Unit normed linear functional on a space of sequences

[docnet]

Homework Statement

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Relevant Equations

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Let $X=\{(x_i)\in \ell^\infty | x_i=1 \text{ for } i\leq n, x_i=0 \text{ for } i>n, n\in \mathbb{N}\}$.

For every sequence $(x_i)$ in $X$, $(x_i)$ has the property that the first $n$ entries are 1s and the rest are 0s. So, every sequence in $X$ trivially converges to $X$ and hence $X$ is a Banach space.

Define $f:X\to \mathbb{R}$ given by $f((x_i))=(x_i)\cdot v$ where $v=(\frac{1}{2},\frac{1}{4},\frac{1}{8},...)$. $f$ is a linear functional by construction.

$$\displaystyle{\lim_{n \to \infty}} \sum_{j=1}^n \frac{1}{2^j}=1.$$
Hence $f$ is bounded and the supremum of $f$ over $X$ is $1$.

$||(x_i)||_\infty\leq 1$, so $f(x_i)_{||(x_i)||\leq 1}=1$.

Lastly, $n$ is finite, so there does not exist $(x_i)\in X$ such that $f((x_i))= 1$.

I want to make a similar argument for a space of sequences over the closed 1-D unit ball that converge to 0, using the same linear functional.

 

[andrewkirk]

You haven't told us what you want to prove.
I am guessing that you want to prove that $f(Y)$ has supremum 1 but never achieves that supremum, where $Y$ is the set of all sequences with elements from the interval $(-1,1)$.
As I understand it, your argument for the case of $X$ (rather than $Y$) consists of the 3rd last and 2nd last paras in your OP.
If so then the first of those applies trivially to the case of $Y$.
The second doesn't apply, because sequences in $Y$ needn't have any zero elements at all. So you just need to find a different argument to show that $||f(y)||<1$ for all $y\in Y$.

Hint: Given $y=(y_1,y_2, ...) \in Y$ try to find an expression $g(y_1)$ in terms of just the first element $y_1$ such that
$$f(y)\le g(y_1)<1$$

EDIT: Just noticed that you actually never proved that 1 is the supremum of $f(X)$. You just proved 1 is an upper bound. To prove it's a supremum, you need to show you can come arbitrarily close to 1 with $f(x)$ for $x\in X$. Then you need to reuse or adapt that argument to the case that replaces $X$ by $Y$.

 

[docnet]

Hint: Given $y=(y_1,y_2, ...) \in Y$ try to find an expression $g(y_1)$ in terms of just the first element $y_1$ such that
$$f(y)\le g(y_1)<1$$

EDIT: Just noticed that you actually never proved that 1 is the supremum of $f(X)$. You just proved 1 is an upper bound. To prove it's a supremum, you need to show you can come arbitrarily close to 1 with $f(x)$ for $x\in X$. Then you need to reuse or adapt that argument to the case that replaces $X$ by $Y$.

Thanks andrewkirk.

W.l.o.g., for $(x_i)\in X$ define $$f((x_i))=\sum_{j=1}^n \frac{1}{2^j}= 1-\frac{1}{2^n}.$$ We can take $n$ arbitrarily large by choosing $(x_i)$ to have an arbitrarily large number, $n$, of $1$s. As $n$ increases, $\frac{1}{2^n}$ approaches $0$. Therefore, $f((x_i))$ approaches 1. $f((x_i))$ gets arbitrarily close to 1 as $n$ increases, but never exceeds 1, so the supremum of $f((x_i))$ for $(x_i)\in X$ is 1.

To have $f(y)\le g(y_1)<1$ and $g=f$ would seem to require that $(y_i)$ is a monotonously decreasing sequence. Unfortunately, I don't see a way to move forward with my line of thought. Could you please share another hint?

Alternatively, I seem to have found another way.
$Y$ is the space of sequences in the unit ball that converge to $0$. $y_i\leq 1$ for all $i\in\mathbb{N}$ for all $(y_i)$ in $Y$. So, $f(Y)\leq 1$, by the geometric series.

There exists a set of of sequences $(y_i)^*\in Y$ with $1$s in the first, w.l.o.g., $n$ positions.

