Let $C_{00}$ be the subspace of $l^2$,and $K=\{x \in C_{00}|\sum_{j=1}^{\infty} \frac{1}{j}x_j=1\}$ and $Y= \{y \in C_{00}||\sum_{j=1}^{\infty}\frac{1}{j}y_j=0\}$
Prove that $K$ is a closed and convex and there does not exist $x_0 \in K$ such that $||x_0||_2=\inf_{x \in K}||x||_2$ and that $Y$ is a closed subspace of $C_{00}$ and $Y \neq (Y^{\bot})^{\bot}$.
Can someone help me with this please?
Thank you in advance!
If we define :
$\mathcal l^2=\{(x_i)_i\mid \sum\limits_{i=1}^{\infty}|x_i|^2<+\infty\}$
$C_0=\{x\in\mathcal l^2\mid x_i\to 0\}\quad$ (but $||x||_2<+\infty\Rightarrow x_i\to 0\quad$ so $\quad C_0=\mathcal l^2$)
$C_{00}=\{x\in\mathcal l^2\mid \exists\ i_0\in\mathbb N,\forall i\ge i_0,\ x_i=0\}$
$S_n(a)=\sum\limits_{j=1}^{n}a_j$
Note: if we were working in $\mathcal l^\infty$ then $C_0$ would not be the entire space, but I introduce it because it is easier to understand with it.
Not closed
We can notice that $\overline{C_{00}}=C_0$ so $C_{00}$ is not closed and it will be the same for $K$ and $Y$.
For instance $X_n=(X_{n,0},\frac{1}{2},\frac{1}{3},...,\frac{1}{n},0,0,0,...)$ with $X_{n,0}=2-S_{n}(1/j^2)$
$\forall n\in\mathbb N$ we have $X_n\in K$ and it has a limit $X=(2-\frac{\pi^2}{6},\frac{1}{2},\frac{1}{3},...,\frac{1}{n},...)\in C_0$
$S_\infty(X)=1$ too, but nonetheless $X\notin K$ because it is not a sequence with finite support.
This is the same for $Y$, just use the same sequence modified for $X_{n,0}=1-S_n(1/j^2)$
Orthogonal
Note that $\mathcal l^2$ being Hilbertian, $Y^{\bot\,\bot}=\overline Y$ so since $Y$ is not closed, we have $Y\neq Y^{\bot\,\bot}$.
Convex
For $(x,y)\in K^2$ and $\lambda\in[0,1]\quad$ and let say $x$ has support $\{1,..,n\}$ and $y$ has support $\{1,..,m\}$.
It is easy to notice that $z=\lambda x+(1-\lambda)y\ $ is also a finite sequence of support $\subset\{1,..,\max(n,m)\}$, so $z\in C_{00}$
Since $S_\infty(\cdot)$ is linear and we operate on finite sums anyway ($S_\infty=S_{\max(n,m)}$) it is also immediate that $S_\infty(z)=\lambda S_\infty(x)+(1-\lambda)S_\infty(y)=\lambda\times 1+(1-\lambda)\times 1=1$, so $z\in K$.
Note that $Y$ is convex too by the same argument.
No minimum
By Cauchy-Schwarz in $\mathcal l^2$ we have $|<X,1/j>|\le||X||_2||1/j||_2=\frac{\pi}{\sqrt 6}||X||_2$ and the equality is achieved for a sequence $X=x_0$ when $x_0$ and $(1/j)_j$ are colinear.
Note that for an $X_n\in K$ then $<X_n,1/j>\;=1$ so $||X_n||_2\ge\frac{\sqrt 6}{\pi}$ thus such an $x_0$ effectively realizes a minimum on $K$.
But to state that $||x_0||_2=\inf_{x \in K}||x||_2$ we have to show that we can find a sequence $X_n$ of elements of $K$ such $||X_n||_2\to||x_0||_2$.
$x_0=(x_j)_j,\ x_j=\frac{6}{\pi^2j}$ and we have effectively $<x_0,1/j>\;=1$ and $||x_0||=\frac{\sqrt 6}{\pi}$.
The construction is very similar to what we've done for closure.
$X_n=(X_{n,0},\frac{6}{2\pi^2},\frac{6}{3\pi^2},...,\frac{6}{n\pi^2},0,0,...)$ with $X_{n,0}=1-\frac{6}{\pi^2}(S_n(1/j^2)-1)$.
It is easy to verify that $<X_n,1/j>\;=1$ and thus $X_n\in K$.
Also $X_{n,0}\to1-\frac{6}{\pi^2}(\frac{\pi^2}{6}-1)=\frac{6}{\pi^2}$
$||X_n||_2=\sqrt{(X_{n,0})^2+(\frac{6}{\pi^2})^2(S_n(1/j^2)-1)}\sim\sqrt{(\frac{6}{\pi^2})^2S_n(1/j^2)}\to\frac{\sqrt 6}{\pi}=||x_0||_2$
Unfortunately $x_0\notin K$ and we cannot find one in $K$ with such a low norm because it would have to be colinear to $(1/j)_j$ which is an infinite sequence.