Challenge Math Challenge - September 2021

[julian]

An idea of a proof of the Bohr-Mollerup theorem:

Spoiler

Let

\begin{align*}
\Gamma_n (x) = \frac{n! n^x}{x (x+1) \cdots (x + n)}
\end{align*}

\begin{align*}
\log \Gamma_n (x) = \log (n!) + x \log n - \sum_{m=0}^n \log (x + m)
\end{align*}

We have

\begin{align*}
\frac{d^2}{dx^2} \log \Gamma_n (x) = \sum_{m=0}^n \frac{1}{(x + m)^2} \geq 0 .
\end{align*}

meaning that it is a convex function. As the limit of a sequence of convex functions is convex we have that the log of $\Gamma (x)$ is convex. $\Gamma (0) = 1$. Also,

\begin{align*}
\Gamma (x + 1) = \lim_{n \rightarrow \infty} \frac{n! n^x n}{(x + 1) (x+2) \cdots (x + n + 1)} = \lim_{n \rightarrow \infty} \frac{n! n^x}{x (x+1) \cdots (x + n)} \cdot x \frac{n}{x + n + 1} = x \cdot \Gamma (x)
\end{align*}

So $\Gamma (x)$ satisfies the required conditions specified in the problem.

Note that

\begin{align*}
\lim_{x \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x) = 0 .
\end{align*}

To attempt to prove uniqueness consider

\begin{align*}
d (x) = \log G(x) - \log \Gamma (x)
\end{align*}

or

\begin{align*}
e^{d (x)} = G (x) / \Gamma (x)
\end{align*}

where $G (x)$ has the same properties as listed in the problem, which we just proved for $\Gamma (x)$.

We easily have that $d(0) = 0$. Also $d (x+1) = d (x)$ and so $d (x)$ is periodic.

Write

\begin{align*}
\frac{d^2}{dx^2} d (x) = \frac{d^2}{dx^2} \log G(x) - \frac{d^2}{dx^2} \log \Gamma (x) .
\end{align*}

Suppose that $d (x)$ is not constant. If we can then show that there exists an $x_0$ such that

\begin{align*}
\frac{d^2}{dx^2} d (x_0) < 0 ,
\end{align*}

then, using the periodicity of $d (x)$ we would have

\begin{align*}
\frac{d^2}{dx^2} d (x_0) & = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} d (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G (x_0 + n) - \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G(x + n) \geq 0
\end{align*}

producing a contradiction. We would then have that $d (x)$ is constant. Since $d(0) = 0$, $d (x) =0$ for all $x$. Implying $G (x) = \Gamma (x)$.

Say we always had that

\begin{align*}
\frac{d^2}{dx^2} d (x) \geq 0 ,
\end{align*}

then this would rule out periodic functions that are convex in places and concave in other places. But there are most likely going to be counter examples to these functions. I think I can think of some already.

I am very tired now. Tomorrow.

[SPLOILER]

 

[MathematicalPhysicist]

Spoiler: Problem 1

$$f(x)=f(x+1)/x=f(x+2)/(x(x+1))=\ldots = f(x+n)/(x(x+1)\ldots (x+n))$$

Spoiler: #1

#1) "Um, I'll take what is the Bohr-Mollerup Theorem for 200 points, Alex."

I want you honest opinions here, what do you think of my new signature? Simple yet amazing truth but is it too long?

Spoiler

I've seen a proof of this theorem in Askey's Special Functions red book.
And I am perplexed, why can you assume that $n$ is a positive integer and $0<x<1$?
It's on page: 35.

 

[julian]

I was hoping @fresh_42 would point out what periodic functions I'm not considering and maybe give a suggestion as to complete the proof I have started.

 

[fresh_42]

Spoiler: Problem 1

$$f(x)=f(x+1)/x=f(x+2)/(x(x+1))=\ldots = f(x+n)/(x(x+1)\ldots (x+n))$$

Spoiler

I've seen a proof of this theorem in Askey's Special Functions red book.
And I am perplexed, why can you assume that $n$ is a positive integer and $0<x<1$?
It's on page: 35.

$n>0$ can be assumed since we consider a limit with $n \to \infty$.
You first show $f(x+n)=(x+n-1)(x-n-2)\cdots (x+1)xf(x).$
Note that $f$ is defined by properties. And these properties are all we are allowed to use. Using too many $\Gamma 's$ makes the proof unreadable!

Next, we have to show with the help of this formula, that for any $y=x+N\in (N,N+1]$ holds $f(y)=\Gamma(y)$ where a) remember that $\Gamma(y)$ is defined as a limit here (see problem statement, and not as the gamma function) and which is b) the first essential part of the proof.

Then we can take log's and show (remember that $\Gamma$ is a limit): $L_n\leq f(x) \leq R_n$ with $L_n,R_n\to \Gamma(x)$.

This is at least the proof structure of my solution. I am skeptical if $\Gamma 's$ are used to make it misty because everybody has the faculty in mind.

The theorem says:
Gauß's definition of the gamma function as a limit is a necessary consequence of the functional requirements.

It's necessity which is the crucial point, not that it suffices.

 

[fresh_42]

I was hoping @fresh_42 would point out what periodic functions I'm not considering and maybe give a suggestion as to complete the proof I have started.

Well, I tried to provide hints in my previous post. I would be happy if you used less $G's$ and $\Gamma 's$ in your prove. This only creates confusion.

 

[julian]

Well, I tried to provide hints in my previous post. I would be happy if you used less $G's$ and $\Gamma 's$ in your prove. This only creates confusion.