For every $(y_i)^*\in Y$ and, w.l.o.g., for every $i>n$, it's true that $y_i \geq 0$.
$$f((x_i))=1-\frac{1}{2^n}\leq f((y_i)^*)\leq 1.$$
Hence, the supremum of $f$ over there set $Y$ is $\sup_Y f((y_i))=1$.

 

[docnet]

Sorry if I didn't explain earlier. I'm trying to show that $f$ is a linear functional over $X$ such that the supremum of $f$ is 1 and $f((x_i))<1$ for all $(x_i)$. I'm trying to do the same with $Y$ as well.

 

[andrewkirk]

W.l.o.g., for $(x_i)\in X$ define
$$f((x_i))=\sum_{j=1}^n \frac{1}{2^j}= 1-\frac{1}{2^n}.$$

That is not the definition you gave above.
But looking back at the definition above, it appears to be circular.
Anyway, I assumed you meant:
$$f((x_i))=\sum_{i=1}^\infty \frac{x_i}{2^i}$$
Is that what you meant? If so, it's not the same as what is in the bit I quoted at top of this post.

Assuming that is what you meant, start by trying using the fact that $\forall i:\ x_i<1$ to show that the RHS is less than $\sum_{i=1}^\infty \frac{1}{2^i}$.

Also, I don't know what you mean by "the supremum of $f$ is 1". Sets of real numbers have suprema. Functionals don't. I assume you mean the supremum of $f(X)$ or of $f(Y)$ (both of which are sets of real numbers.

If you keep the terminology and concepts straight, you're less likely to run into difficulties.

 

[docnet]

That is not the definition you gave above.
But looking back at the definition above, it appears to be circular.
Anyway, I assumed you meant:
$$f((x_i))=\sum_{i=1}^\infty \frac{x_i}{2^i}$$
Is that what you meant? If so, it's not the same as what is in the bit I quoted at top of this post.

Assuming that is what you meant, start by trying using the fact that $\forall i:\ x_i<1$ to show that the RHS is less than $\sum_{i=1}^\infty \frac{1}{2^i}$.

Also, I don't know what you mean by "the supremum of $f$ is 1". Sets of real numbers have suprema. Functionals don't. I assume you mean the supremum of $f(X)$ or of $f(Y)$ (both of which are sets of real numbers.

If you keep the terminology and concepts straight, you're less likely to run into difficulties.

Yes, I mean that, sorry. Being inconsistent with math terminology is a skill that I'm working on improving. I see my mistake in post #1. It says $f((x_i))=f((x_i)\cdot v)$. What I want to say is $f((x_i))=(x_i)\cdot v$, where $v=\left(\frac{1}{2},\frac{1}{4},...\right)$.

Yes, I think we both agree that without loss of generality with respect to $n$, $$f((x_i))=(x_i)\cdot v=\sum_{i=1}^\infty\frac{x_i}{2^i}=\sum_{i=1}^n\frac{1}{2^i}$$
since $x_i=1$ for all $i\leq n$ and $x_i=0$ for all $i>n$.

It seems correct to say that the RHS is less than 1 because $n$ is finite.

Edit: Remove a sentence because of a grammatical error

 

[docnet]

Forgive me, I'm noticing more mistakes as I read back my posts.

It could be why no one else wants to help. I'm making too many mistakes and they don't see the point in trying to point them all out.

The goal is to prove that $f$ is a linear functional with a supremum of 1 over $Y$ and a maximum of less than 1 over $Y$. $Y$ is the space of sequences in $[-1,1]$ that converge to $0$. $X$ is a subset of $Y$.

 

[fresh_42]

It could be why no one else wants to help. I'm making too many mistakes and they don't see the point in trying to point them all out.

To be honest, you lost me in the first line. The LHS is an unparameterized object $X$ and the RHS has an unquantified parameter $n.$ I had simply no idea what you were talking about, especially after reading ...

So, every sequence in $X$ trivially converges to $X$ and hence $X$ is a Banach space.

Since @andrewkirk answered, I was happy that he apparently knew and stopped reading the posts.

