Find f s. t. ||f||=1 and f(x) < 1 with ||x||=1

[docnet]

Homework Statement

I'm trying to come up with ways to find a continuous $f$ such that $\sup_{||x||\leq 1}||f||=1$ and $f(x)\neq 1$ with $||x||=1$.

Relevant Equations

My understanding of this problem is a bit infantile. I want some math veterans to haze my poor understanding into shape.

Let $f$ be a continuous function defined in $\mathbb{R}^n$. $||\cdot ||$ is the standard Euclidean metric. Then here are my suggested ways to choose $f$:

1. Choose any continuous $f$ that satisfies
$$1=\sup_{||x||\leq 1}||f||\neq \max_{||x||\leq 1}||f||$$ because the inequality ensures $f(x)\neq 1$ with $||x||\leq 1$. Can you think of any specific examples?

2. Choose any continuous $f$ that satisfies $$1=\max_{||x||< 1}||f||$$ and $$1>\max_{||x||=1}f.$$ A simple example would be $f=1-||x||$.

Thank you.

edit: continuous $f$

Also, I'm wondering if category 1 is invalid, i.e., if there do not exist functions that meet category 1.

 

[Office_Shredder]

Item 1 does not exist.

The key point is for the supremum to be 1 but the maximum to be less than 1, there must exist a sequence $x_n$ such that $f(x_n)\to 1$ (probably worth proving if it's not obvious) What do you know about sequences in the closed unit balls? Are they guaranteed to have any interesting properties?

This is posted in the homework section so I've given an appropriate amount of guidance for a homework question, but it's not totally obvious from your post if this is actually assigned homework. For example the homework question you posted is clearly not word for word a homework assignment.

 

[docnet]

Item 1 does not exist.

Thank you!

The key point is for the supremum to be 1 but the. maximum to be less than 1,, there must exist a sequence $x_n$ such that $f(x_n)\to 1$ (probably worth proving if it's not obvious) What do you know about sequences in the closed unit balls? Are they guaranteed to have any interesting properties?

The sequence must converge to a limit point in the closed unit ball? And this gives contradiction?

This is posted in the homework section so I've given an appropriate amount of guidance for a homework question, but it's not totally obvious from your post if this is actually assigned homework. For example the homework question you posted is clearly not word for word a homework assignment.

You're right, it's not homework. I'm helping to train a GPT model to get the solution by itself. And my analysis knowledge is still shaky compared to yours

 

[Office_Shredder]

The sequence must converge to a limit point in the closed unit ball? And this gives contradiction?

No, for example the sequence $x_k=(1,0)$ when $k$ is odd and $(0,1)$ when $k$ is even does not converge.

But it is true that every sequence contained in a closed bounded set has a convergent subsequence. And that subsequence will give you your contradiction.

 

[docnet]

No, for example the sequence $x_k=(1,0)$ when $k$ is odd and $(0,1)$ when $k$ is even does not converge.

Oops, I think I misunderstood your hint.

But it is true that every sequence contained in a closed bounded set has a convergent subsequence. And that subsequence will give you your contradiction.

Is this closer to being right?

For some sequence contained in the closed unit ball, there exists a convergent subsequence, $x_k$, such that $f(x_k)$ converges to $1$. This means the maximum of $f$ is greater than or equal to $1$, which contradicts the condition $$1=\sup_{||x||\leq 1}||f||>\max_{||x||\leq 1}||f||.$$

 

[fresh_42]

But it is true that every sequence contained in a closed bounded set has a convergent subsequence. And that subsequence will give you your contradiction.

Is that true if $f$ is not continuous? The subsequence is clearly there, but the limit doesn't get into the function in non-continuous cases. I thought of something like the Gauß curve on $[-1,1]$ with $f(0)=1$ and converging to zero at the boundaries, plus setting $f(\pm 1)=0.5$ or something like this. Not sure, whether I have overlooked something.

Edit: I know that the OP started his attempt with "let $f$ be continuous" but the original statement said nothing about continuity.

 

[docnet]

Is that true if $f$ is not continuous? The subsequence is clearly there, but the limit doesn't get into the function in non-continuous cases. I thought of something like the Gauß curve on $[-1,1]$ with $f(0)=1$ and converging to zero at the boundaries, plus setting $f(\pm 1)=0.5$ or something like this. Not sure, whether I have overlooked something.

