Challenge Micromass' big August challenge

[disregardthat]

Thank you so much, @disregardthat. I see that you live up to your nickname.
Incidentally, Zorn's lemma (or any other version of A.C.) is unnecessary, see the hint for 2.19(c) ("the lemma") in Billingsley and the corresponding "Notes on the Problems" in the back of the book.

I don't have the book, but I would be interested to see what the hint says. My attempts not using Zorn's lemma didn't lead anywhere.

 

[TeethWhitener]

Here's my weak attempt at advanced number 7. I broke it into a bunch of separate cases. I assume that $r,s>0$ and $\binom{n}{k}=0$ for $n<k$.

First, for the trivial case of $r=s$, we have $r_k = s_k$ for all $k$, and thus
$$\binom {r}{s} = \prod_k {\binom{r_k}{s_k}} = 1$$
So from now on, we'll let $r>s$. For $r,s<p$, we have that $r = r_0$ and $s=s_0$ with all other $r_k, s_k = 0$ and again, the two sides of the congruence are equal even without the modulo. For $r=p$, the left side is $\binom{p}{n}$ and since $p$ is prime, neither $n!$ nor $(p-n)!$ divide $p$ (by the fundamental theorem of arithmetic, neither has a necessary factor of $p$ to cancel out the $p$ in the numerator of the binomial coefficient). So the left side is $0 \bmod p$. The right side is also $0 \bmod p$ since $r_0 = 0$ and $s_0 >0$.

Now we're left with the case when $r>p$. This is where things get a little dodgy. For the time being, I'm going to be completely non-rigorous and ignore my gut and simply manipulate the symbols in the hopes of some deeper insight. The modular expression is equivalent to
$$\binom{r}{s} - \prod_k {\binom{r_k}{s_k}} = 0 \bmod p$$
or
$$\frac{r!\prod_k{s_k!(r_k-s_k)!}-s!(r-s)!\prod_k{r_k!}}{s!(r-s)!\prod_k{s_k!(r_k-s_k)!}} = 0 \bmod p$$
which is satisfied if the numerator is divisible by $p$. Now $r!$ is automatically divisible by $p$ since $r>p$, and if one (or both) of $s$ and $(r-s)$ is greater than or equal to $p$, then the entire numerator is divisible by $p$ and we're done. The remaining case is when both $s$ and $(r-s)$ are less than $p$. We turn to the expression
$$\binom{r}{s} - \prod_k {\binom{r_k}{s_k}} = 0 \bmod p$$
Since $s, (r-s) <p$, we have that $\binom{r}{s}=0\bmod p$ (from $r!$ in the numerator). On the other hand, $s = s_0$ and all the other $s_k =0$. In addition, since $(r-s)<p$, we have that $p<r<2p$, so $r=p+r_0$. This means that $p+r_0-s_0 < p$, or in other words, $r_0<s_0$. This makes the $k=0$ term of the product equal to zero, so the second term on the left side is also $0 \bmod p$.

This would complete the proof, except for the sleight of hand in the expression
$$\frac{r!\prod_k{s_k!(r_k-s_k)!}-s!(r-s)!\prod_k{r_k!}}{s!(r-s)!\prod_k{s_k!(r_k-s_k)!}} = 0 \bmod p$$
Here the issue is that $(r_k-s_k)$ might be negative, naively making the factorial infinite. I'm not sure how to make this rigorous, but I'm worried it involves promoting the factorials to gamma functions and looking at the limiting behavior, in which case, I'll take a pass.

 

[TeethWhitener]

I can kind of patch the hole from my post above (still not wildly confident about it though). Starting with
$$\binom{r}{s}-\prod_k{\binom{r_k}{s_k}} = 0\bmod p$$
multiply by $s!(r-s)!$ to get
$$r!-s!(r-s)!\prod_k{\binom{r_k}{s_k}} = 0\bmod p$$
Then the rest of the proof for the case where $r>p$ and one or both of $s,(r-s)>p$ is the same: since both terms in the difference are divisible by p, the whole thing is $0\bmod p$. This way all the nasty negative factorials stay hidden in the binomial coefficients.

 

[fresh_42]

Well, I'm going to give an outstanding problem a chance (previous advanced #1).

We start by writing $G/H=\{g_iH| 1 \leq i \leq p\}$ with $\left. p \, \right| |G|$ as the smallest of all prime divisors.
The left multiplication $L_x : G/H \rightarrow G/H \; , \; L_x(g_iH) = xg_iH$ defines a group homomorphism from $G$ to $Sym(G/H)$ whose kernel is thus a normal subgroup of $G$ and its image a subgroup of $Sym(G/H)$. This means $im L$ is of finite order and the greatest possible divisor is $p$, because $|Sym(G/H)|=p!$
Since $im L \cong G/ker L$ and $p$ is the smallest prime in $|G|$, all greater primes have to be canceled out by $|ker L|$.
But there is only one $p$ available in $p! \;$, so $|im L| \in \{1,p\}$

$im L = \{1\}$ would imply $L_x(g_iH)=xg_iH=g_iH$ for all $x$ of $G$ and all $i$. However, this would imply $g^{-1}xg \in H$ for all $g \in G$ and $x\in H$, which is obviously wrong for at least one element $x=g_i$.

Thus $im L \cong \mathbb{Z}_p \cong G/ker L$. Since we haven't specified our cosets $g_iH$ so far, we may choose an element $a \in G$ such that $L_a$ generates $im L$ and $a \notin H$. ($a \in H \Rightarrow g_iH = L_{g_i}(H) = L_{a^n}(H) = H$ which cannot be the case for all $i$.)

Now for $x \in ker L$ there is $xH = L_x(H) = id_{G/H} (H) = H$ which means $ker L \subseteq H$ and therefore $ker L = H$ because $|G/H|\,=\,p\,\,=\, |im L| =\,|G/ ker L |$. As a kernel of a group homomorphism $H$ has to be normal.

Let's now consider a group $G$ of order $|G|=pq$ with primes $p \leq q$.
By the previous we have a short exact sequence $H=\mathbb{Z}_q \rightarrowtail G \twoheadrightarrow \mathbb{Z}_p = G/H$
Let $\mathbb{Z}_q = <b\, | \, b^q = 1> \triangleleft \; G$ be the normal subgroup and $\mathbb{Z}_p = <a\, | \, a^p = 1>$
We have seen that $L_x(\mathbb{Z}_q ) = a^n \mathbb{Z}_q$ for some $n$, i.e. we may write $G = \mathbb{Z}_q \times \mathbb{Z}_p$ as sets. Furthermore we have $a\cdot b=b^m\cdot a$ for some $m$, because right and left cosets of $\mathbb{Z}_q = <b>$ are equal. In addition $m$ and the prime $q$ have to be coprime. Thus we have an equation $1=rm+sq$.