The only reason I considered $\Gamma_n (x)$'s was to prove for finite $n$ that these functions are convex, which would then imply that the limit function $\lim_{n \rightarrow \infty} \Gamma_n (x)$ is convex. I was proving that the limit function stated in the problem satisfies the properties stated in the problem.

I then wanted to prove that it is is the only function that satisfies the properties (uniqueness proof part) and so I necessarily had to introduce an arbitrary function $G (x)$ that also satisfies the properties stated in the problem.

(I made a few edits to my partial solution to make it a bit clearer and correct some typos)

 

[julian]

I think that a U-shaped function over a unit interval, repeated would produce a continuous periodic function satisfying $\frac{d^2 d (x)}{dx^2} \geq 0$ as so would spoil the uniqueness part of my proof. I will later attempt to modify my proof to incorporate such functions.

 

[fresh_42]

Would you mind answering the following remarks?

An idea of a proof of the Bohr-Mollerup theorem:

Let

\begin{align*}
\Gamma_n (x) = \frac{n! n^x}{x (x+1) \cdots (x + n)}
\end{align*}

\begin{align*}
\log \Gamma_n (x) = \log (n!) + x \log n - \sum_{m=0}^n \log (x + m)
\end{align*}

We have

\begin{align*}
\frac{d^2}{dx^2} \log \Gamma_n (x) = \sum_{m=0}^n \frac{1}{(x + m)^2} \geq 0 .
\end{align*}

Note to self: there is no restriction on the domain. $x$ is arbitrary here.

meaning that it is a convex function. As the limit of a sequence of convex functions is convex we have that the log of $\Gamma (x)$ is convex. $\Gamma (0) = 1$.

This is why I dislike the use of $\Gamma(x)$ here. We need $\Gamma(1)=1$ to be a solution.

Also,

\begin{align*}
\Gamma (x + 1) = \lim_{n \rightarrow \infty} \frac{n! n^x n}{(x + 1) (x+2) \cdots (x + n + 1)} = \lim_{n \rightarrow \infty} \frac{n! n^x}{x (x+1) \cdots (x + n)} \cdot x \frac{n}{x + n + 1} = x \cdot \Gamma (x)
\end{align*}

So $\Gamma (x)$ satisfies the required conditions specified in the problem.

Note that

\begin{align*}
\lim_{x \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x) = 0 .
\end{align*}

To attempt to prove uniqueness consider

\begin{align*}
d (x) = \log G(x) - \log \Gamma (x)
\end{align*}

Come on! $d$ as a function name where you use differentials, too? Really?

My problem, however, is another one. You have shown (or tried to) that the limit fulfills the functional equations. How can you (logically) use this limit again (hidden in $\Gamma(x)$) to prove uniqueness? All you are allowed to use are the functional equations, not the form of a single solution.

$\Gamma(x)$ shouldn't occur anymore from here on.

Or do you want to show that any other solution is arbitrarily close to the given one? In which metric? Pointwise? It would help a lot if you would explain what you want to do.

or

\begin{align*}
e^{d (x)} = G (x) / \Gamma (x)
\end{align*}

where $G (x)$ has the same properties as listed in the problem, which we just proved for $\Gamma (x)$.

We easily have that $d(0) = 0$. Also $d (x+1) = d (x)$ and so $d (x)$ is periodic.

Write

\begin{align*}
\frac{d^2}{dx^2} d (x) = \frac{d^2}{dx^2} \log G(x) - \frac{d^2}{dx^2} \log \Gamma (x) .
\end{align*}

Suppose that $d (x)$ is not constant. If we can then show that there exists an $x_0$ such that

\begin{align*}
\frac{d^2}{dx^2} d (x_0) < 0 ,
\end{align*}

then, using the periodicity of $d (x)$ we would have

\begin{align*}
\frac{d^2}{dx^2} d (x_0) & = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} d (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G (x_0 + n) - \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log \Gamma (x_0 + n)
\nonumber \\
& = \lim_{n \rightarrow \infty} \frac{d^2}{dx^2} \log G(x + n) \geq 0
\end{align*}

producing a contradiction. We would then have that $d (x)$ is constant. Since $d(0) = 0$, $d (x) =0$ for all $x$. Implying $G (x) = \Gamma (x)$.

Say we always had that

\begin{align*}
\frac{d^2}{dx^2} d (x) \geq 0 ,
\end{align*}

then this would rule out periodic functions that are convex in places and concave in other places. But there are most likely going to be counter examples to these functions. I think I can think of some already.

The idea is correct. Maybe you should consider the following: Assume $d(x)$ is not constant, then there are $x,y>0$ with e.g. $g(y)>g(x).$ By periodicty we may assume $x<y<x+1.$ Now consider the second divided difference
$$
-\dfrac{\log f(x)}{x-y}+\dfrac{\log f(y)}{(y-x)(y-x-1)}-\dfrac{\log f(x+1)}{x-y+1}
$$

 

[MathematicalPhysicist]

$n>0$ can be assumed since we consider a limit with $n \to \infty$.
You first show $f(x+n)=(x+n-1)(x-n-2)\cdots (x+1)xf(x).$
Note that $f$ is defined by properties. And these properties are all we are allowed to use. Using too many $\Gamma 's$ makes the proof unreadable!