Edit: I know that the OP started his attempt with "let $f$ be continuous" but the original statement said nothing about continuity.

Hmm can I ask why you specified $f(\pm 1)=0.5$ if $f$ is allowed to be $0$ at the boundaries $x= \pm 1$?

Also $f$ is assumed to be continuous because it's in the dual space of $\mathbb{R}^n$. Sorry about that!.

 

[docnet]

Sorry, I edited response #7 because of my grammar issues.

 

[fresh_42]

It seems like the gauss curve you mentioned would be an example of $f$ that would work.

Oops, I should edit the original statement so that $f$ is supposed to be a function from the dual space of $\mathbb{R}^n$, and so it's supposed to be continuous. if $f$ is allowed to be discontinuous, something like the gauss curve with $f(0)=0$ would work as well?

Be cautious with my posts today. It's not my day, and I tend to make mistakes. But the longer I think about it, then why isn't $f(x)=-x^2+1$ on $[-1,1]$ an example?

 

[docnet]

Be cautious with my posts today. It's not my day, and I tend to make mistakes. But the longer I think about it, then why isn't $f(x)=-x^2+1$ on $[-1,1]$ an example?

It's an awesome example.

Can you help me prove why there is no continuous function $f$ such that $1=\sup_{||x||\leq 1}||f||\neq \max_{||x||\leq 1}||f||$?

Sketch of proof using sequences:

For some sequence contained in the closed unit ball, there exists a convergent subsequence, $x_k$, such that $f(x_k)$ converges to $1$. This means the maximum of $f$ is greater than or equal to $1$, which contradicts the condition $$1=\sup_{||x||\leq 1}||f||>\max_{||x||\leq 1}||f||.$$

Sketch of proof using the compactness of $[-1,1]$:

$[-1,1]$ is compact, so the image of $[-1,1]$ under $f$ is compact. So the image is closed and bounded, so the image of $f$ will take a maximum and minimum value. $\sup_{x\ in [-1,1]}f$ is a limit point in the set, and since the set is closed, the sup is the max.

Are either of these valid plans for a proof?

Thank you.

 

[fresh_42]

It's an awesome example.

Can you help me prove why there is no continuous function $f$ such that $1=\sup_{||x||\leq 1}||f||\neq \max_{||x||\leq 1}||f||$?

Sketch of proof using sequences:

For some sequence contained in the closed unit ball, there exists a convergent subsequence, $x_k$, such that $f(x_k)$ converges to $1$. This means the maximum of $f$ is greater than or equal to $1$, which contradicts the condition $$1=\sup_{||x||\leq 1}||f||>\max_{||x||\leq 1}||f||.$$

Sketch of proof using the compactness of $[-1,1]$:

$[-1,1]$ is compact, so the image of $[-1,1]$ under $f$ is compact. So the image is closed and bounded, so the image of $f$ will take a maximum and minimum value. $\sup_{x\ in [-1,1]}f$ is a limit point in the set, and since the set is closed, the sup is the max.

Are either of these valid plans for a proof?

Thank you.

I'm not sure. I stopped my own thoughts at "the continuous images of compact sets are compact". Wouldn't that say that continuous functions on compact sets are closed? And why should they? Continuity means the pre-images of closed sets are closed, why the images? On the other hand, I can't imagine a counterexample. I'm too tired for topology on a day like this.

We attribute the following theorem to Weierstraß, the English Wikipedia calls it "extreme value theorem".$^1$

Every continuous function defined on a compact set is bounded there and takes its minimum and maximum.

This is the situation here, so the theorem already says that the supremum is actually a maximum. You can find a proof on Wikipedia https://en.wikipedia.org/wiki/Extreme_value_theorem. It uses the theorem of Bolzano-Weierstraß that provides a convergent subsequence.


_____
$^1$ I think I should refrain from ranting against British imperialism when it comes to naming theorems if any British are involved compared to cases where they are not involved. There are far more cases than this one.

 

[docnet]

I'm not sure. I stopped my own thoughts at "the continuous images of compact sets are compact". Wouldn't that say that continuous functions on compact sets are closed? And why should they? Continuity means the pre-images of closed sets are closed, why the images? On the other hand, I can't imagine a counterexample. I'm too tired for topology on a day like this.