Now $\varphi (a)$ with $\varphi (a)(b) = a^{-1}b^ma$ and $\psi(a)$ with $\psi(a)(b) = ab^ra^{-1}$ define two group homomorphisms of $\mathbb{Z}_q$ which are inverses of themselves. Thus $\varphi(a)$ is an automorphism of $\mathbb{Z}_q$ and $\varphi : \mathbb{Z}_p \rightarrow Aut(\mathbb{Z}_q) \cong \mathbb{Z}^*_q \cong \mathbb{Z}_{q-1}$ a group homomorphism.
This establishes a semi-direct product $G \cong \mathbb{Z}_q \rtimes_{\varphi} \mathbb{Z}_p$.

If $\left. p \right| (q-1)$ we get $\frac{q-1}{p}$ many automorphisms in the semi-direct product, which all give rise to isomorphic groups $G$. (I'm not quite sure here. Maybe there are as many isomorphism classes as there are prime factors in $\frac{q-1}{p}$.)

If $p \nmid (q-1)$ then the entire subgroup $\mathbb{Z}_p$ is mapped to the identity in $Aut( \mathbb{Z}_q)$ which covers the case of a direct product $G \cong \mathbb{Z}_q \times \mathbb{Z}_p$.

 

[micromass]

Well, I'm going to give an outstanding problem a chance (previous advanced #1).

We start by writing $G/H=\{g_iH| 1 \leq i \leq p\}$ with $\left. p \, \right| |G|$ as the smallest of all prime divisors.
The left multiplication $L_x : G/H \rightarrow G/H \; , \; L_x(g_iH) = xg_iH$ defines a group homomorphism from $G$ to $Sym(G/H)$ whose kernel is thus a normal subgroup of $G$ and its image a subgroup of $Sym(G/H)$. This means $im L$ is of finite order and the greatest possible divisor is $p$, because $|Sym(G/H)|=p!$
Since $im L \cong G/ker L$ and $p$ is the smallest prime in $|G|$, all greater primes have to be canceled out by $|ker L|$.
But there is only one $p$ available in $p! \;$, so $|im L| \in \{1,p\}$

$im L = \{1\}$ would imply $L_x(g_iH)=xg_iH=g_iH$ for all $x$ of $G$ and all $i$. However, this would imply $g^{-1}xg \in H$ for all $g \in G$ and $x\in H$, which is obviously wrong for at least one element $x=g_i$.

Thus $im L \cong \mathbb{Z}_p \cong G/ker L$. Since we haven't specified our cosets $g_iH$ so far, we may choose an element $a \in G$ such that $L_a$ generates $im L$ and $a \notin H$. ($a \in H \Rightarrow g_iH = L_{g_i}(H) = L_{a^n}(H) = H$ which cannot be the case for all $i$.)

Now for $x \in ker L$ there is $xH = L_x(H) = id_{G/H} (H) = H$ which means $ker L \subseteq H$ and therefore $ker L = H$ because $|G/H|\,=\,p\,\,=\, |im L| =\,|G/ ker L |$. As a kernel of a group homomorphism $H$ has to be normal.

This is ok.

Let's now consider a group $G$ of order $|G|=pq$ with primes $p \leq q$.
By the previous we have a short exact sequence $H=\mathbb{Z}_q \rightarrowtail G \twoheadrightarrow \mathbb{Z}_p = G/H$
Let $\mathbb{Z}_q = <b\, | \, b^q = 1> \triangleleft \; G$ be the normal subgroup and $\mathbb{Z}_p = <a\, | \, a^p = 1>$
We have seen that $L_x(\mathbb{Z}_q ) = a^n \mathbb{Z}_q$ for some $n$, i.e. we may write $G = \mathbb{Z}_q \times \mathbb{Z}_p$ as sets. Furthermore we have $a\cdot b=b^m\cdot a$ for some $m$, because right and left cosets of $\mathbb{Z}_q = <b>$ are equal. In addition $m$ and the prime $q$ have to be coprime. Thus we have an equation $1=rm+sq$.

Now $\varphi (a)$ with $\varphi (a)(b) = a^{-1}b^ma$ and $\psi(a)$ with $\psi(a)(b) = ab^ra^{-1}$ define two group homomorphisms of $\mathbb{Z}_q$ which are inverses of themselves. Thus $\varphi(a)$ is an automorphism of $\mathbb{Z}_q$ and $\varphi : \mathbb{Z}_p \rightarrow Aut(\mathbb{Z}_q) \cong \mathbb{Z}^*_q \cong \mathbb{Z}_{q-1}$ a group homomorphism.
This establishes a semi-direct product $G \cong \mathbb{Z}_q \rtimes_{\varphi} \mathbb{Z}_p$.

If $\left. p \right| (q-1)$ we get $\frac{q-1}{p}$ many automorphisms in the semi-direct product, which all give rise to isomorphic groups $G$. (I'm not quite sure here. Maybe there are as many isomorphism classes as there are prime factors in $\frac{q-1}{p}$.)

If $p \nmid (q-1)$ then the entire subgroup $\mathbb{Z}_p$ is mapped to the identity in $Aut( \mathbb{Z}_q)$ which covers the case of a direct product $G \cong \mathbb{Z}_q \times \mathbb{Z}_p$.

You should work out the case $p~\vert~q-1$ in some more detail. But also, your proof fails in the case $p=q$, I"ll let you find out where.

 

[fresh_42]

You should work out the case $p~\vert~q-1$ in some more detail. But also, your proof fails in the case $p=q$, I"ll let you find out where.

Yep, found it. You caught me cheating on this semi-/ direct part.
I think I have it now. I'll type it in tomorrow.

...
(1st commandment: You shall not post when tired.)

 

[fresh_42]

(group question, continued ...)

The starting point: A group $G$ of order $|G|=pq$ with primes $p \leq q$.
By the previous part we have a short exact sequence $H=\mathbb{Z}_q = <b \,|\, b^q=1> \triangleleft \;\, G \twoheadrightarrow \mathbb{Z}_p = <a \,|\, a^p=1> = G/H$

Since $\mathbb{Z}_q$ is normal in $G$ we have $a\mathbb{Z}_q = \mathbb{Z}_q a$ and may write $b=a^{-1}b^m a$ for some $m$.
We can now define a group homomorphism $\varphi : \mathbb{Z}_p \longrightarrow Aut(\mathbb{Z}_q) \cong \mathbb{Z}^*_q \cong \mathbb{Z}_{q-1}$ by $\varphi(a)(b)=aba^{-1}=b^m$ and get a semi-direct product $G \cong \mathbb{Z}_q \rtimes_\varphi \mathbb{Z}_p$ because $a \cdot b = \varphi(a)(b) \cdot a$
[Remark: Conversely are all such semi-direct products groups of order $pq$. And of course is the direct product also semi-direct with $\varphi = id_{\mathbb{Z}_q}$.]

