1)(e_3+e_4)$, and on $(x_4-x_3+x_2 - x_1)(e_4+e_5)$, and so on. So it vanishes on the sum of the first $n$ vectors constructed in that way, and (by continuity) on the limit of those sums as $n\to\infty$.
 

Fermat

Active member
Nov 3, 2013
188
The function $f$ vanishes on $x_1(e_1 + e_2)$. It also vanishes on $(x_2-x_1)(e_2+e_3)$, on $(x_3-x_2+x_1)(e_3+e_4)$, and on $(x_4-x_3+x_2 - x_1)(e_4+e_5)$, and so on. So it vanishes on the sum of the first $n$ vectors constructed in that way, and (by continuity) on the limit of those sums as $n\to\infty$.
Thanks opalg, I think I reasoned this way but rejected it because it doesn't use ther fact that x in X satisfies\(\displaystyle x_{1}+x_{3}+....=x_{2}+x_{4}+.....\) so are you not effectively proving Y is dense in \(\displaystyle l_1\)?
 

Fermat

Active member
Nov 3, 2013
188
Do you have an answer to that? (I'm not being impatient but looking at my response you might mistake it for 'I'm satisfied, thanks')
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Thanks opalg, I think I reasoned this way but rejected it because it doesn't use ther fact that x in X satisfies \(\displaystyle x_{1}+x_{3}+....=x_{2}+x_{4}+.....\) so are you not effectively proving Y is dense in \(\displaystyle l_1\)?
That fact comes in at the end, when you have to use the continuity of $f$ to deduce that $f(x)=0$. At that stage, you will have found an element \(\displaystyle z_n = \sum_{k=1}^{n+1} \alpha_ke_k \in \ker(f)\), where $\alpha_k = x_k$ for $k\leqslant n$ and $\alpha_{n+1} = x_n-x_{n-1} + x_{n-2} - \ldots .$ You have to show that the $l_1$-norm $\|z_n-x\|_1$ goes to $0$ as $n\to\infty.$
 

Fermat

Active member
Nov 3, 2013
188
That fact comes in at the end, when you have to use the continuity of $f$ to deduce that $f(x)=0$. At that stage, you will have found an element \(\displaystyle z_n = \sum_{k=1}^{n+1} \alpha_ke_k \in \ker(f)\), where $\alpha_k = x_k$ for $k\leqslant n$ and $\alpha_{n+1} = x_n-x_{n-1} + x_{n-2} - \ldots .$ You have to show that the $l_1$-norm $\|z_n-x\|_1$ goes to $0$ as $n\to\infty.$
The norm has only one term\(\displaystyle : |x_{n}-x_{n-1}+....|->0\) since x is in X.
 
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