I was trying to understand Reisz Theorem -

How can we show that there exists such a $y_{0}$?

I thought that $dist(x,Y) \leq \frac{dist(x,Y)}{r}$

and now $dist(x,Y) = inf\{(||x-y|| | y \in Y)\}$

Now I think it is using defn of infimum

but i am not getting how

seems to me like $dist(x,y) \leq ||x - y_{0}|| \leq \frac{dist(x,Y)}{r}$

but how $dist(x,y) \leq ||x - y_{0}||$

?

Perhaps @MartinSleziak,you can give a look when you are around!

Anybody can give something that can bound above and below $\frac{1}{n( \log n)^2}$

@BAYMAX They are using that $$\operatorname{dist}(x,Y)<\frac{\operatorname{dist}(x,Y)}r.$$

If you have any number which is strictly larger than infimum, $U>\inf A$, then you have $a_0\in A$ such that $$\inf A \le a_0 < U.$$

thanks @MartinSleziak that helped!

Hi@MartinSleziak

Is that screenshot from Limaye's book?

yes

how do u know ?

so famous!

book

I never mentioned page no. and title

I have tried to search in Google

hm..nice

I was trying @rei

question

and discussion in main

gave me this $e^{-n} - e^{-1}< log(n) < n $

and we can do the necessary calculation for

$\frac{1}{n(log(n))^2}$

Without saying whether you need a very tight bound, it's king of open ended question. For example, $\frac1{n^2}<\frac1{n\log^2n}<\frac1n$.

oh

I've copied wrong search query, in fact it wasthis one which lead me to a book by Limaye with similar formulations.

Its the second click point one! which we are now referring!

i will be reading the theorems next to Reisz and will ask you doubts :)

I guess by you mean you = users of this room, right?

I wish more users stared visit this room. Then it could actually become useful.

yes I know you feel pressurised by doing the stuffs,sorry

I do not have a problem - but I might be not always available.

yes,I see

I hope users will visit this room exponentially!

I too have created a room on dynamical ssytems

which one seldom visits

In every field Cauchy Sequence converges?

or is every field complete?

ohk..here

I do not think you can talk about Cauchy sequence in arbitrary field.

As the WP article you linked mentions, you can define Cauchy sequence if you have metric on that field.

metric on that field

can we associate a norm on a field

$Y$ is complete in $X$ then in particular $Y$ is closed in $X$.

Perhaps we saw this when we dealt some examples before determining which spaces are complete!

In fact, the above message was starred: chat.stackexchange.com/rooms/info/14150/…

yes!

the link in the comment you made some yrs later doesn't exist

here

the wiki proof

@user8469759 You know that you linked in the question to the video at timestamp 50:33 but in the post you mention minute 23, right? Perhaps you could change t=3033s in the link to t=1380s or something like that. (Depending on the spot where you want to link to.)

can people bother in doing a simple click...

anyway I'm fixing it

Personally, I think that if somebody is asking others for help, it seems polite to make their life as easy as it gets.

Of course, it is quite likely that somebody else would fix the link for you, if you didn't. There are magical fairies on this site, as somebody said on meta.

I have fixed it

anyway there's something I don't understand about that definition

the PVM is not a measure

it's somehow related but it's not the same thing

because a measure maps measurable sets in the extended R in general

this one maps into bounded operators

A reminder - just in case it is needed:

My problem with understanding what you wrote is that you wrote $\limsup_{n\to\infty} f_n(x)$ on one site - which is a number.

And then you wrote something which seems like close to formula, but actually isn't. (If you wrote some quantifiers, they should be followed by somethings which has a truth value. And $f_k(x)$ is a number.)

I would agree with something like this: $\limsup\limits_{n\to\infty} a_n > b$ implies $\forall_{n\in\mathbb N} \exists_{k\ge n} a_k >b$

$\forall_{n\in\mathbb N} \exists_{k\ge n} a_k >b$ implies $\limsup\limits_{n\to\infty} a_n \ge b$

Hi Martin, thanks for being helpful. In the picture above, why do we have the equality $A=\bigcup_{m \in \mathbb{N}} \limsup A_n^{1/m}$?

I understood this as $\limsup A_n^{1/m}=\{\forall_{n \in mathbb{N}}\exists_{k\geq n}|\frac{1}{n} \tilde S_n|>\frac{1}{m}\}=\{\limsup |\frac{1}{n} \tilde S_n|>\frac{1}{m}\}$

and Thanks for the chatJax thing. I didn't know it existed. ;)

@MartinSleziak forgot to tag you. sorry

Hi @Anoldmaninthesea. I was away from coputer for some time.