We now consider $ker \varphi$. Since it is a (normal) subgroup of $\mathbb{Z}_p$ and $p$ is prime, $ker \varphi \in \{1,\mathbb{Z}_p \}$.

If $ker \varphi = \mathbb{Z}_p$ then $\varphi (a) = id_{\mathbb{Z}_q}$ , $ab=ba$ and $G$ is abelian and especially $\mathbb{Z}_p \; \triangleleft \; G$. The only groups with these conditions are the direct product $G = \mathbb{Z}_q \times \mathbb{Z}_p$ and the cyclic group $G = \mathbb{Z}_{pq}$ (which I have previously and mistakenly withheld in post #54). $\mathbb{Z}_q \times \mathbb{Z}_p \cong \mathbb{Z}_{pq}$ in case $p \neq q$, but we get two different groups in case $p=q$. [To prove this, one may consider generating elements again whose orders have to be in $\{p,p^2\}$. Since $G$ is abelian, one doesn't have to bother about normality or the order of multiplication. $\mathbb{Z}_{p^2}$ has an element of order $p^2$ whereas $\mathbb{Z}^2_p$ has only elements of order $p$ (and $1$ of course). In the case $p \neq q$ we get a generator of $\mathbb{Z}_{pq}$ from the product of two generators from $\mathbb{Z}_p$ and $\mathbb{Z}_q$.]

This leaves us with the case $ker \varphi = 1$ or a monomorphism $\varphi : \mathbb{Z}_p \hookrightarrow Aut(\mathbb{Z}_q) \cong \mathbb{Z}_{q-1} =: \{\sigma \,|\, \sigma^{q-1} = 1\}$.
Let us consider two possible groups $G_\varphi := \mathbb{Z}_q \rtimes_\varphi \mathbb{Z}_p$ and $G_\psi := \mathbb{Z}_q \rtimes_\psi \mathbb{Z}_p$ with
$\varphi (a)(b) = b^m \,,\, (m,q)=1$ and $\varphi (a) = \sigma^\mu$
$\psi (a)(b) = b^n \,,\, (n,q)=1$ and $\psi (a) = \sigma^\nu$
We extend the generator $\sigma$ of $Aut(\mathbb{Z}_q)$ to a group homomorphism of (all) $G$ by setting $\hat{\sigma} \vert_{\mathbb{Z}_q} = \sigma$ and $\hat{\sigma}\vert_{\mathbb{Z}_p} = id_{\mathbb{Z}_p}$.
This gives us an isomorphism $\hat{\sigma}^{\mu-\nu} : G_\psi \longrightarrow G_\varphi$
$$a \cdot_{\varphi} b = \varphi (a)(b) \cdot_{\varphi} a = \hat{\sigma}^{\mu}(b) \cdot_{\varphi} a = \hat{\sigma}^{\mu-\nu} \psi (a) (b)\cdot_{\varphi} \hat{\sigma}^{\mu-\nu} (a) = \hat{\sigma}^{\mu-\nu} (a \cdot_{\psi} b)$$
and all proper semi-direct products ${\mathbb{Z}_q} \rtimes {\mathbb{Z}_p}$ are isomorphic. [The direct product with $\varphi \equiv 1$ is ruled out here by the condition on the kernel $\ker \varphi = 1$.]
- hopefully I haven't missed something again -

 

[TeethWhitener]

@micromass Was mine just wrong (posts 52, 53)?

 

[micromass]

@micromass Was mine just wrong (posts 52, 53)?

Nope, it was ok. I just missed it.

 

[member 587159]

4 high school questions)

Consider the finite sequence: k! + 2, k! + 3, k! + 4, ..., k! + k-1, k! + k with k>1. Then it's obvious that there are no primes in this sequence. Because, using the definition of the factorial, we have the equivalent sequence: k*(k-1)*(k-2)*...*2*1 + 2, k*(k-1)*(k-2)*...*3*2*1 + 3, ... , k*(k-1)*(k-2)*...*2*1 + k-1, k*(k-1)*(k-2)*...*2*1 + k
and 2 divides the first term, 3 divides the third term, ...

Now take, k = 2017, then we have the sequence:

2017! + 2, 2017! + 3, ... , 2017! + 2016, 2017! + 2017. This sequence has 2016 consectutive integers in it and none of them are prime. So QED.

Note: we can use the same reasoning to show that 'prime gaps' can be as large as we want them to be.

 

[member 587159]

5a high school questions) Find all the 10-digit numbers such that each digit {0,1,2,3,4,5,6,7,8,9} is exactly used once.

If I understand the question correctly, I have to find how many numbers there are satisfying this condition.

So obviously, 0 can't be the first digit of a number. So, for the first digit, we have 9 possibilities. For the second digit, we have 9 possibilities left (now we can choose 0 as well). For the third digit, 8 possibilities, ...

Ultimately, using the product rule, we have 9*9*8*7*6*5*4*3*2*1 = 3265920 possibilities where each digit is exactly used once.

 

[mfb]

I'm quite sure the numbers should satisfy (a) and (b) together, not separately.

The list in the problem statement seems to have a typo in it, and I don't understand why 9 is not listed.

 

[micromass]

I'm quite sure the numbers should satisfy (a) and (b) together, not separately.

This is right.

The list in the problem statement seems to have a typo in it, and I don't understand why 9 is not listed.

Thank you.

 

[mfb]

The 1 is still in there twice.
Unless I missed follow-up posts to this post by @Biker, we don't have all possible points for high school #2 yet.

Challenges for high school:
2-
I will just pretend that the Earth is a perfect sphere.
If you move close to the south pole, You can mark a circle with a circumference of 1 mile around the earth. Now move 1 mile further to the south and and face the north.
Move one mile north and you are on the circle, Move one mile west and you will come back to the same place (assuming west from your initial location) move 1 mile south and you are back to the same place.

and you can be on the north pole, as the Earth is a sphere you will form a triangle that will leads you back to the same place (Disregarding the method I suggested) or you could use the same method but with minor changes at the north pole

Who needs a gps :P

 

[Erland]

I have been trying to solve Advanced Problem no 4 for a long time now, but it seems impossible. Conditions 1 and 5 seem impossible to reconcile. I have been trying an Axiom of Choice (Zorn's Lemma) approach, but I don't get anywhere. Perhaps this is a bad approach...