Next, we have to show with the help of this formula, that for any $y=x+N\in (N,N+1]$ holds $f(y)=\Gamma(y)$ where a) remember that $\Gamma(y)$ is defined as a limit here (see problem statement, and not as the gamma function) and which is b) the first essential part of the proof.

Then we can take log's and show (remember that $\Gamma$ is a limit): $L_n\leq f(x) \leq R_n$ with $L_n,R_n\to \Gamma(x)$.

This is at least the proof structure of my solution. I am skeptical if $\Gamma 's$ are used to make it misty because everybody has the faculty in mind.

The theorem says:
Gauß's definition of the gamma function as a limit is a necessary consequence of the functional requirements.

It's necessity which is the crucial point, not that it suffices.

And why can we assume that $x<1$?
@fresh_42 I am referring to the proof in Askey's red book on special functions.

I can copy the proof from the book, but should I bother?

 

[fresh_42]

And why can we assume that $x<1$?

You have to show it. I laid out a way to do so.

@fresh_42 I am referring to the proof in Askey's red book on special functions.

I can copy the proof from the book, but should I bother?

Wait for Julian's solution.

 

[Not anonymous]

And here I am getting lost. What is $K$ meant to be? As I see it, we can simply say $K=n$ and $j$ is a counterexample. It is neither the maximal nor the minimal index that violates the condition. You always say "some".

Just as I feared, my answer caused more confusion than clarification with its use of indices. I tried to keep the use of different indices to the minimum but couldn't avoid using $i, j, K, K', n$ with the way I tried to explain. Admittedly, my proof is too verbose and not the most elegant. I had in fact tried to get a more formulaic proof such as by transforming all labels by subtracting the average (similar to your hint, though without $k$ multiplying the average), but didn't think deep enough along that line to get to a proof; and I am not a quick problem solver. The proof I gave is more procedural and came more naturally when I tried to solve, Also, I shouldn't have called it proof by induction, though my method shares some similarities with an inductive proof. Let me try explain my proof in less mathematical terms, through a sequence of logical statements. If there is some error or gap in these statements, please do let me know so that I can learn and correct, and will attempt to solve using the hint you have provided later.

1. For any sequence $A=(a_1, ..., a_n)$, we define $S_k(A) = \sum_{i=1}^k a_i (k=1,...,n)$. We say an index $j$ is a "violating index" if $S_j(A) > j-1$.
2. If the sequence does not have a violating index, then we have already got a sequence that defines the cut in a straightforward way - cut the necklace between $a_n$ and $a_1$ to get the desired string.
3. If $A$ corresponds to labels of the $n$ consecutive beads starting from an arbitrarily chosen bead in the necklace, all we can say is that if it has one or more violating indices, those indices are upper-bounded by $n-1$, since $S_n(A) = n-1$ and so $n$ can never be a violating index. We may denote by $K$ this upper bound, and in this start case, $K=n-1$ (minor difference from earlier posted solution - there I used $K=n$ to denote 1 + upper bound)
4. Now, if $A$ has 1 or more violating indices upper bounded by $K$, we take the smallest violating index $j$ and based on that define a rotated sequence $B = (a_{j+1}, a_{j+2, ..., a_n, a_1, ..., a_j})$. As proved in my earlier post, this rotated sequence cannot have any violating index that falls in the "tail" subsequence $(a_1, ..., a_j)$ (that subsequence that moved from being at the start of $A$ to the end of $B$) or even $a_n$. Any violating index, if present, should lie in the head subsequence $(a_{j+1}, ..., a_{n-1})$, i.e. in that part of the earlier sequence $A$ that has now been shifted left. Hence, the upper bound on violating indices has also shifted left when we formed $B$, so the upper bound index $K'$ (corresponding to $B$) must be smaller than $K$ of $A$.
5. The above observations imply that we will always able to rotate any sequence with a violating index to form another sequence whose violating indices will have a smaller upper bound. Thus, we can start from any randomly chosen bead in the necklace and by performing a finite number of rotations (at most $n-1$, to be more specific), we can reach a sequence that has no violating offset. This sequence gives the desired string and the cut to be performed on the necklace is chosen accordingly.

 

[fresh_42]

Just as I feared, my answer caused more confusion than clarification with its use of indices. I tried to keep the use of different indices to the minimum but couldn't avoid using $i, j, K, K', n$ with the way I tried to explain.

This is ok as long as you rigorously define them. "Some" is insufficient.

Admittedly, my proof is too verbose and not the most elegant. I had in fact tried to get a more formulaic proof such as by transforming all labels by subtracting the average (similar to your hint, though without $k$ multiplying the average), but didn't think deep enough along that line to get to a proof;

The idea with the sum $S(k)$ I mentioned is basically similar to what you did, only without bothering the many possible locations $i,j,K,K',n.$

Observe that $S(0)=S(n)=0$. Then some index has to be the maximum (or minimum, depending on the signs in $S(k).$) The maximum now defines the split, and all you have to do is, to show, that this choice of $x_n$ works.

Very similar to your argument.

and I am not a quick problem solver. The proof I gave is more procedural and came more naturally when I tried to solve, Also, I shouldn't have called it proof by induction, though my method shares some similarities with an inductive proof. Let me try explain my proof in less mathematical terms, through a sequence of logical statements. If there is some error or gap in these statements, please do let me know so that I can learn and correct, and will attempt to solve using the hint you have provided later.