We attribute the following theorem to Weierstraß, the English Wikipedia calls it "extreme value theorem".$^1$

Every continuous function defined on a compact set is bounded there and takes its minimum and maximum.

This is the situation here, so the theorem already says that the supremum is actually a maximum. You can find a proof on Wikipedia https://en.wikipedia.org/wiki/Extreme_value_theorem. It uses the theorem of Bolzano-Weierstraß that provides a convergent subsequence.


_____
$^1$ I think I should refrain from ranting against British imperialism when it comes to naming theorems if any British are involved compared to cases where they are not involved. There are far more cases than this one.

I hope you feel better soon and thank you for your efforts to help me understand. Just to make sure I'm understanding , the 'subsequence' mentioned in Bolzano-Weierstraß theorem is necessarily of infinite length, right?

 

[FactChecker]

Consider $f(y) = 1-y^2$ and set $y \equiv |x|$. It will have its maximum at $x=0$ and equal 0 if $|x| = 1$.

 

[fresh_42]

I hope you feel better soon and thank you for your efforts to help me understand.

Thanks. I think I only didn't sleep well and had some bad dreams. Nothing serious, just a case of concentration deficits. Plus I know that topology can be very tricky. To me, it is the most counterintuitive branch of mathematics.

Just to make sure I'm understanding , the 'subsequence' mentioned in Bolzano-Weierstraß theorem is necessarily of infinite length, right?

The point is that we only have a converging sequence in the codomain $(f(d_k))_{k\in \mathbb{N}},$ say
$$\displaystyle{\lim_{k \to \infty}}f(d_k)=M=\sup_{\|x\|\leq 1}\|f(x)\|=1,$$
but we need a convergent sequence in the domain of $f.$ $(d_k)_{k\in \mathbb{N}}$ might not do it, because we don't know anything about the $d_k$ except that they are an infinite sequence and live in a compact set. That's where Bolzano-Weierstraß kicks in. It guarantees a converging subsequence, therewith a limit, and with the closeness of $\{x\, : \,\|x\|\leq 1\}$ a limit within that ball. Hence, we have a converging subsequence of $(d_k)_{k\in \mathbb{N}},$ say $\displaystyle{\lim_{k \to \infty}d_{n_k}}=d,$ and continuity of $f$ allows us to swap the limit from outside to inside the function:
$$
\lim_{k \to \infty}f(d_{n_k})= f\left(\lim_{k \to \infty}d_{n_k}\right)=f(d).
$$
Remains to make sure that $\displaystyle{\lim_{k \to \infty}f(d_{n_k})=M.}$ But $f(d_{n_k})$ is a subsequence of $f(d_{n})$ that converges to $M$ and a subsequence of a convergent series cannot have a distinct limit.

 

[docnet]

Now I see why @FactChecker gave up trying to explain (kidding) it's not an simple proof to explain to someone who isn't willing to spend effort and time to learn it.

The point is that we only have a converging sequence in the codomain $(f(d_k))_{k\in \mathbb{N}}$

Is the existence of such a convergent sequence guaranteed by $f$ being continuous? It seems a bit unnatural to define the sequence in the image in terms of the sequence in the domain.

with the closeness of $\{x\, : \,\|x\|\leq 1\}$ a limit within that ball.

The closedness of the domain seems like a crucial part of the proof because $\lim_{n\to \infty}d_{n_k}=d$ being within the ball allows us to argue that $f(d)=M$ is in the image.

Hence, we have a converging subsequence of $(d_k)_{k\in \mathbb{N}},$ say $\displaystyle{\lim_{k \to \infty}d_{n_k}}=d,$

At this stage, we haven't assumed that this $d$ makes $f(d)=1$, right? The Bolzano-Weierstraß theorem just seems to say there is a convergent subsequence in the domain, and doesn't seem to specify what the subsequence converges to. If so, I have no issues with it.

continuity of $f$ allows us to swap the limit from outside to inside the function:
$$
\lim_{k \to \infty}f(d_{n_k})= f\left(\lim_{k \to \infty}d_{n_k}\right)=f(d).
$$
Remains to make sure that $\displaystyle{\lim_{k \to \infty}f(d_{n_k})=M.}$ But $f(d_{n_k})$ is a subsequence of $f(d_{n})$ that converges to $M$ and a subsequence of a convergent series cannot have a distinct limit.

