Show that a set has no "unique nearest point" property

[psie]

Homework Statement

Let $c_0$ be the real vector space of all real sequences that converge to $0$. Let $S$ be the subset of $c_0$ consisting of all elements $(x_n)$ such that $\sum _{n=1}^{\infty }2^{-n}x_n=0$. Show that $S$ is a closed subspace of $c_0$ and that no point of $c_0\setminus S$ has a closest point in $S$.

Relevant Equations

Linear functionals, kernel, sequence spaces, norms, etc.

From Bridges' Foundations of Real and Abstract Analysis.

I'm given the following hint. Given $a=(a_n)\in c_0\setminus S$, set $\alpha=\sum _{n=1}^{\infty }2^{-n}a_n$ and show that $d(a,S)\leq|\alpha|$. Let $x=(x_n)\in S$, suppose that $\lVert a-x\rVert\leq|\alpha|$, and obtain a contradiction.

I can show that $S$ is a closed subspace of $c_0$ by considering the linear functional $$\phi : c_0 \to \Bbb{R} : (x_n) \mapsto \sum_{n=1}^\infty \frac{1}{2^n}x_n,$$ and show that it is bounded and hence its kernel is closed, which is $S$. But I'm stuck at the second part. I do not really understand the hint. I do not know how to show $d(a,S)\leq|\alpha|$, why we can suppose $\lVert a-x\rVert\leq|\alpha|$ and what would be contradicted. Grateful for any help.

 

[pasmith]

What is $d(a,S)$?

 

[psie]

What is $d(a,S)$?

It is $d(a,S)=\inf\{\lVert a-x\rVert :x\in S\}$, where the norm is $\lVert a\rVert=\sup_n |a_n|$.

 

[pasmith]

If the summation was $\sum_{n=1}^\infty 2^{1-n}x_n$, then we could say that $$b_n = \begin{cases} a_1 - \alpha & n = 1 \\ a_n & n > 1 \end{cases}$$ is a member of $S$ which is a distance $|\alpha|$ from $a$. Unfortunately because the summation is $\sum_{n=1}^\infty 2^{-n}a_n$, we have instead to set $$b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1 \end{cases}$$ which is a distance $2|\alpha|$ from $a$.

If there exists an element $x \in S$ which is closest to $a$, then it satisfies $$0 < d(a,x) = d(a,S) \leq |\alpha|$$ (if the question is correct) because $x \neq a$. So when looking for such an element it makes no sense to look for one with $d(a,x) > |\alpha|$.

 

[psie]

If there exists an element $x \in S$ which is closest to $a$, then it satisfies $$0 < d(a,x) = d(a,S) \leq |\alpha|$$ (if the question is correct) because $x \neq a$.

I don't understand your claim. Shouldn't it be $0<d(a,x)=d(a,S)\leq 2|\alpha|$?

 

[psie]

The following article was quite helpful (start reading from "A counterexample in the Banach space $c_0$"). This treats a similar case as in the exercise above, except they use the summation $\sum_{n=0}^\infty 2^{-n}x_n$. Does it make any difference if we use $\sum_{n=1}^\infty 2^{-n}x_n$ versus $\sum_{n=0}^\infty 2^{-n}x_n$?

 

[WWGD]

The following article was quite helpful (start reading from "A counterexample in the Banach space $c0$"). This treats a similar case as in the exercise above, except they use the summation $\sum_{n=0}^\infty 2^{-n}x_n$. Does it make any difference if we use $\sum_{n=1}^\infty 2^{-n}x_n$ versus $\sum_{n=0}^\infty 2^{-n}x_n$?

Certainly no difference in convergence in general, nor convergence to $0$. A finite set of terms won't make a difference either way.

 

[pasmith]

I don't understand your claim. Shouldn't it be $0<d(a,x)=d(a,S)\leq 2|\alpha|$?

By definition, $d(a,x) = 0$ if and only if $a = x$. Here we know $a \neq x$ since $x \in S$ while $a \notin S$.

 

[psie]

By definition, $d(a,x) = 0$ if and only if $a = x$. Here we know $a \neq x$ since $x \in S$ while $a \notin S$.

This explains why $0<d(a,S)$, but not why $d(a,S)\leq |\alpha|$. The example you gave with $$b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1, \end{cases}$$ shows that there is an element of $S$ with a distance $2|\alpha|$ from $a$. So the distance $d(a,S)$ should be at most $2|\alpha|$. What am I misinterpreting?

 

[WWGD]

I suspect your $d$is a distance function, but not a metric. Take, e.g., $S:=\{(x,0): x \in \mathbb R\}$ , i.e., the x-axis , and $T:=\{(x,y): x,y\ in \mathbb R: xy=1\}$, i.e., the curve $y=1/x; x>0$ Then $d(S,T)=0$, as the points in $y=1/x$ get indefinitely close to the x-axis as x goes to $\infty$, so that $d(S,T)=0$, but $S\neq T$. Yes, the use of $d$ is confusing, maybe something like $Dist(x,y)$ would be better here.