1. For any sequence $A=(a_1, ..., a_n)$, we define $S_k(A) = \sum_{i=1}^k a_i (k=1,...,n)$. We say an index $j$ is a "violating index" if $S_j(A) > j-1$.
2. If the sequence does not have a violating index, then we have already got a sequence that defines the cut in a straightforward way - cut the necklace between $a_n$ and $a_1$ to get the desired string.
3. If $A$ corresponds to labels of the $n$ consecutive beads starting from an arbitrarily chosen bead in the necklace, all we can say is that if it has one or more violating indices, those indices are upper-bounded by $n-1$, since $S_n(A) = n-1$ and so $n$ can never be a violating index. We may denote by $K$ this upper bound, and in this start case, $K=n-1$ (minor difference from earlier posted solution - there I used $K=n$ to denote 1 + upper bound)
4. Now, if $A$ has 1 or more violating indices upper bounded by $K$, we take the smallest violating index $j$ and based on that define a rotated sequence $B = (a_{j+1}, a_{j+2, ..., a_n, a_1, ..., a_j})$. As proved in my earlier post, this rotated sequence cannot have any violating index that corresponds to the "tail" subsequence $(a_1, ..., a_j)$ (that subsequence that moved from being at the start of $A$ to the end of $B$). Any violating index, if present, should lie in the head subsequence $(a_{j+1}, ..., a_n)$, i.e. in that part of the earlier sequence $A$ that has now been shifted left. Hence, the upper bound on violating indices has also shifted left when we formed $B$, so the upper bound index $K'$ (corresponding to $B$) must be smaller than $K$ of $A$.
5. The above observations imply that we will always able to rotate any sequence with a violating index to form another sequence whose violating indices will have a smaller upper bound. Thus, we can start from any randomly chosen bead in the necklace and by performing a finite number of rotations (at most $n-1$, to be more specific), we can reach a sequence that has no violating offset. This sequence gives the desired string and the cut to be performed on the necklace is chosen accordingly.

 

[julian]

I'm extremely tired right now so go easy on me. Let me clarify the method of proof I'm attempting:

----

First I establish that the limit function:

\begin{align*}
\lim_{n \rightarrow \infty} \frac{n! n^x}{x (x+1) \cdots (x + n)}
\end{align*}

satisfies all the conditions stated in the question:

Let $f$ be a function defined on $(0, \infty)$ such that for all $f (x) > 0$ for all $x > 0$

\begin{align*}
\log f (x) \text{ is convex} , \quad f (x+1) = x \cdot f (x) , \quad f(1) = 1 .
\end{align*}

Next I prove that the limit function (which I denote by $\Gamma (x)$) is the only function that satisfies the above conditions. I do this the standard way. I introduce an arbitrary function that satisfies the above conditions and then I form the difference between the logs:

\begin{align*}
d (x) = \log G (x) - \log \Gamma (x)
\end{align*}

and then prove that $d (x) = 0$ for all $x > 0$ (which here is done by first proving that $d (x)$ is a constant). A lot of the time you do this by proof by contradiction. I'm allowed to use properties of the function $\Gamma (x)$ in doing this, if fact you must somehow use properties of $\Gamma (x)$ in doing this.

----

I've used the functional conditions to obtain properties of $d (x)$ and I'm attempting to use these together with asymptotic behaviour of $d^2 \Gamma (x) / dx^2$ (and the assumption that $G (x)$ is convex) to obtain a contradiction thereby proving $d (x)$ to be constant.

(Note: most of the time you consider the difference between the functions rather than the difference between their logs, it doesn't matter here).

 

[julian]

Hi @fresh_42. Here are my comments in relation to post #38.

(a)

I said:

“We have

\begin{align*}
\frac{d^2}{dx^2} \log \Gamma_n (x) = \sum_{m=0}^n \frac{1}{(x + m)^2} \geq 0 .
\end{align*}”

This is because in the question you say the function is defined on $(0 , \infty)$. Are you saying I should have mentioned this in my proof?

(b)

$\Gamma (0) =1$ is a typo.

(c)

$d (x)$ doesn't stand for differential. I choose $d$ to denote the function because I was considering the difference between two functions.

I think I have shown the limit function satisfies the functional conditions. I forgot to mention that the limit function is $> 0$ for $x \in (0 , \infty)$.

(d)

@fresh_42 said:

“Or do you want to show that any other solution is arbitrarily close to the given one? In which metric? Pointwise? It would help a lot if you would explain what you want to do. ”

That is kind of the idea. I'm doing the standard thing of introducing an arbitrary function, $G (x)$, that satisfies the above functional conditions and then forming the difference between the log of $G (x)$ and the log of $\Gamma (x)$:

\begin{align*}
d(x) = log⁡G(x) − log ⁡\Gamma (x)
\end{align*}

and then attempting to prove that $d(x)=0$ for all $x>0$ (which here is done by first proving that $d(x)$ is a constant).Given that the limit function ($=: \Gamma (x)$) is the only function that satisfies the functional conditions, how can you talk about other solutions converging to it? (It has been proven other ways that it is the only function satisfying the functional conditions).

So I'm not attempting to prove that other solutions converge to the limit function ($=: \Gamma (x)$). I'm attempting to outright prove that the limit function is the only solution to the functional conditions. I'm (attempting) to do this by proving that if any alternative solution exists you would get a contradiction.

(e)

I know about polynomial interpolations, including Newton's polynomial and how to establish that its coefficients are the $n-$th divided difference via algebra and proof by induction. I know that in the limit that where point become nodes converge you get $\frac{1}{n!} d^n f (x) / dx^n$ for the $n-$th coefficient. I'm really tired right now. I'll have to think about it later.LAST remark: you are a a lot of the time you are just saying that my explanation could be clearer?

