Proving Y is Dense in X: Let's See!

[Fermat1]

Let X be the subset of $$ l_{1}$$ consisting of those sequences $$(x_{n})$$such that

$$$$$$x_{1}+x_{3}+x_{5}+...=x_{2}+x_{4}+x_{6}+...$$

Let Y be the linear span of $${{e_{n}+e_{n+1}:n=1,2,...}}$$ where $$e_{n}$$ 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:$$t_{1}(e_{1}+e_{2})+...+t_{k}(e_{k}+e_{k+1})$$$$$$ 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 $$f(e_{n}+e_{n+1})=0$$ for all n. Can anyone help me finish to show f(x)=0 for any x in X?

Thanks

 

[Opalg]

Let X be the subset of $$ l_{1}$$ consisting of those sequences $$(x_{n})$$such that

$$x_{1}+x_{3}+x_{5}+...=x_{2}+x_{4}+x_{6}+...$$

Let Y be the linear span of $${{e_{n}+e_{n+1}:n=1,2,...}}$$ where $$e_{n}$$ 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:$$t_{1}(e_{1}+e_{2})+...+t_{k}(e_{k}+e_{k+1})$$ 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 $$f(e_{n}+e_{n+1})=0$$ 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 $n\to\infty$.

 

[Fermat1]

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$$x_{1}+x_{3}+...=x_{2}+x_{4}+...$$ so are you not effectively proving Y is dense in $$l_1$$?

 

[Fermat1]

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]

Thanks opalg, I think I reasoned this way but rejected it because it doesn't use ther fact that x in X satisfies $$x_{1}+x_{3}+...=x_{2}+x_{4}+...$$ so are you not effectively proving Y is dense in $$l_1$$?

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 $$z_n = \sum_{k=1}^{n+1} \alpha_ke_k \in \ker(f)$$, 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 $n\to\infty.$

 

[Fermat1]

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 $$z_n = \sum_{k=1}^{n+1} \alpha_ke_k \in \ker(f)$$, 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$$: |x_{n}-x_{n-1}+...|->0$$ since x is in X.