 

[micromass]

I have been trying to solve Advanced Problem no 4 for a long time now, but it seems impossible. Conditions 1 and 5 seem impossible to reconcile. I have been trying an Axiom of Choice (Zorn's Lemma) approach, but I don't get anywhere. Perhaps this is a bad approach...

It does require the axiom of choice. But it's not equivalent to it.

If it's not found by the beginning of september, I'll repost the question in the new challenge thread with a hint.

 

[S.G. Janssens]

It does require the axiom of choice.

I think it requires a result that depends on AC, but not AC directly? At least that is how I know the proof of this problem. (I will stick by the rules and not say more.)

 

[micromass]

I think it requires a result that depends on AC, but not AC directly? At least that is how I know the proof of this problem. (I will stick by the rules and not say more.)

That's correct.

 

[mfb]

Hmm, the approach I was investigating does not need the AC, so I guess there is some counterexample to my definition. Can someone find one?

4. Let $X$ denote the set of all bounded real-valued sequences. Prove that a generalized limit exists for any such sequence. A generalized limit is any function function $L:X\rightarrow \mathbb{R}$ such that
1) $L((x_n + y_n)_n) = L((x_n)_n) + L((y_n)_n)$
2) $L((\alpha x_n)_n) = \alpha L((x_n)_n)$
3) $\liminf_n x_n \leq L((x_n)_n)\leq \limsup_n x_n$
4) If $x_n\geq 0$ for all $n$, then $L((x_n)_n)\geq 0$.
5) If $y_n = x_{n+1}$, then $L((x_n)_n) = L((y_n)_n)$
6) If $x_n\rightarrow x$, then $L((x_n)_n) = x$.

My first idea was to define a modified sequence where the nth element is the average of the first n elements, and then look for that limit.
$$y_n = \frac{1}{n} \sum_{i=1}^n x_n$$
$$L((x_n)_n) = \lim_{n \to \infty} y_n$$

It should be clear that this satisfies all 6 criteria if that limit exists. That catches all convergent sequences and many divergent sequences, but it does not catch sequences like (1,-1,-1,1,1,1,1, then 8 times -1, 16 times 1 and so on) where the limit does not exist. Let's repeat this step:

$$z_n = \frac{1}{n} \sum_{i=1}^n y_n = \frac{1}{n} \sum_{i=1}^n \frac{1}{i} \sum_{k=1}^i x_i$$

Now you can come up with a series where this does not converge either, and I can add a third step, ... but here is the idea: for each sequence, repeat this step as often as necessary to get convergence, and then define this limit as generalized limit of the sequence.

I didn't find a proof that every sequence has to lead to convergence after a finite number of steps.

 

[Erland]

mfb I have been thinking along the same lines. But I didn't consider more than one step, also that the averages need not converge... So you're smarter than me :)

 

[micromass]

I didn't find a proof that every sequence has to lead to convergence after a finite number of steps.

Right, and you can't find such a proof. It's a good idea though, but it will fail to provide a generalized limit on every sequence. Note that generalized limits are not unique either!

 

[mfb]

If there is no such proof (and if you say that I trust you), then there should be a counterexample, but I don't find one.

 

[micromass]

If there is no such proof (and if you say that I trust you), then there should be a counterexample, but I don't find one.

It might actually be so that AC might need to be invoked to find a counterexample. I know for sure your statement is false with AC though, but I will not show it until this question is solved.

 

[micromass]

This doesn't spoil anything, but it's a counterexample to the claim made by mfb: http://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-cesàro-mean

 

[mfb]

This doesn't spoil anything, but it's a counterexample to the claim made by mfb: http://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-cesàro-mean

Thanks. So the point is to blow up the length of the sub-sequences fast enough. Probably faster than e^e^x or something like that (didn't check it in detail).

 

[fresh_42]

Thanks. So the point is to blow up the length of the sub-sequences fast enough. Probably faster than e^e^x or something like that (didn't check it in detail).

This sounds like transforming the issue until you end up transcending something. (At least this would give us a reason for AC.)
I was thinking about another idea: All difficulties arise either from linearity or (if this is solved) from $\underline{\lim} \leq L \leq \overline{\lim}$.
How about putting AC in a Hamel basis (if needed at all) and define $L(x_n)$ as a certain derivation? (Don't know whether this is a real idea or simply a crude thought.)

 

[mfb]

Hamel basis is a good keyword.

We can solve parts of the linearity by directly using this linearity in the definition. Find a generalized limit in [-1,1] for all sequences where $\liminf_n x_n = -1$ and $\limsup_n x_n = 1$. Then scale sequence and generalized limit linearly to find the generalized limit of sequences which have other liminf and limsup. This doesn't work for convergent series, where we set the generalized limit to the regular limit separately.

Now the interesting question: does this procedure survive condition (1), the limit of the sum of two sequences? I guess this depends on the definition. The critical point here is a cancellation of accumulation points, e.g. (0,2,-1,2,0,2,-1,...) + (0,-1,0,0,0,-1,0,0,0,-1,...) and (0,2,-1,2,0,2,-1,...) + (0,-1,0,0,0,-1,0,0,0,-1,...) have different limsup but need the same limit (condition 5 plus condition 1).

 

[fresh_42]

I'll have a try on #4, although I admit that my solution hides AC in a general result, which I won't prove here. And my first idea hasn't been that far from what I have (not exactly a derivation, but something linear instead). The whole problem is to connect the algebraic property of linearity with the topological property of the $\lim \inf$ and $\lim \sup$.

I'll start with a description of Hewitt, Stromberg, Real and Abstract Analysis, Springer in which the reals are defined as the cosets of a Cauchy sequence $\mathfrak{C}$ modulo null sequences $\mathfrak{N}$. Both are subsets of all bounded real sequences $\mathfrak{B}$.

With $\mathfrak{N} \subset \mathfrak{C} \subset \mathfrak{B}$ we get $\mathbb{R} \cong \mathfrak{C} / \mathfrak{N} \subset \mathfrak{B} / \mathfrak{N} =: \mathfrak{A}$ a real (topological, infinite dimensional, commutative) $\mathbb{R}-$algebra $\mathfrak{A}$ (with $1$).