 

[pasmith]

This explains why $0<d(a,S)$, but not why $d(a,S)\leq |\alpha|$. The example you gave with $$b_n = \begin{cases} a_1 - 2\alpha & n = 1 \\ a_n & n > 1, \end{cases}$$ shows that there is an element of $S$ with a distance $2|\alpha|$ from $a$. So the distance $d(a,S)$ should be at most $2|\alpha|$. What am I misinterpreting?

I am assuming that the question is correct in its assertion that $d(a,S) \leq |\alpha|$.

 

[Bosko]

I'm given the following hint. Given $a=(a_n)\in c_0\setminus S$, set $\alpha=\sum _{n=1}^{\infty }2^{-n}a_n$ and show that $d(a,S)\leq|\alpha|$. Let $x=(x_n)\in S$, suppose that $\lVert a-x\rVert\leq|\alpha|$, and obtain a contradiction.

Let $a=(a_1,a_2,a_3,...)\in c_0\setminus S$ be given and according to the metric

It is $d(a,S)=\inf\{\lVert a-x\rVert :x\in S\}$, where the norm is $\lVert a\rVert=\sup_n |a_n|$.

We can check that $\alpha$ is finite and estimate range of values it can take
$$\left|\ \alpha \right| ~
= \left| \sum_{n=1}^{\infty} \frac{a_n}{2^n} \right|
\leq \sum_{n=1}^{\infty} \frac{\|a\|}{2^n}
= \|a\| \sum_{n=1}^{\infty} \frac{1}{2^n}
= \|a\|$$

Let $x=(a_1-\alpha,a_2-\alpha,a_3-\alpha,...)$ is given where
$$ \sum_{n=1}^{\infty} \frac{x_n}{2^n}
= \sum_{n=1}^{\infty} \frac{a_n-\alpha}{2^n}
= \sum_{n=1}^{\infty} \frac{a_n}{2^n} - \sum_{n=1}^{\infty} \frac{\alpha}{2^n}
=\alpha-\alpha
=0$$
and
$$d(a,x)
=\| a-x\|=\sup_{n \in \mathbb{N}} |a_n-x_n|
=\sup_{n \in \mathbb{N}} |a_n-a_n+\alpha|
= |\alpha|$$
but ( Edit: $-\alpha$ instead of $\alpha$ in lim ...)
$$\lim_{n \rightarrow +\infty} x_n = -\alpha \neq 0 \Rightarrow x \not\in S $$
Can you do something about $x \not\in S$ ?

 

[psie]

We can check that $\alpha$ is finite and estimate range of values it can take
$$\left|\ \alpha \right| ~
= \left| \sum_{n=1}^{\infty} \frac{a_n}{2^n} \right|
\leq \sum_{n=1}^{\infty} \frac{\|a\|}{2^n}
= \|a\| \sum_{n=1}^{\infty} \frac{1}{2^n}
= \|a\|$$

This computation shows $d(a,S)\leq \lVert a\rVert$, but how does one show $d(a,S)\leq |\alpha|$?

 

[Bosko]

This computation shows $d(a,S)\leq \lVert a\rVert$, but how does one show $d(a,S)\leq |\alpha|$?

TIP:
If $\sum_{i=1}^{\infty} 1/2^i=1$ and $\sum_{i=n+1}^{\infty} 1/2^i=1/2^n$
how much is
$$\sum_{i=1}^{n} \frac{1}{2^i}=?$$

Can you make array of sequences
$x_1=(a_1-\text{"something"},a_2,a_3,...)$
$x_2=(a_1-\text{"something"},a_2-\text{"something"},a_3,...)$
....
$x_n=(a_1-\text{"something"},a_2-\text{"something"},a_3-\text{"something"},...,a_{n-1}-\text{"something"},a_n-\text{"something"},a_{n+1},...)$
...
so that they belong to $S$ ?

Check #2 post > https://www.physicsforums.com/threa...e-nearest-point-property.1059670/post-7055518
What @pasmith doing ?

 

[Bosko]

TIP:
If $\sum_{i=1}^{\infty} 1/2^i=1$ and $\sum_{i=n+1}^{\infty} 1/2^i=1/2^n$
how much is
$$\sum_{i=1}^{n} \frac{1}{2^i}=?$$

From
$$\sum_{i=1}^{n} \frac{1}{2^i}=1-\frac{1}{2^n}=\frac{2^n-1}{2^n}$$
it follows
$$\frac{2^n}{2^n-1} \sum_{i=1}^{n} \frac{1}{2^i}=1$$
Multiplying both sides by $-\alpha$ we get
$$- \frac{2^n\alpha}{2^n-1} \sum_{i=1}^{n} \frac{1}{2^i}=-\alpha$$
For n=1 sequence $x_1$ is $(a_1-2\alpha,a_2,a_3,a_4,...)$ and the distance from $a$ to $x_1$ is $2\alpha$.
For n=2 sequence $x_2$ is $(a_1-4\alpha/3,a_2-4\alpha/3,a_3,a_4,...)$ and the distance from $a$ to $x_2$ is $4\alpha/3$.
For n=3 sequence $x_3$ is $(a_1-8\alpha/7,a_2-8\alpha/7,a_3-8\alpha/7,a_4,...)$ and the distance from $a$ to $x_3$ is $8\alpha/7$.
...
The distance from $a$ to $x_n$ is $2^n\alpha/(2^n-1)$