 

[fresh_42]

Hi @fresh_42. Here are my comments in relation to post #38.

(a)

I said:

“We have

\begin{align*}
\frac{d^2}{dx^2} \log \Gamma_n (x) = \sum_{m=0}^n \frac{1}{(x + m)^2} \geq 0 .
\end{align*}”

This is because in the question you say the function is defined on $(0 , \infty)$. Are you saying I should have mentioned this in my proof?

Forget it. I just saw the first differential and wondered where $\log n$ has gone to. It's late here, too.

(b)

$\Gamma (0) =1$ is a typo.

(c)

$d (x)$ doesn't stand for differential. I choose $d$ to denote the function because I was considering the difference between two functions.

I think I have shown the limit function satisfies the functional conditions. I forgot to mention that the limit function is $> 0$ for $x \in (0 , \infty)$.

Yes, you did. Modulo the typo and my disability to do two differentiations in mind.

(d)

@fresh_42 said:

“Or do you want to show that any other solution is arbitrarily close to the given one? In which metric? Pointwise? It would help a lot if you would explain what you want to do. ”

That is kind of the idea. I'm doing the standard thing of introducing an arbitrary function, $G (x)$, that satisfies the above functional conditions and then forming the difference between the log of $G (x)$ and the log of $\Gamma (x)$:

\begin{align*}
d(x) = log⁡G(x) − log ⁡\Gamma (x)
\end{align*}

and then attempting to prove that $d(x)=0$ for all $x>0$ (which here is done by first proving that $d(x)$ is a constant).

Sure. But it is an unlucky choice. You could have used $\Delta$ or $\operatorname{dist}.$

Given that the limit function ($=: \Gamma (x)$) is the only function that satisfies the functional conditions, how can you talk about other solutions converging to it? (It has been proven other ways that it is the only function satisfying the functional conditions).

So I'm not proving that other solutions converge to the limit function ($=: \Gamma (x)$). I'm attempting to outright prove that the limit function is the only solution that satisfies the functional conditions. I'm (attempting) to do this by proving that if any alternative solution exists you would get a contradiction.

(e)

I know about polynomial interpolations, including Newton's polynomial and how to establish that its coefficients are the $n-$th divided difference via algebra and proof by induction. I know that in the limit that where point become nodes converge you get $\frac{1}{n!} d^2 f (x) / dx^2$ for the $n-$th coefficient. I'm really tired right now. I'll have to think about it later.LAST remark: you are a a lot of the time that my explanation could be clearer?

Yes, sorry. That is why your proofs cannot be read without a lot of brainwork or with paper and pencil in hand. And I have had one of the less bright days today.

Reading a proof (which wasn't my version) requires a lot of care in order to get possible loopholes. And if it isn't written for dummies, then it is sometimes quite exhausting. You probably have seen a few textbooks in your life, yet. Did you ever notice that some authors have a writing style that suits you better than others?

Last remark: How do you know that $G(x)$ is twice differentiable? Newton's interpolations avoid that problem.

 

[julian]

I don't know if Randall R. Holmes has written any books but he has written notes. I've gone through a lot of notes and algebra and Galois theory. It is all laid out clearly, not overly wordy, and it's like he often he knows where you brain is going to go next and steps in.

 

[jbergman]

I don't understand what 1 is asking? Are you looking for a proof that f(x) is what you said it was?

 

[fresh_42]

I don't understand what 1 is asking? Are you looking for a proof that f(x) is what you said it was?

Yes. Given a function with properties 1,2,3, prove that it is this limit.

 

[jbergman]

5. A group $G$ together with a topology, such that the mapping on $G\times G$ (equipped with the product topology) to $G$ given by $(x,y)\longmapsto xy^{-1}$ is continuous, is called a topological group (e.g. a Lie group). Prove

1. $G$ is a topological group if and only if inversion and multiplication are continuous.
2. The mappings $x\longmapsto xg$ and $x\longmapsto gx$ are homeomorphisms for each $g\in G.$
3. Each open subgroup $U\leq G$ is closed, and each closed subgroup $U\leq G$ of finite index is open. If $G$ is compact, then each open subgroup is of finite index.
4. Let $H \leq G$ be a subgroup equipped with the subspace topology, $K\trianglelefteq G$ a normal subgroup, and $G/K$ equipped with the quotient space topology. Then $H$ and $G/K$ are again topological groups and the projection $\pi\, : \,G \twoheadrightarrow G/K$ is open.
5. $G$ is Hausdorff if and only if $\{1\}$ is a closed set in $G.$ $G/K$ is Hausdorff for a normal subgroup $K\trianglelefteq G,$ if and only if $K$ is closed in $G.$ If $G$ is totally disconnected, then $G$ is Hausdorff.
6. Let $G$ be a compact topological group and $\{X_j \;(j\in I)\}\subseteq G$ a family of closed subsets such that for all $i ,j\in I$ there is a $k\in I$ with $X_k\subseteq X_i\cap X_j.$ Then we have for any closed subset $Y\subseteq G$
   $$Y\cdot \left(\bigcap_{i\in I}X_i\right) = \bigcap_{i\in I} YX_i $$

1. If $G$ is a topological group then, $inv(g) = (e,g) \longmapsto eg^{-1} = g^{-1}$ is continuous. Also, $(g, inv(h)) \longmapsto gh$ is continuous, since it is the composition of continuous maps. The other direction is trivial because $gh^{-1}$ is again the composition of continuous maps and hence gives the required continuous map for a topological group.