Proofs like these are surprising to me because it's all so unintuitive but it works. Is this a well-known result? Thanks a lot for putting in the work to explain it to a simpleton like me. I'll keep thinking about this one for a while :)

By the way, have you thought about getting on TRT? A lot of 20+ year old guys seem to rave about its positive effects on concentration.

 

[Office_Shredder]

I'm not sure. I stopped my own thoughts at "the continuous images of compact sets are compact". Wouldn't that say that continuous functions on compact sets are closed? And why should they? Continuity means the pre-images of closed sets are closed, why the images? On the other hand, I can't imagine a counterexample. I'm too tired for topology on a day like this.

Suppose $f$ is continuous and $f(A)=B$, where $A$ is compact. Let $U_i$ be an open infinite cover of $B$ and consider $V_i=f^{-1}(U_i)$. This is an open subcover of $A$ so has a finite subcover which I will call $V'_i$. Let $U'_i$ be the subset of $U_i$ for which $f^{-1}(U'_i)=V'_i$. Then $\cup U'_i$ must contain $B$, as $f^{-1}(\cup U'_i)=\cup V'_i$ which covers $A$. This proves every open cover has a finite subcover and hence continuous functions map compact sets to compact sets.

 

[FactChecker]

Now I see why @FactChecker gave up trying to explain (kidding) it's not an simple proof to explain to someone who isn't willing to spend effort and time to learn it.

Sorry. I didn't see that you already had a similar example for what you call Case #2. I thought that you were trying to prove that was not possible.

 

[fresh_42]

Is the existence of such a convergent sequence guaranteed by $f$ being continuous? It seems a bit unnatural to define the sequence in the image in terms of the sequence in the domain.

Do you mean a sequence converging to the supremum or a convergent subsequence that exists by Bolzano-Weierstraß?

Neither is because of continuity. The former is because of the definition of a supremum. We either have a maximum and a constant sequence, or we can find a value closer to that with which we started and so forth. The latter is as in the old joke about how a mathematician catches a hare: he divides the terrain into half and looks in which half the hare is. Then he divides that part into half and looks where the hare is, and so on. The crucial point here is the boundness of the unit ball.

The closedness of the domain seems like a crucial part of the proof because $\lim_{n\to \infty}d_{n_k}=d$ being within the ball allows us to argue that $f(d)=M$ is in the image.

Indeed. Without it, like in an open ball $\{x\, : \,\|x\|<1\}$ we could find that $\|d\|=1$ and fail to find a pre-image of $M.$

At this stage, we haven't assumed that this $d$ makes $f(d)=1$, right? The Bolzano-Weierstraß theorem just seems to say there is a convergent subsequence in the domain, and doesn't seem to specify what the subsequence converges to. If so, I have no issues with it.

Right. But as it is a convergent sequence $(d_{n_k})$ within a convergent sequence $(d_n)$ it cannot have two distinct limits. The norms of the sequence members are irrelevant, except that they are all within the domain of $f$ since we are searching for a limit $d$ such that $f(d)=M.$ The norm of $d$ is also irrelevant as the examples show where $d$ is placed at the origin of the ball.

Proofs like these are surprising to me because it's all so unintuitive but it works.

Look up the German Wikipedia pages (and eventually use Chrome to translate the page.) At least the proof of Bolzano-Weierstraß has some nice images that describe the situation:
https://de.wikipedia.org/wiki/Satz_von_Bolzano-Weierstraß#Beweisskizze

We need the boundness to avoid that $d$ slips into the infinities, and we need closeness to avoid that $d$ escapes into loopholes like $\mathbb{R}\backslash \mathbb{Q}$ or $[-1,1] \backslash (-1,1).$

Is this a well-known result?

Yes, it has a name, with or without mentioning Weierstraß. I guess that results with names can generally be considered well-known. But this result is important because it provides a solution to the problem of finding an extreme value. This is an important part of many other proofs. And the reason why the joke with the hare works. And it guarantees that a numerical solution by an algorithm converges.