 

[docnet]

To be honest, you lost me in the first line. The LHS is an unparameterized object $X$ and the RHS has an unquantified parameter $n.$ I had simply no idea what you were talking about, especially after reading ...


Since @andrewkirk answered, I was happy that he apparently knew and stopped reading the posts.

Hi @fresh_42, thanks for your message. I read that explanation about $X$ being a Banach space in a Google search, but the source wasn't credible and it turned out to be wrong (and not even relevant to the problem as I realized). I'm sorry for it. I'm trying to learn math for myself and it's not for work or school, so I'm glad for any input you guys have.

Here's the argument I'm trying to make, in more detail, and put in a numbered list:

1. Define $X$ as the space of sequences, $(x_i)$, over $\{0,1\}$ of the form $(1,...,1,0,...)$.

Assume that for each $n\in \mathbb{N}$, there exists one $(x_i)\in X$ with $n$ consecutive 1s and then only 0s after.

For example:

For $n=1$ there exists $(1,0,0,0,0..)\in X$.

For $n=2$, there exists $(1,1,0,0,0....)\in X$.

For $n=3$, there exists $(1,1,1,0,0...)\in X$, and so on, for each $n$.

2. For each $n\in\mathbb{N}$, there exists exactly one $(x_i)\in X$, such that
$$f((x_i))=(x_i)\cdot \left(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\right)=\frac{1}{2}+\frac{1}{4}+\cdots$$

and there are $n$ summation terms on the RHS. Each summation term is related to the previous term by the common ratio of $\frac{1}{2}$. The sum is a truncated geometric series.

3. We can get as many summation terms as we want, by choosing the appropriate $(x_i)$. This is because of our assumption in 1.

4. The infinite geometric series with the common ratio of $\frac{1}{2}$ converges to 1. This means that the truncated series can be taken as close to $1$ as we want, i.e., the supremum of $f$ on $X$ is 1.

5. $(x_i)$ cannot have an infinite number of $1$s, because it's impossible to string together the two infinite sequences $(1,...)$ and $(0,...)$ end-to-end. I don't know the name of the property that makes it so, maybe the property of $\aleph_0$.

6. This means that there does not exist a $(x_i)\in X$ such that $f((x_i))=1$, as this would require $(x_i)$ to have an unbounded number of consecutive $1$s.

 

[fresh_42]

Hi @fresh_42, thanks for your message. I read that explanation about $X$ being a Banach space in a Google search, but the source wasn't credible and it turned out to be wrong (and not even relevant to the problem as I realized). I'm sorry for it. I'm trying to learn math for myself and it's not for work or school, so I'm glad for any input you guys have.

Here's the argument I'm trying to make, in more detail, and put in a numbered list:

1. Define $X$ as the space of sequences, $(x_i)$, over $\{0,1\}$ of the form $(1,...,1,0,...)$.

Assume that for each $n\in \mathbb{N}$, there exists one $(x_i)\in X$ with $n$ consecutive 1s and then only 0s after.

For example:

For $n=1$ there exists $(1,0,0,0,0..)\in X$.

For $n=2$, there exists $(1,1,0,0,0....)\in X$.

For $n=3$, there exists $(1,1,1,0,0...)\in X$, and so on, for each $n$.

2. For each $n\in\mathbb{N}$, there exists exactly one $(x_i)\in X$, such that
$$f((x_i))=(x_i)\cdot \left(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\right)=\frac{1}{2}+\frac{1}{4}+\cdots$$

and there are $n$ summation terms on the RHS. Each summation term is related to the previous term by the common ratio of $\frac{1}{2}$. The sum is a truncated geometric series.

3. We can get as many summation terms as we want, by choosing the appropriate $(x_i)$. This is because of our assumption in 1.

4. The infinite geometric series with the common ratio of $\frac{1}{2}$ converges to 1. This means that the truncated series can be taken as close to $1$ as we want, i.e., the supremum of $f$ on $X$ is 1.

So far so good. That was the understanding I started with, too. But then came ...

So, every sequence in $X$ trivially converges to $X$ and hence $X$ is a Banach space.