2. The kernel of both maps is empty since group operations are invertible. We can exhibit an inverses $x \longmapsto xg^{-1}$ and $x \longmapsto g^{-1}x$ respectively.

 

[julian]

Just want to make a comment, before moving on to proving that the limit function is the unique solution of the functional conditions.

I wanted to prove that the limit function:

\begin{align*}
\lim_{n \rightarrow \infty} \frac{n! n^x}{x (x+1) \cdots (x + n)} =: \Gamma (x)
\end{align*}

satisfies all the conditions stated in the question. In particular I wanted to prove that the limit function is log-convex. So I began by proving the sequence of functions:

\begin{align*}
\Gamma_n (x) = \frac{n! n^x}{x (x+1) \cdots (x + n)}
\end{align*}

are log-convex. You then can then apply Theorem 1.6 of Artin's notes:

http://inis.jinr.ru/sl/vol1/UH/_Ready/Mathematics/Advanced calculus/Artin E. The Gamma function (1931)(L)(23s).pdf

which states that "A convergent sequence of log-convex functions has a log-convex limit function, provided the limit is positive". These conditions are satisfied and so the limit function is log-convex. Verifying the other functional conditions was straightforward.

I now want to prove that the limit function ($\Gamma (x)$) is the only function that satisfies the functional conditions, but first I need to give some definitions and theorems

Some definitions and theorems:

Given data points $(x_0 , f (x_0)) , (x_1 , f (x_1)) , (x_2, f (x_2))$, the first divided difference is defined by

\begin{align*}
f [x_0,x_1] = \frac{f (x_1) - f (x_0)}{x_1 - x_0} = f [x_1,x_0]
\end{align*}

The second divided difference is defined by

\begin{align*}
f [x_0,x_1,x_2] & = \frac{f [x_1,x_2] - f [x_0,x_1]}{x_2 - x_0}
\nonumber \\
& = \frac{\frac{f (x_2) - f (x_1)}{x_2 - x_1} - \frac{f (x_1) - f (x_0)}{x_1 - x_0}} {x_2 - x_0}
\nonumber \\
& = \frac{f (x_0)}{(x_1 - x_0) (x_2 - x_0)} - \frac{f (x_1)}{(x_2 - x_1) (x_1 - x_0)} + \frac{f (x_2)}{(x_2 - x_1) (x_2 - x_0)} \quad (*)
\nonumber \\
& = \frac{(x_2 - x_1) f (x_0) + (x_0 - x_2) f (x_1) + (x_1 - x_0) f (x_2)}{(x_0 - x_1) (x_1 - x_2) (x_2 - x_0)}
\end{align*}

The last line demonstrates that the second divided difference is invariant under permutation of the arguments $x_0 , x_1 , x_2$.

If we permute $x_0$ and $x_1$ and use $f [x_1 , x_0] = f [x_0 , x_1]$ then we have;

\begin{align*}
f [x_0,x_1,x_2] & = \frac{f [x_0 , x_2] - f [x_0 , x_1]}{x_2 - x_1}
\end{align*}

The function $f (x)$ is convex (on the interval $(a,b)$) if for every number $x_0$ of the interval, $f [x_0 , x_2]$ is a monotonically increasing function of $x_2$. This means that for any pair of numbers $x_2 > x_1$ distinct from $x_0$ the inequality $f [x_0 , x_2] \geq f [x_0 , x_1]$ holds; in other words $f [x_0,x_1,x_2] \geq 0$. Since the value of $f [x_0,x_1,x_2]$ is not changed by permutation of the arguments, the convexity of $f$ is equivalent to the inequality

\begin{align*}
f [x_0,x_1,x_2] \geq 0
\end{align*}

for all triples of distinct numbers on the interval (as explained in Artin's notes, though I'm using slightly different notation). This demonstrates that the "second divided difference" does encode whether a function is convex or not.

We will need the mean value theorem for divided differences:

https://en.wikipedia.org/wiki/Mean_value_theorem_(divided_differences)

For the case $n = 2$ it states that For any $3$ pairwise distinct points $x_0 , x_1 , x_2$ in the domain of a twice differentiable function $f$ there exists an interior point $\xi$

\begin{align*}
\xi \in (\min \{ x_0 , x_1 , x_2 \} , \max \{ x_0 , x_1 , x_2 \})
\end{align*}

where the $2$nd derivative of $f$ equals $2!$ times the $2$nd divided difference at this point:

\begin{align*}
f [x_0,x_1,x_3] = \frac{f'' (\xi)}{2!}
\end{align*}

Proof of uniqueness:

I now prove that the limit function ($\Gamma (x)$) is the only function that satisfies the functional conditions. I do this the standard way. I introduce an arbitrary function, $G (x)$, that satisfies the functional conditions and then I form the difference between the logs:

\begin{align*}
r (x) = \log G (x) - \log \Gamma (x)
\end{align*}

and then prove that $r (x) = 0$ for all $x > 0$. We do this by considering the second divided difference of the above function:

\begin{align*}
r [x_0 , x_1 , x_2] = \log G [x_0 , x_1 , x_2] - \log \Gamma [x_0 , x_1 , x_2]
\end{align*}

and by using the functional conditions, the properties of $\Gamma (x)$, and the mean value theorem for divided differences.