The linear function $L$ can now be viewed as a linear functional on $\mathfrak{A}$.
I will use the sub-additivity $\overline{\lim}((x_n)+(y_n)) \leq \overline{\lim} (x_n)+\overline{\lim} (y_n)$ and $\underline{\lim}(x_n) = - \overline{\lim}(-x_n)$

Now here's the trick and the point where the real work is done:
If we define $L(x_n)\vert_{\mathfrak{C} / \mathfrak{N}} := \lim(x_n)$ then $L(x_n) = \lim(x_n) \leq \overline{\lim}(x_n)$ holds on $\mathfrak{C} / \mathfrak{N} \subset \mathfrak{A} \;$, is linear and we may apply the theorem of Hahn-Banach. (Its proof uses Zorn's Lemma and therefore AC.)
This theorem guarantees us the existence of a linear functional $L$ on $\mathfrak{A}$ which extends our $L$ defined on $\mathfrak{C} / \mathfrak{N} \;$.

Linearity is already given by Hahn-Banach.
With that we have $L(x_n) \leq \overline{\lim}(x_n)$ by the theorem and
$$L(x_n)=-L(-x_n) \geq -\overline{\lim}(-x_n) = \underline{\lim}(x_n)$$
For the cutting process in point 5) we define a sequence $z_n$ by $z_1=x_1$ and $z_n=x_{n} - y_{n-1}=0$ for $n > 1$.
Then $L(z_n)=L(x_1,0, \dots) = 0 = L(x_n) - L(0,y_1,y_2, \dots) = L(x_n) - L(y_n)$ and 5) is equivalent to $L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

For $0 \leq x_n$ we have $0 \leq \underline{\lim} (x_n) \leq L(x_n)$; and for $\lim (x_n) = x$ we have $x = \underline{\lim} (x_n) \leq L(x_n) \leq \overline{\lim} (x_n) = x$ and get equality everywhere. Therefore 4) and 6) hold.

 

[Erland]

Nice fresh42, I should have thought of Hahn-Banach...

 

[micromass]

Here is the solution for #2 of the previously unsolved advanced challenges. I got this from the excellent probability book from Feller, which contains a lot of other gems.

Take the random distributions of $r$ balls in $n$ cells assuming that each arrangement has probability $1/n^r$. We seek the probability$p_m(r,n)$ that exactly $m$ cells are empty.

Let $A_k$ be the event that cell number $k$ is empty. In this even all $r$ balls are placed in the remaining $n-1$ cells and this can be done in $(n-1)^r$ different ways. Similarly, there are $(n-2)^r$ arrangments leaving two preassigned cells empty. By the inclusion-exclusing rule we get that the probability that all cells are occupied is
p_0(r,n) = \sum_{\nu = 0}^n (-1)^\nu \binom{n}{\nu} \left(1-\frac{\nu}{n}\right)^r
Consider a distribution in which exactly $m$ cells are empty. These $m$ cells can be chosen in $\binom{n}{m}$ ways. The $r$ balls are distributed among the remaining $n-m$ cells so that each of these cells is occupied. The number of such distributions is $(n-m)^r p_0(r,n-m)$. Dividing by $n^r$ we find for the probability that exactly $m$ cells remain empty
p_m(r,n)=\binom{n}{m}\sum_{\nu=0}^{n-m} (-1)^\nu \binom{n-m}{\nu} \left(1-\frac{m+\nu}{n}\right)^r

Now we wish to find an asymptotic distribution. First note that
(n-\nu)^r&lt;n(n-1)...(n-\nu+1)&lt;n^r
Thus we have
n^\nu\left(1 - \frac{\nu}{n}\right)^{\nu+r}&lt;\nu!\binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r &lt; n^\nu\left(1-\frac{\nu}{n}\right)^r
For $0<t<1$, it is clear that $t<-\log(1-t)<\frac{t}{1-t}$ and thus that
\left(n e^{-(\nu+r)/(n-\nu)}\right)^\nu &lt;\nu!\binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r &lt; \left(ne^{-r/n}\right)^\nu
Now we put $\lambda = ne^{-r/n}$ and we assume that $r$ and $n$ increase in such a way that $\lambda$ remained constrained to a finite interval $0<a<\lambda<b$. For each fixed $\nu$ the ratio of the extreme members in the inequality tends to unity and so
0\leq \frac{\lambda^\nu}{\nu!} - \binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r\rightarrow 0
This relation holds trivially when $\lambda\rightarrow 0$ and hence it remains true whenever $r$ and $n$ increase such a way that $\lambda$ remains bounded. Now
e^{-\lambda} - p_0(r,n) =\sum_{\nu=0}^{+\infty} (-1)^\nu\left(\frac{\lambda^\nu}{\nu!} - \binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r\right)
and the right hand side tends to zero. We therefore have that for each $m$ fixed
p_m(r,n) - e^{-\lambda}\frac{\lambda^m}{m!}\rightarrow 0

 

[Erland]

There seems to be an error in the given proof of Advanced Problem 4:

I'll have a try on #4, although I admit that my solution hides AC in a general result, which I won't prove here. And my first idea hasn't been that far from what I have (not exactly a derivation, but something linear instead). The whole problem is to connect the algebraic property of linearity with the topological property of the $\lim \inf$ and $\lim \sup$.

I'll start with a description of Hewitt, Stromberg, Real and Abstract Analysis, Springer in which the reals are defined as the cosets of a Cauchy sequence $\mathfrak{C}$ modulo null sequences $\mathfrak{N}$. Both are subsets of all bounded real sequences $\mathfrak{B}$.

With $\mathfrak{N} \subset \mathfrak{C} \subset \mathfrak{B}$ we get $\mathbb{R} \cong \mathfrak{C} / \mathfrak{N} \subset \mathfrak{B} / \mathfrak{N} =: \mathfrak{A}$ a real (topological, infinite dimensional, commutative) $\mathbb{R}-$algebra $\mathfrak{A}$ (with $1$).

The linear function $L$ can now be viewed as a linear functional on $\mathfrak{A}$.
I will use the sub-additivity $\overline{\lim}((x_n)+(y_n)) \leq \overline{\lim} (x_n)+\overline{\lim} (y_n)$ and $\underline{\lim}(x_n) = - \overline{\lim}(-x_n)$

Now here's the trick and the point where the real work is done:
If we define $L(x_n)\vert_{\mathfrak{C} / \mathfrak{N}} := \lim(x_n)$ then $L(x_n) = \lim(x_n) \leq \overline{\lim}(x_n)$ holds on $\mathfrak{C} / \mathfrak{N} \subset \mathfrak{A} \;$, is linear and we may apply the theorem of Hahn-Banach. (Its proof uses Zorn's Lemma and therefore AC.)
This theorem guarantees us the existence of a linear functional $L$ on $\mathfrak{A}$ which extends our $L$ defined on $\mathfrak{C} / \mathfrak{N} \;$.

