# [SOLVED]Show Y is dense in X

#### Fermat

##### Active member
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
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
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
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
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
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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