It was proven in a previous post, using the functional conditions, that $r$ is periodic: $r (x+1) = r (x)$ and that $r (1) = 0$. Suppose that $r$ is not constant, then there are $x,y > 0$ with $r(y) > r(x)$. By periodicity we may assume $x < y < x + 1$. Let $x_0 = x$, $x_1 = y$, and $x_2 = x + 1$ in equation $(*)$ where we are taking $f$ to be the function $r$. Then using the periodicity of $r$ we establish the inequality

\begin{align*}
r [x , y , x + 1] & = \frac{r(x)}{y - x} - \frac{r (y)}{(x + 1 - y) (y - x)} + \frac{r (x + 1)}{x + 1 - y}
\nonumber \\
& = \frac{r (x) - r (y)}{(y - x) (x + 1 - y)} < 0
\end{align*}

As $\Gamma (x)$ is log-convex on $(0, \infty)$ we have

\begin{align*}
\log \Gamma [x_0 , x_1, x_2] \geq 0 .
\end{align*}

for all triples of distinct numbers. Same statement applies to the function $G (x)$.

By the periodicity of $r$ we have that

\begin{align*}
r [x , y , x + 1] & = r [x + n , y + n , x + n + 1]
\nonumber \\
& = \log G [x + n , y + n , x + n + 1] - \log \Gamma [x + n , y + n , x + n + 1] .
\end{align*}

Rearranged we have

\begin{align*}
r [x , y , x + 1] + \log \Gamma [x + n , y + n , x + n + 1] = \log G [x + n , y + n , x + n + 1] \qquad (**)
\end{align*}

By the mean value theorem for divided differences, we have that:

\begin{align*}
\log \Gamma [x + n , y + n , x + n + 1] & = \frac{1}{2} \frac{d^2}{dx^2} \log \Gamma (\xi) \qquad \text{for some } \xi \in (x + n , x + n + 1)
\nonumber \\
& \leq \frac{1}{2} \max_{z \in (x + n , x + n + 1)} \frac{d^2}{dx^2} \Gamma (z)
\end{align*}

In a previous post we demonstrated that $\lim_{x \rightarrow \infty} \frac{d^2}{dx^2} \Gamma (x) \rightarrow 0$. By choosing $n$ large enough the LHS of $(**)$ becomes negative and we have a contradiction since $\log G [x + n , y + n , x + n + 1]$ is supposed to be non-negative. Therefore, $r (x)$ must be a constant. Given that $r (1) = 0$ this means that $r (x) = 0$ for all $x \in (0 , \infty)$ and so $\Gamma (x)$ is indeed the unique solution to the functional conditions.

 

[fresh_42]

1. If $G$ is a topological group then, $inv(g) = (e,g) \longmapsto eg^{-1} = g^{-1}$ is continuous. Also, $(g, inv(h)) \longmapsto gh$ is continuous, since it is the composition of continuous maps. The other direction is trivial because $gh^{-1}$ is again the composition of continuous maps and hence gives the required continuous map for a topological group.

2. The kernel of both maps is empty since group operations are invertible. We can exhibit an inverses $x \longmapsto xg^{-1}$ and $x \longmapsto g^{-1}x$ respectively.

Kernel is an unlucky term in this context. The idea can be summarized by $\mu^{-1}_{g}=\mu_{g^{-1}}$ with $\mu_g(x)=xg.$

 

[jbergman]

5. A group $G$ together with a topology, such that the mapping on $G\times G$ (equipped with the product topology) to $G$ given by $(x,y)\longmapsto xy^{-1}$ is continuous, is called a topological group (e.g. a Lie group). Prove

1. (solved by @jbergman ) $G$ is a topological group if and only if inversion and multiplication are continuous.
2. (solved by @jbergman ) The mappings $x\longmapsto xg$ and $x\longmapsto gx$ are homeomorphisms for each $g\in G.$
3. Each open subgroup $U\leq G$ is closed, and each closed subgroup $U\leq G$ of finite index is open. If $G$ is compact, then each open subgroup is of finite index.
4. Let $H \leq G$ be a subgroup equipped with the subspace topology, $K\trianglelefteq G$ a normal subgroup, and $G/K$ equipped with the quotient space topology. Then $H$ and $G/K$ are again topological groups and the projection $\pi\, : \,G \twoheadrightarrow G/K$ is open.
5. $G$ is Hausdorff if and only if $\{1\}$ is a closed set in $G.$ $G/K$ is Hausdorff for a normal subgroup $K\trianglelefteq G,$ if and only if $K$ is closed in $G.$ If $G$ is totally disconnected, then $G$ is Hausdorff.
6. Let $G$ be a compact topological group and $\{X_j \;(j\in I)\}\subseteq G$ a family of closed subsets such that for all $i ,j\in I$ there is a $k\in I$ with $X_k\subseteq X_i\cap X_j.$ Then we have for any closed subset $Y\subseteq G$
   $$Y\cdot \left(\bigcap_{i\in I}X_i\right) = \bigcap_{i\in I} YX_i $$

Moving on to the first part 5.3

If $U$ is an open subgroup in $G$ and $c$ is a limit point of $U$, then the function $f(g) = gc^{-1}$ is a homeomorphism as we showed in 5.2, that also maps $c$ to $e$.

Since $c$ is a limit point of $U$, every open set containing $c$, contains points in $U$.

Now if we take an open set of the identity $V$, then $f^{-1}(V)$ is an open set containing $c$ and points in $U$.