Linearity is already given by Hahn-Banach.
With that we have $L(x_n) \leq \overline{\lim}(x_n)$ by the theorem and
$$L(x_n)=-L(-x_n) \geq -\overline{\lim}(-x_n) = \underline{\lim}(x_n)$$
For the cutting process in point 5) we define a sequence $z_n$ by $z_1=x_1$ and $z_n=x_{n} - y_{n-1}=0$ for $n > 1$.
Then $L(z_n)=L(x_1,0, \dots) = 0 = L(x_n) - L(0,y_1,y_2, \dots) = L(x_n) - L(y_n)$ and 5) is equivalent to $L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

For $0 \leq x_n$ we have $0 \leq \underline{\lim} (x_n) \leq L(x_n)$; and for $\lim (x_n) = x$ we have $x = \underline{\lim} (x_n) \leq L(x_n) \leq \overline{\lim} (x_n) = x$ and get equality everywhere. Therefore 4) and 6) hold.

The problem lies in this line:

$L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

It is not clear what you mean by $\Vert \, . \Vert_{\infty}$. Normally, it would mean that $\Vert \, (x_n) _n\Vert_{\infty}=\sup_n|x_n|$, but to make sense here, I suppose it means $\Vert \, (x_n)_n\Vert_{\infty}=\overline\lim |x_n|$.
Anyway, it is certainly wrong that $\mathfrak{C} / \mathfrak{N}$ is dense in $\mathfrak{A}$ with respect to this. For example, let $x_n=(-1)^n$ ($n=1,2,\dots$). Then, there is no Cauchy sequence $(y_n)_n$ in $\mathbb R$ such that $\Vert (y_n-x_n)_n\Vert_\infty< 1$, for both these possible definitions of $\Vert \, . \Vert_{\infty}$.
Thus, $\mathfrak{C} / \mathfrak{N}$ is not dense in $\mathfrak{A}$.

So, the proof that 5) holds for the extended functional fails. And in fact, there is no obvious reason that an extended functional, which is proved to exist by Hahn-Banach's Theorem, should satisfy 5) just because the original (unextended) functional does.

Advanced Problem 4 must therefore still be considered as open (at least for us) until this is fixed.

 

[micromass]

@fresh_42 do you have an answer to this? If not I will put it back as an open problem.

 

[fresh_42]

How about an algebraic argument?

Let $S$ denote the linear operator that inserts a zero in the first position, i.e. $S(y_1,y_2, \dots ) = S(0,y_1,y_2, \dots )$

Then $L_0 : S.(\mathfrak{C}/\mathfrak{N}) \rightarrow \mathbb{R}$ with $L_0(0,(x_n)) := L((0,(x_n))$ is a linear functional on $\{(0,x_n) \vert (x_n) \in \mathfrak{C}/\mathfrak{N}\}$ which is identical to $L$ on Cauchy sequences ($L$ as defined in the solution above) since $L(0,x_n) = L(x_n)$ for Cauchy sequences $(x_n)$.

Especially $L_0$ is majorized by $L$, i.e. $L_0(0,(x_n)) \leq L(x_n)$. Thus $L_0$ can be linearly extended on $S.\mathfrak{A}$ by Hahn-Banach. (The fact, that the first component is zero, makes all sets linear subspaces.)
Thus $L_0(0,(y_n)) \leq L(y_n)$ for all $(0,(y_n)) \in S.\mathfrak{A}$. But this can only be true if it is an equation, because we can take $(-y_n)$ and linearity to turn it into $L_0(0,(y_n)) \geq L(y_n)$. Hence $L_0(0,(y_n)) = L(y_n)$ for all $(y_n) \in \mathfrak{A}$ and $L_0$ can be expanded by $L_0((y_n)) := L(y_n)$. The new $L_0$ then automatically fulfills all properties.

Remark: I hope there is no circular conclusion in it. And I assume there is a topological proof, too, as $S$ is linear, has norm $1$ and is thus uniformly continuous. One could probably even take a $\mathfrak{C}/\mathfrak{N}-$ Hamel basis of $\mathfrak{A}$ to extend $LS$, because $1 \in \mathfrak{A}$ can be chosen as first basis vector. The important property of Hahn-Banach is the majorizing sublinear function.

 

[Erland]

Ok, it's late night in Sweden and I'm tired, but I still can't see that you solved the problem. Your notation is a little confusing, but as far as I can see, you actually did nothing but defined $L_0$ on $\mathfrak A$ by $L_0(y_1,y_2,y_3,...)=L(y_2,y_3,y_4,...)$, and I don't see that $L_0$ satisfies 5).

Please, correct me if I am wrong...

 

[fresh_42]

Ok, it's late night in Sweden and I'm tired, but I still can't see that you solved the problem. Your notation is a little confusing, but as far as I can see, you actually did nothing but defined $L_0$ on $\mathfrak A$ by $L_0(y_1,y_2,y_3,...)=L(y_2,y_3,y_4,...)$, and I don't see that $L_0$ satisfies 5).

Please, correct me if I am wrong...

I'm afraid we share the same time zone ...

The argument goes as follows:
(I only have to consider the case $L(0,y_1,y_2, \dots )= L(y_1,y_2, \dots)$ due to linearity (see my first post on this). And this equation is already true for Cauchy sequences.)

I take the function $L$ as previously defined, i.e. it has all properties except (5).
This means it is linear and defined on all bounded sequences (modulo the ideal of null sequences) which I denote as $\mathfrak{A}$.
$\mathbb{R} \cong \mathfrak{C}/\mathfrak{N}$ are the Cauchy sequences (modulo null sequences), which are a subspace of $\mathfrak{A}$.

Now $\{0\} \times \mathfrak{A}$ defines a vector space with $\{0\} \times \mathfrak{C}/\mathfrak{N}$ as a subspace.
$L_0(0,(x_n)) := L(0,(x_n)) = L((x_n)) \leq L((x_n))$ defines a linear functional on this subspace.

I now apply the theorem of Hahn-Banach (without property (5)) to $L_0$ which is majorized by the linear and therewith sublinear functional $L$ on $\{0\} \times \mathfrak{C}/\mathfrak{N}$ to get a linear extension of $L_0$ on $\{0\} \times \mathfrak{A}$.

Thus $L_0(0,(y_n)) \leq L((y_n))$ for all $(0,y_n) \in \{0\} \times \mathfrak{A}$.
Therefore $L_0(0,-(y_n)) \leq L(-(y_n))$ and $L_0(0,(y_n)) \geq L((y_n))$ because $L_0$ and $L$ are linear.
I then have $L_0(0,(y_n)) = L((y_n))$ on all $(0,y_n) \in \{0\} \times \mathfrak{A}$ which means on all $(y_n) \in \mathfrak{A}$.