As far as I can say $$\limsup\limits_{n\to\infty} A_n^{1/m} = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^{1/m}.$$

This would give $$\limsup\limits_{n\to\infty} A_n^{1/m} = \{\forall_{n\ge 1} \exists_{k\ge n} \left|\frac1k \widetilde{S_k}\right|> \frac1m\}.$$

So I would write $\frac1k \widetilde{S_k}$ rather than $\frac1n \widetilde{S_n}$.

So your question basically amounts to checking that for a positive sequence $(a_n)$ you have $$\limsup_{n\to\infty} a_n >0 \iff \exists_{m\in\mathbb N} \exists_{n\in\mathbb N} \forall_{k\ge n} a_k>1/m.\tag{*}$$

Where you have $a_n=|\frac1n\widetilde S_n|$.

@MartinSleziak no problem. You're write I should have written with k instead of n, my mistake.

*right

np

So, despite that small typo, my reasoning was correct?

So it seems that we wrote basically the same thing.

Hi guys can I ask a question about reflexive in functional analysis?

When I look at the last part of this message, it looks a bit as if you think that $$\limsup_{n\to\infty} a_n > \frac1m \iff \forall_n \exists_{k\ge n} a_k>1/m.\tag{**}$$

Here I mean that $m$ is fixed.

The equivalence $(**)$ is not true.

The fact that you're dealing with positive sequence should be used somewhere.

But perhaps you did not mean that you're using (**) there.

@AnswerLee Fire away! If we'll see that having two different discussions at the same time is too much, we will stop you.

And if we do not know the answer, it would be probably good to repost your question once again after the discussion about limit superior is finished. (So that your question is better visible and not lost among many other messages.)

Ok that's what I wanted to know. I'm not sure if it changes something, but $\tilde S_n(\omega)$ is a measurable function. If the equivalence is not true, then how can we have the equality?

I'll be right back after dinner.

You have two implications.

There is a definition: A normed space X is reflexive if X**={\hat{x}:x\in X}. But we konw there is a isometric isomorphism between X and $\hat{X}$. Can we directly say A normed space X is reflexive if X**=X.

Combining them and the fact that $a_n>0$ you should get the result. (I hope I did not mix up there something. And the proof of the whole equality $A=\bigcup \ldots$ can be done in an easier way.)

Sorry, that was still a reply to Anoldmaninthesea.

@AnswerLee It is "abuse of language" but AFAICT we commonly identify $x$ with $\hat x$ and then we indeed say that reflexivity means $X=X^{**}$.

Perhaps this is also interesting with your question, but it goes in a slightly different direction: Are there non-reflexive vector spaces isomorphic to their bi-dual? and Wikipedia: James' space.

Thank you so much!

So it is important that we are using the canonical embedding.

Thanks for your help Martin! But sometimes I am very confused. If we equip X^{} with weak star topology. Can we still say $X=X^{}$? Because I am not sure there is a weak star topology on X.

X^{**} I mean.

@AnswerLee Well, if we say $X=X^{**}$, then we mean both as the normed linear space (with the usual norm).

At least that's what I would expect.

BTW isn't weak* topology a topology on $X^*$? (I mean on the dual space, not on the double dual.)

$X^{**}$ can also equip with weak star topology i think.

@MartinSleziak Thanks. ;)

@AnswerLee I guess we could look at X** as on dual of X* and get there weak* topology in that way. However, if we then view X as a subspace of X**, then the topology on X we get will be the usual weak topology on X.

At least I think so - I hope I did not miss some problem with that.

X** has a natural normed topology(strong topology). If we view X as a subspace of X**, shouldn't be the normed topology on X?

Here is a question. We know ball X is normed closed on X. Is ball X normed closed on X**?

@AnswerLee By "ball X" you mean that unit closed ball of the space $X$, i.e., $\overline B=\{x\in X; \|x\|\le 1\}$?

Yes

This probably might not be closed if the space is not reflexive, let me think a bit.

@MartinSleziak Sure. But in Conway's book it is true. I am so confused!

@AnswerLee I can have a look, if you give me theorem number.

You mean A Course in Functional Analysis, right?

@MartinSleziak Yes page 132 4.2 Theorem (b)-->(a)

"Now ball X is norm closed in X**; hence..."

Hm....

@MartinSleziak Yes! That is where I am stuck!

We are also given that X is Banach space, this might help.

Let us try to prove this.