Since $U$ also contains the identity we can take the intersection of an open set in $U$ containing the identity and $V$, call it $W$

However, that means that $hc^{-1}\in U$ for some point $h \in W$ which implies that $U$ contains all of its limit points and is closed, because $h^{-1}$ inverse is also in $U$ which shows that $c^{-1}$ is in $U$ hence $c$ is in $U$.

 

[fresh_42]

Moving on to the first part 5.3

If $U$ is an open subgroup in $G$ and $c$ is a limit point of $U$, then the function $f(g) = gc^{-1}$ is a homeomorphism as we showed in 5.2, that also maps $c$ to $e$.

Since $c$ is a limit point of $U$, every open set containing $c$, contains points in $U$.

Now if we take an open set of the identity $V$, then $f^{-1}(V)$ is an open set containing $c$ and points in $U$.

Since $U$ also contains the identity we can take the intersection of an open set in $U$ containing the identity and $V$, call it $W$

However, that means that $hc^{-1}\in U$ for some point $h \in W$ which implies that $U$ contains all of its limit points and is closed, because $h^{-1}$ inverse is also in $U$ which shows that $c^{-1}$ is in $U$ hence $c$ is in $U$.

A bit complicated.

A shorter version is:
Let $U\leq G$ be an open subgroup, and $g\in G-U.$ Then $gU\subseteq G-U$ is open because left multiplication is a homeomorphism, and
$$
G-U = \bigcup_{g\in G-U}g = \bigcup_{g\in G-U}gU
$$
is a union of open sets, i.e. open, i.e. $U$ is closed.

I will wait for the other parts.

 

[jbergman]

Continuing 5.3.

If $U$ is closed then each coset is also closed, again because left multiplication is a homeomorphism.

Since $U$ has finite index, $G$ is the union of $U$'s cosets. Also, cosets are disjoint, so the complement of any individual coset $gU$ is $G\backslash gU$. But $G\backslash gU$ is just the union of the other cosets which implies it is closed and it's complement must be open, hence $gU$ must be open since it equals the complement.

Still thinking about the compact case.

 

[fresh_42]

Continuing 5.3.

If $U$ is closed then each coset is also closed, again because left multiplication is a homeomorphism.

Since $U$ has finite index, $G$ is the union of

finitely many

$U$'s cosets. Also, cosets are disjoint, so the complement of any individual coset $gU$ is $G\backslash gU$. But $G\backslash gU$ is just the union of the

finitely many

other cosets which implies it is closed and it's complement must be open, hence $gU$ must be open since it equals the complement.

Still thinking about the compact case.

Well, compact means finite subcover, finite index means finitely many cosets of the open subgroup. Now combine these. The finite index is crucial. In formulas it is

Let $A\leq G$ be a closed subgroup of finite index. Then
$$
G-A = \bigcup_{g\in G-A}g = \bigcup_{g\in G-A}gA = \bigcup_{g=1}^n gA
$$
is a finite union of closed sets, hence $G-A$ is closed and $A$ is open.

 

[jbergman]

Thanks, I had the finite in there earlier, but dropped them in my edits.

 

[jbergman]

finitely many
finitely many
Well, compact means finite subcover, finite index means finitely many cosets of the open subgroup. Now combine these. The finite index is crucial. In formulas it is

Let $A\leq G$ be a closed subgroup of finite index. Then
$$
G-A = \bigcup_{g\in G-A}g = \bigcup_{g\in G-A}gA = \bigcup_{g=1}^n gA
$$
is a finite union of closed sets, hence $G-A$ is closed and $A$ is open.

Thanks for the hint. Let $A\leq G$ be an open subgroup of the compact space $G$.

Assume $A$ is not of finite index. Then the cosets of $A$ are again open because left multiplication is a homeomorphism and
$$
G = \bigcup_{g\in G} gA
$$
But since the cosets, $gA$ are distinct and open, they form an open cover of $G$ of which there is no finite subcover which is a contradiction. Hence $A$ must be of finite index.

 

[fresh_42]

Thanks for the hint. Let $A\leq G$ be an open subgroup of the compact space $G$.

Assume $A$ is not of finite index. Then the cosets of $A$ are again open because left multiplication is a homeomorphism and
$$
G = \bigcup_{g\in G} gA
$$
But since the cosets, $gA$ are distinct and open, they form an open cover of $G$ of which there is no finite subcover which is a controduction. Hence $A$ must be of finite index.

Right. Just two remarks. You can write this without contradiction. Simply consider all $gA$. They are an open cover. Therefore, there is a finite subcover. Since all cosets are disjoint, the cosets in the subcover all possible cosets, i.e. finite index. I'm not 100% sure whether this convention is the same in English literature. Here we use $U,V$ or even sometimes $O$ for the open sets, and $A,B$ for the closed ones. Of course this doesn't change anything, but it is easier to read.

 

[jbergman]

Right. Just two remarks. You can write this without contradiction. Simply consider all $gA$. They are an open cover. Therefore, there is a finite subcover. Since all cosets are disjoint, the cosets in the subcover all possible cosets, i.e. finite index. I'm not 100% sure whether this convention is the same in English literature. Here we use $U,V$ or even sometimes $O$ for the open sets, and $A,B$ for the closed ones. Of course this doesn't change anything, but it is easier to read.

Yes, in most of the books I have see $U,V$ for open sets. I am less familiar with the notation for closed sets.

 

[jbergman]

Yes, in most of the books I have see $U,V$ for open sets. I am less familiar with the notation for closed sets.

BTW, you have some nice insight articles! I am going to have to look through them. I've want to get a handle on representation theory.