Now applying Hahn-Banach again to $\{0\} \times \mathfrak{A} \subset \mathfrak{A}$ and $L_0(0,(y_n)) = L((y_n))$ gives with the same argument / trick as above a linear functional $L'_0$ on $\mathfrak{A}$ with $L'_0(0,(y_n)) = L_0(0,(y_n)) = L((y_n))$ and $L'_0((y_n)) = L((y_n))$.

(Maybe the last argument can stand alone without the previous $L_0$ but as you said, it's too late.)

 

[Erland]

How do you get $L'_0((y_n))=L((y_n))$?

 

[fresh_42]

How do you get $L'_0((y_n))=L((y_n))$?

The same way as before.
$L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ on $\{0\} \times \mathfrak{A}$.
Hahn-Banach gives $L'_0((y_n)) \leq L((y_n))$ on $\mathfrak{A}$.
Now $L'_0(-(y_n)) \leq L(-(y_n))$ and by linearity $L'_0((y_n)) \geq L((y_n))$, i.e. $L'_0((y_n)) = L((y_n))$ on $\mathfrak{A}$.
Since $L'_0$ extends $L$ (5) still holds.

Edit: But you have been absolutely right. I tried to get to $L'_0$ in one step and had trouble with an egg-chicken kind of problem in defining it (and which I didn't even realize in the first place). The trick with the subspace $\{0\} \times \mathfrak{A} \subset \mathfrak{A}$ seems to be an easy way out of this circle. In addition I have to thank you for the insights into the power of this theorem. I've found quite a bit different versions and applications on the way. And I now know that Banach is the keyword of first choice whenever topology meets linearity.

 

[Erland]

Perhaps I am slow-witted, but I just don't get it.

The same way as before.
$L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ on $\{0\} \times \mathfrak{A}$.
Hahn-Banach gives $L'_0((y_n)) \leq L((y_n))$ on $\mathfrak{A}$.

How can you go from $L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ to $L'_0((y_n)) \leq L((y_n))$? I suppose you use that $L'_0(0,(y_n))=L_0(0,(y_n))$, but then, you don't have the same upper estimation functional before and after you apply Hahn-Banach (unless $L$ already satisfies 5). To get an upper estimate by $L((y_n))$ after Hahn-Banach, you should have $L(0,(y_n))$ before Hahn-Banach, while in fact you have $L((y_n))$, which is not the same if $L$ does not satisfy 5).

 

[fresh_42]

I'm not sure I understand you.
First I construct an $L$ on $\mathfrak{C}/\mathfrak{N}$ which satisfies all properties, (5) including.
Then I expand $L$ on $\mathfrak{A}$ say $L_1$.
$L_1$ satisfies all properties except (5).
Now I take $L_1$ on $\{0\} \times \mathfrak{C}/\mathfrak{N} \subset \{0\} \times \mathfrak{A}$ with $L_1 (0,(y_n)) \leq L(y_n)$.
Then I expand $L_1$ on $\{0\} \times \mathfrak{A}$, say $L_2$ which still satisfies $L_2(0,(y_n)) \leq L_1(0,(y_n)) = L(y_n)$.
And now I expand $L_2$ from $\{0\} \times \mathfrak{A}$ on $\mathfrak{A}$, say $L_3$.
It still satisfies $L_3(y_n) \leq L(y_n)$ and its restriction on $\{0\} \times \mathfrak{A}$ is $L_2$

At every step linearity guarantees that "$\leq$" is actually an equation and $L_3(y_n) = L(y_n) = L_2(0,(y_n)) = L_3(0,(y_n))$ by construction.

 

[Erland]

I'm not sure I understand you.
First I construct an $L$ on $\mathfrak{C}/\mathfrak{N}$ which satisfies all properties, (5) including.
Then I expand $L$ on $\mathfrak{A}$ say $L_1$.
$L_1$ satisfies all properties except (5).
Now I take $L_1$ on $\{0\} \times \mathfrak{C}/\mathfrak{N} \subset \{0\} \times \mathfrak{A}$ with $L_1 (0,(y_n)) \leq L(y_n)$.
Then I expand $L_1$ on $\{0\} \times \mathfrak{A}$, say $L_2$ which still satisfies $L_2(0,(y_n)) \leq L_1(0,(y_n)) = L(y_n)$.
And now I expand $L_2$ from $\{0\} \times \mathfrak{A}$ on $\mathfrak{A}$, say $L_3$.
It still satisfies $L_3(y_n) \leq L(y_n)$ and its restriction on $\{0\} \times \mathfrak{A}$ is $L_2$

At every step linearity guarantees that "$\leq$" is actually an equation and $L_3(y_n) = L(y_n) = L_2(0,(y_n)) = L_3(0,(y_n))$ by construction.

You write "It still satisfies $L_3(y_n) \leq L(y_n)$", meaning that it did that before. But, as far as I can see, $L_2$ does not need to satisfy $L_2(y_n) \leq L(y_n)$ for $(y_n)\in \{0\} \times \mathfrak{A}$. What it satisfies is $L_2(0,(y_n)) \leq L(y_n)$.

A correct application of Hahn-Banach would lead from $L_2(0,(y_n)) \leq L(y_n)$ to something like $L_3(x,(y_n))\le L(y_n)$, for $x\in \mathbb R$, as I understand it.

Btw, I have no problem with your clever trick of making an equality into an equation. It is the original inequality I doubt.

 

[fresh_42]

$L_2$ isn't even defined on $\mathfrak{A}$ and you are right. I only have $L_2(0,y_n) = L(y_n)$.
But $L_3$ is defined on all $\mathfrak{A}$ with $L_2$ being its restriction on the subspace $\{0\} \times \mathfrak{A}$.
$L$ always serves as the (sub)linear majorizing function, and linearity to get rid of the $\leq$.
So in the end $L_3(0,y_n) = L_2(0,y_n) = L(y_n) = L_3(y_n)$.

It is the original inequality I doubt.

Why? For Cauchy sequences we have convergence and $L$ is the usual limit which doesn't change, if we put a zero at the start.

 

[Erland]

So in the end $L_3(0,y_n) = L_2(0,y_n) = L(y_n) = L_3(y_n)$.

I don't understand why $L_3(y_n)$ should be equal to the other three expressions.

 

[fresh_42]

I don't understand why $L_3(y_n)$ should be equal to the other three expressions.