... that didn't make sense anymore. All of a sudden, $X$ had three different meanings in one sentence!

a) If $X$ is a set, then it cannot be a limit; at least not here.
b) Say $(x_i)_n:=(\underbrace{1,1,\ldots,1}_{n-\text{times}},0,0,\ldots).$ Then $\displaystyle{\lim_{n \to \infty}(x_i)_n}=(1,1,1,\ldots) \not\in X.$
c) How is $X$ a Banach space if it isn't even a vector space, let alone complete (see b))?

5. $(x_i)$ cannot have an infinite number of $1$s, because it's impossible to string together the two infinite sequences $(1,...)$ and $(0,...)$ end-to-end. I don't know the name of the property that makes it so, maybe the property of $\aleph_0$.

I'd say point b) from above. I don't know what you mean by stringing them together. "String together" is called concatenate, btw. You can concatenate two elements from $X$. You'll just be no longer within $X.$ The same is true for vector space operations. They are possible, just not within $X.$

6. This means that there does not exist a $(x_i)\in X$ such that $f((x_i))=1$, as this would require $(x_i)$ to have an unbounded number of consecutive $1$s.

Yes, $f((x_i)_n)=1-2^{-n}<1.$ What is your quest?

I want to make a similar argument for a space of sequences over the closed 1-D unit ball that converge to 0, using the same linear functional.

What is the one-dimensional unit ball for you? The ordinary definition would be the interval $[0,1]$ or its open version $(0,1).$ To make it connected, it would be the circle $C=\{e^{2\pi i \varphi }\,|\,0\leq \varphi \leq 1\}.$ The combination one-dimensional and ball is confusing. Anyway, neither $X,$ nor $[0,1]$ or $C$ is a vector space, so $f$ cannot be linear.

Do you want to consider the vector space with $(x_i)_n$ as basis vectors? Real or complex? With a norm, which? Is this even relevant if you want to consider another space anyway? Which one?

 

[docnet]

Thank you! @fresh_42, you're the best. Know this: your contributions are so much appreciated by me and by many others.

So, every sequence in $X$ trivially converges to $X$ and hence $X$ is a Banach space.

I Google-searched for a short definition of a Banach space and I somehow interpreted it as "every sequence in $X$ converges to a limit point within $X$" without reading carefully. I don't know why I included it in the first post. I think I hoped that including the definition might somehow turn it into a Banach space.

This whole quest is about addressing a resurfacing curiosity about an earlier post because the theory was a bit too abstract/unfamiliar for me. I got busy and stopped thinking about it for a while, so I wanted to construct a simpler example to build some intuition. I thought this $X$ would have properties similar to $\ell^1$ but it turns out to be different.

a) If $X$ is a set, then it cannot be a limit; at least not here.
b) Say $(x_i)_n:=(\underbrace{1,1,\ldots,1}_{n-\text{times}},0,0,\ldots).$ Then $\displaystyle{\lim_{n \to \infty}(x_i)_n}=(1,1,1,\ldots) \not\in X.$
c) How is $X$ a Banach space if it isn't even a vector space, let alone complete (see b))?

I fully agree with all three points. a) doesn't make sense here, b) is true, and c) is a great point because Banach spaces must be complete, vector spaces.

I'd say point b) from above. I don't know what you mean by stringing them together. "String together" is called concatenate, btw. You can concatenate two elements from $X$. You'll just be no longer within $X.$ The same is true for vector space operations. They are possible, just not within $X.$

By concatenating two elements from $X$, would you get another sequence? I believed it would be impossible to concatenate $(1,1,...,1,0,0...)$ and $(1,...,1,0,0,0,...)$ because they're both infinite sequences and heuristically, you'd have to define something like "the mid-point of the concatenated sequence where 1 gets glued to 0."

Yes, $f((x_i)_n)=1-2^{-n}<1.$ What is your quest?

Honestly, I'm not posting here because of a school- or work-related responsibility but only because I enjoy it. Learning math provides a sense of reprieve from the boredom in my life.