By the way, have you thought about getting on TRT? A lot of 20+ year old guys seem to rave about its positive effects on concentration.

It's far better today. You could have complained that I posted although I knew that I had a weak day, but I somehow got sucked in. Mathematics is a never-ending sequence of the constantly posed question "why". One has to convince oneself over and over again that the next step in a chain of conclusions is correct. Sometimes you simply get lost somewhere in a trapdoor.

But, yes, age plays a significant role. Here are two sad, however, prominent examples where age was the problem:
https://www.physicsforums.com/threads/has-the-riemann-hypothesis-been-proven.955832/
https://www.physicsforums.com/threa...-the-history-of-physics.1059261/#post-7005132

Not that I would compare myself with either of the two, never ever, but it shows that it could be worse.

 

[docnet]

Do you mean a sequence converging to the supremum or a convergent subsequence that exists by Bolzano-Weierstraß?

Neither is because of continuity. The former is because of the definition of a supremum. We either have a maximum and a constant sequence, or we can find a value closer to that with which we started and so forth. The latter is as in the old joke about how a mathematician catches a hare: he divides the terrain into half and looks in which half the hare is. Then he divides that part into half and looks where the hare is, and so on.

The convergent sequence in the codomain that you give to begin the proof. Sorry if it was a dumb question but I didn't want to take this chance for granted.

Indeed. Without it, like in an open ball $\{x\, : \,\|x\|<1\}$ we could find that $\|d\|=1$ and fail to find a pre-image of $M.$

If the ball is were open, would Bolzano-Weierstraß theorem fail to give a convergent subsequence in the open ball? And if we have a convergent subsequence such that $\|d\|=1$, the rest of the proof would not be successful since $f(d)$ would not be in the image, right?

Look up the German Wikipedia pages (and eventually use Chrome to translate the page.) At least the proof of Bolzano-Weierstraß has some nice images that describe the situation:
https://de.wikipedia.org/wiki/Satz_von_Bolzano-Weierstraß#Beweisskizze

We need the boundness to avoid that $d$ slips into the infinities, and we need closeness to avoid that $d$ escapes into loopholes like $\mathbb{R}\backslash \mathbb{Q}$

What would happen if $d$ is irrational?

It's far better today. You could have complained that I posted although I knew that I had a weak day, but I somehow got sucked in. Mathematics is a never-ending sequence of the constantly posed question "why". One has to convince oneself over and over again that the next step in a chain of conclusions is correct. Sometimes you simply get lost somewhere in a trapdoor.

Oh no, I'm just happy that someone is making an effort to answer my questions. It's an extremely valuable opportunity for me to learn, and I could never complain about that.

But, yes, age plays a significant role. Here are two sad, however, prominent examples where age was the problem:
https://www.physicsforums.com/threads/has-the-riemann-hypothesis-been-proven.955832/
https://www.physicsforums.com/threa...-the-history-of-physics.1059261/#post-7005132

Not that I would compare myself with either of the two, never ever, but it shows that it could be worse.

Doing STEM at the age of 89 is a feat in itself

 

[fresh_42]

If the ball is were open, would Bolzano-Weierstraß theorem fail to give a convergent subsequence in the open ball?

Yes. The limit could be on the boundary and therewith outside of the open ball.

And if we have a convergent subsequence such that $\|d\|=1$, the rest of the proof would not be successful since $f(d)$ would not be in the image, right?

Not sure I understand this. The entire proof is about finding a pre-image of the supremum of function values because only a pre-image within the domain of the function is a maximum. We have to find $d\in\{x\, : \,\|x\|=1\}$ such that $f(d)=M=1.$ If $d$ wasn't an element of the domain, than it wouldn't be a maximum, only a supremum. Whether $\|d\|=1$ or $\|d\|<1$ doesn't matter, only $\|d\|\leq 1$ matters, and $f(d)=M$ of course.

What would happen if $d$ is irrational?

Nothing, because we are in $\mathbb{R}^n.$ But we can have rational sequences that are Cauchy sequences but their "limit" could be e.g. $\sqrt{2}\not\in \mathbb{Q}.$ Such a sequence does not converge in $\mathbb{Q}$ although it converges in $\mathbb{R}.$ Topological closeness makes sure that the limit is within the set of considerations.