Because $L_3$ is constructed as extension of $L_2 \leq L$ from $(0,\mathfrak{A})$ on $\mathfrak{A}$.
The restriction makes it equal to $L_2$ which is still equal to $L$ and the expansion makes it equal to $L$. Equality is actually never lost in the process.

Maybe I've overlooked a logical pit, although I can't see one and I admit it looks a bit like cheating.
I'm still convinced that there might be an easier way to show it. Perhaps with linearity and continuity of the shift-operator and a limit procedure.
Also Hewitt, Stromberg remark on their proof of Hahn-Banach that it could be proven with linear extensions of a Hamel basis to the cost of the upper bounding sublinear function. The latter requires the existence of a minimal element by Zorn's Lemma. Since we have equality in our case, a simple linear extension could be enough. But I'm not very used to Hamel basis so I might not see the traps there.

 

[Erland]

Because $L_3$ is constructed as extension of $L_2 \leq L$ from $(0,\mathfrak{A})$ on $\mathfrak{A}$.
The restriction makes it equal to $L_2$ which is still equal to $L$ and the expansion makes it equal to $L$. Equality is actually never lost in the process.

But you didn't have equality in the first place! $L_2$ is not equal to $L$ (restricted to $\{0\}\times \mathfrak A$). You have $L_2(0,(y_n))=L(y_n)$, not $L_2(y_n)=L(y_n)$ and not $L_2(0,(y_n))=L(0,(y_n))$ (at least I don't see you proved it), because $(y_n)$ and $(0,(y_n))$ is not the same thing.

 

[fresh_42]

But you didn't have equality in the first place! $L_2$ is not equal to $L$ (restricted to $\{0\}\times \mathfrak A$). You have $L_2(0,(y_n))=L(y_n)$, not $L_2(y_n)=L(y_n)$ and not $L_2(0,(y_n))=L(0,(y_n))$ (at least I don't see you proved it), because $(y_n)$ and $(0,(y_n))$ is not the same thing.

That is right. But I don't see where I would need it.

I have
a) $L(0,y) = L(y)$ on $C/N \quad$ [property of limits]
b) $L_1(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_1 \vert_{C/N} = L \leq L$ for $C/N \subset A$]+[linearity]
c) $L_1(0,y) = L(0,y) = L(y)$ on $(0,C/N) \subset A \quad$ [first equation in $A$, second in $C/N$]
d) $L_2(0,y) = L(y)$ on $(0,A) \quad$ [Hahn-Banach with $L_2 \vert_{(0,C/N)} = L_1 \leq L$ for $(0,C/N) \subset (0,A)$]+[linearity]
e) $L_3(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_3 \vert_{(0,A)} = L_2 \leq L$ for $(0,A) \subset A$]+[linearity]

Now $L_3(y) = L(y)$ on $A$ and $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.
Since both are equal to $L(y) \;$, $\; L_3(0,y) = L_3(y)$ for all $y \in A$.

 

[Erland]

That is right. But I don't see where I would need it.

I have
a) $L(0,y) = L(y)$ on $C/N \quad$ [property of limits]
b) $L_1(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_1 \vert_{C/N} = L \leq L$ for $C/N \subset A$]+[linearity]
c) $L_1(0,y) = L(0,y) = L(y)$ on $(0,C/N) \subset A \quad$ [first equation in $A$, second in $C/N$]
d) $L_2(0,y) = L(y)$ on $(0,A) \quad$ [Hahn-Banach with $L_2 \vert_{(0,C/N)} = L_1 \leq L$ for $(0,C/N) \subset (0,A)$]+[linearity]
e) $L_3(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_3 \vert_{(0,A)} = L_2 \leq L$ for $(0,A) \subset A$]+[linearity]

Now $L_3(y) = L(y)$ on $A$ and $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.
Since both are equal to $L(y) \;$, $\; L_3(0,y) = L_3(y)$ for all $y \in A$.

But as you write it now, you define $L_1=L$ on $C/N$ and extend with Hahn-Banach so that $L_1=L$ on all of $A$, so there is no point in defining $L_1$ at all. Likewise with the definitions of $L_2$ and $L_3$. All of them are defined as equal to the previous one at some subspace, and then it turns out that it is equal to the previous one at all of $A$. So, $L_3=L_2=L_1=L$, and nothing is gained.

But the real error, this time, is in d). It does not follow from your application of Hahn-Banach there that $L_2(0,y)=L(y)$ on $(0,A)$, it just follows that $L_2(0,y)=L(0,y)$ for all $y\in A$. But the conclusion of d) is essential for the argument, in the line $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.

In your previuos version you had $L_2(0,y)=L(y)$ for $y\in A$, but then, you couldn't obtain $L_2(0,y)=L(0,y)$ instead. You simply cannot prove that $L_3(0,y)=L_3(y)$ anywhere outside $C/N$ (and likewise for $L_1$ and $L_2$).

 

[fresh_42]

Thank you. A closer look on how to apply Hahn-Banach finally showed me the trap I've fallen into. It doesn't matter whether there is a difference between my previous or my last version, which I did not intend to be. At some point I trick myself while losing the $0$. It doesn't matter where.
So let me think about one of my other ideas with Hamel basis and continuity, resp. a limit procedure.
Sorry for your inconvenience. At least it showed that the most "obvious" results are often the hardest to prove. (Of course mostly only until one has found the solution.)

 

[Erland]

Ok, we all make mistakes...

My idea is to look at the proof of Hahn-Banach's Theorem and see if it could be modified, for this case, so that 5) holds. No success yet...

 

[fresh_42]

Ok, we all make mistakes...

My idea is to look at the proof of Hahn-Banach's Theorem and see if it could be modified, for this case, so that 5) holds. No success yet...

The problem with my approach is, that $L$ is already fixed. However, $L$ is linear, continuous and behaves well on $\mathbb{R}$ and $||L||_{\infty} = 1$. There must be a way to show it. I mean what more can I hope for?
Another approach could be by starting with the shift invariance, linearity and $||\, .\, ||_{\infty}$ as sublinear upper bound, apply Hahn-Banach and deduce the rest. Perhaps it's easier this way round.

 

[Erland]

Ok, I did a Google search and found a paper by Abraham Robinson, from which it is clear that generalized limits can be constructed by using nonstandard analys (several of them, thus solving also Advanced Problem 7 a, in the September Challenge, and probably b, c, and d can also be solved by this method). This proof would be really simple, if one is aquainted to nonstandard analysis, that is . Unfortunately, I cannot give the proof here, since that would be cheating.

And the axiom of choice shows up indirectly here too, in the construction of a nonstandard extension.