What is the one-dimensional unit ball for you? The ordinary definition would be the interval $[0,1]$ or its open version $(0,1).$ To make it connected, it would be the circle $C=\{e^{2\pi i \varphi }\,|\,0\leq \varphi \leq 1\}.$ The combination one-dimensional and ball is confusing. Anyway, neither $X,$ nor $[0,1]$ or $C$ is a vector space, so $f$ cannot be linear.

Do you want to consider the vector space with $(x_i)_n$ as basis vectors? Real or complex? With a norm, which? Is this even relevant if you want to consider another space anyway? Which one?

Thank you

Can we consider $\ell^1$ and $f=(.9,.99,.999,....)\in \ell^\infty$? I'm having a dilemma showing that $||f||=1$ in the sup norm. It seems like, heuristically, we have to define an infinite-dimensional vector with 1 in the last entry and 0s elsewhere to get the sup by a dot product, which doesn't seem possible in this world.

 

[andrewkirk]

Can we consider $\ell^1$ and $f=(.9,.99,.999,....)\in \ell^\infty$? I'm having a dilemma showing that $||f||=1$ in the sup norm.

To prove that, all you need to do is prove that:
(1) 1 is an upper bound for all the $f_n$. Can you show that, for any n, $f_n\le 1$? Start by writing $f_n$ as an expression in only $n$, rather than using "...." as above, which is vague and leads directly to confusion.
(2) We can find $f_n$ arbitrarily close to 1. That is, for any $\epsilon>0$, we can find $f_n$ such that $|1-f_n|<\epsilon$. Again you will use the explicit formula for $f_n$ that you constructed in part (1).

(1) proves 1 is an upper bound. (2) proves there can be no upper bound less than 1, so 1 must be a least upper bound, ie supremum.

 

[docnet]

To prove that, all you need to do is prove that:
(1) 1 is an upper bound for all the $f_n$. Can you show that, for any n, $f_n\le 1$? Start by writing $f_n$ as an expression in only $n$, rather than using "...." as above, which is vague and leads directly to confusion.
(2) We can find $f_n$ arbitrarily close to 1. That is, for any $\epsilon>0$, we can find $f_n$ such that $|1-f_n|<\epsilon$. Again you will use the explicit formula for $f_n$ that you constructed in part (1).

(1) proves 1 is an upper bound. (2) proves there can be no upper bound less than 1, so 1 must be a least upper bound, ie supremum.

Hey! thanks for your reply.

1. Explicit formula for $f_n$:
$$f_n=1-.1^n.$$

For any $n$, $f_n\leq 1.$

Short proof:

Base case: $f_1=1-.1^1=.9\leq1$.

Assume that $f_k=1-.1^k\leq1$.

Then, $f_{k+1}=1-.1^{k+1}=1-(.1^k)(.1)=(.1)(1-.1^k)+.9$.

Apply the induction hypothesis to $(1-.1^k)$.

$f_{k+1}=(.1)(1-.1^k)+.9\leq(.1)(1)+.9=1$.

So for any $n$, $f_n\leq 1.$

2. We can find $f_n$ arbitrarily close to $1$. For any $\epsilon>0$, we can find $|f_n-1|<\epsilon.$

$$\begin{align*}|f_n-1|&<\epsilon\\
\Rightarrow 1-.1^n &<\epsilon\\
1-\epsilon&<.1^n\\
\ln(1-\epsilon)&<n\ln(.1)\\
n&<\frac{\ln(1-\epsilon)}{-\ln(.1)}\end{align*}.$$

Checking if $n$ satisfies $|f_n-1|<\epsilon$:

$$1-.1^n<1-.1^{\frac{\ln(1-\epsilon)}{-\ln(.1)}}=1-10^{\frac{\ln(1-\epsilon)}{\ln(10)}}=1-(1-\epsilon)=\epsilon.$$

3. Combining 1 and 2, we have that $1$ is the supremum of $f_n$.

 

[docnet]

Oops, there was a typo in part 1. "So for any $n$, $f_n< 1$"" should have $\leq$ instead of $<$.

I read somewhere the implication

$$1 \text{ is the supremum of } f_n \Longleftrightarrow \sup_{|||x||_{\ell^1}\leq 1}||\langle f,x\rangle ||=1$$

stands by definition, but I want to prove it.