@MartinSleziak But actually we don't need X to be a Banach space. Normed space will do in this theorem.

Suppose $(x_\nu)$ is a net such that $\|x_\nu\|\le 1$ and $\hat x_\nu \to \varphi$ for some $\varphi \in X^{**}$.

Since $\hat x_\nu$ is convergent, it is also Cauchy.

This means that $x_\nu$ is a Cauchy net in the unit ball of $X$.

We have $x_\nu \to x$ for some $x$ from the unit ball.

This means $\hat x_\nu \to \hat x$. And we finally get $\varphi = \hat x$.

@AnswerLee Conway writes: "If $X$ is a Banach space, the following statements are equivalent."

Perhaps we can prove this without completeness, but my attempt above uses the fact that Cauchy net converges.

Yes but no properties of Banach space are used here. So for a normed space these statements will still hold.

In fact, I should have used sequences rather than nets - this was about normed spaces.

Well, in my proof I used that Cauchy $\Rightarrow$ convergent.

So ball X is normed closed in X^**, right?

Unless I made some mistake,the above should be proof at least if X is Banach.

Just some random googling: "The image $\hat X = \{\hat x; x\in X\}$ of the canonical embedding is closed in $X^{**}$ if and only if $X$ is complete." It is from these notes.

But in "An Introduction To Banach Space Theory(Megginson)" this book just a normed space will do.

Which theorem are we looking at?

@MartinSleziak Do you have this copy?

4.2 Theorem in Conway's book

Yes, you have mentioned that one. It has Banach space among assumptions.

Where in Megginson do you see this for normed spaces?

BTW Megginson says this in Proposition 1.11.3: "Furthermore, the subspace $Q(X)$ of $X^{**}$ is closed if and only if $X$ is a Banach space."

Page 245 2.8.2 Theorem

Similar theorem like 4.2 Theorem in Conway's book but different spaces.

But it is only (a) and (d) from the Conway's version.

Take a look at page 246 the equivalent statements will appear.

Ok maybe we need that Banach space. Thanks for your help.

I'm not sure I helped that much. It would be nice if somebody more knowledgeable visited this chat room...

@MartinSleziak BTW. 4.2 Theorem in Conway's book (a)-->(c), we use X=X**. But from the definition of reflexive we can only get \hat{X}=X^**. How to explain X=X**? Is that because there is a isometric isomorphism between X and $\hat{X}$.

Coz I write X=X^** directly on my homework. But my professor asked me why?

I'd guess they are identified with each other.

So that is because there is a isometric isomorphism between X and $\hat{X}$.

> Since $X^{**}$ is always complete, if $X$ is reflexive then it is complete.

The above quote is from this answer.

2.8.2 in Megginson has the assumption that $X$ is reflexive, so we should (somehow) get completeness from that.

That makes sense!

In fact, we have that dual is always complete, right?

I think so.

I don't have in front of me how $\sigma(X^*,X)$ and $\sigma(X^*,X^{**})$ are defined in Conway.

However, if we have a net $\varphi_\nu$ in $X^*$, then $\varphi_\nu$ converges to $\varphi$:

In $\sigma(X^*,X)$ iff $\varphi_\nu(x)\to \varphi(x)$ for each $x$.

In $\sigma(X^*,X^{**})$ iff $\langle \varphi_\nu, f \rangle \to \langle \varphi, f \rangle$ for each $f\in X^{**}$.

But from reflexivity we have that each $f\in X^{**}$ is of the form $f=\hat x$ for some $x\in X$.

$\sigma(X^*,X)$ means weak topology on X and $\sigma(X^*,X^{})$ means weak star topology on X^.

So the latter condition reduces to: "$\hat x(\varphi_\nu) \to \hat x(\varphi)$ for each $x\in X$"

Which is the same as: "$\varphi_\nu(x)\to \varphi(x)$ for each $x$"

Exactly!

BTW here in chat you have to use \*\* to get that rendered correctly.

Get it!

Chat engine interprets ** as the MarkDown for bold.

Ok, it's getting quite late in my timezone and I'll have to get up early tomorrow, so I should get some sleep.

I'll have to go. Good luck with your studying and have a nice day!

Thank you so much!

hi guys

out of curiosity

in the baire thereom (in functional analysis)

namely given a complete metric space X and a sequence of closed set S_n whose union is equal to X then there's a S_n not empty

why is this equivalent to

given a complete metric space X and a sequence of open dense set U_n (sequence) then the intersection is not empty?

I'm struggling to understand why those two are equivalent