Math Challenge - August 2020

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  • #51
Math_QED
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This implies there exists [itex]c\in\mathbb R[/itex] such that [itex]F = I_{[c,\infty]}[/itex].
Explain this line.
 
  • #52
PeroK
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Hmm, I need a little more details here. Explain why the discontinuities arise. I find your argument a bit too handwavy.
Effectively it boils down to the discontinuity of ##\arg z##. I thought that was trivial and well established!
 
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  • #53
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We can kill a fly with a fly swatter. Don't need the rifle.
Suppose [itex]g:\mathbb C^* \to\mathbb C[/itex] is continuous and satisfies the identity [itex]e^{g(z)} = z[/itex]. Since [itex]g[/itex] is right inverse to [itex]\exp[/itex], it is injective, thus [itex]g(\mathbb C^*) \cong \mathbb C^*[/itex]. But now [itex]\exp[/itex] is forced to be injective on [itex]\mathbb C^*[/itex]. Indeed, for any [itex]u,v\neq 0[/itex] we have
[tex]e^u = e^v \Leftrightarrow e^{g(z)} = e^{g(w)} \Leftrightarrow z=w \Rightarrow u=v.[/tex]
But that's impossible.
Ok, here are the details. Firstly, since [itex]g[/itex] is injective, it means [itex]g(\mathbb C^*) \cong \mathbb C^*[/itex] as sets. That means any [itex]u\neq 0[/itex] can be written uniquely as [itex]u=g(z)[/itex]. But we also have a homemorphism with the obvious choice ( this is forced, really) [itex]f(g(z)) := z,\ z\neq 0[/itex]. For continuity of [itex]f:g(\mathbb C^*) \to \mathbb C^*[/itex] suppose [itex]g(z_n) \to g(z)[/itex], then continuity of exponential map implies [itex]e^{g(z_n)} \to e^{g(z)}[/itex] which by assumption is the same as [itex]z_n\to z[/itex] so [itex]f[/itex] is continuous.

What fails if [itex]g[/itex] is not continuous: it wouldn't be a morphism in the category with continuous maps, so the above won't apply.
 
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  • #54
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Explain this line.
By definition of limit. As [itex]x\to -\infty[/itex] we must have [itex]A>0[/itex] such that [itex]x\leq -A[/itex] implies [itex]F(x) = \mathbb P\{X\leq x\}=0[/itex]. Since such [itex]A[/itex] are bounded from below, take the infimum i.e [itex]c := \inf \{x\in\mathbb R \mid X\leq x\}[/itex]. Now, whatever happens for [itex]x>c[/itex] must occur with probability [itex]1[/itex] (otherwise [itex]c[/itex] wouldn't be the infimum). There can be no in-betweens so [itex]F = I_{[c,\infty]}[/itex] is forced due to right continuity of [itex]F[/itex] i.e [itex]F(c+) = 1[/itex].
 
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  • #55
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Suffices to show [itex]C(f) = \{x\in X \mid f\text{ is continuous at }x\}[/itex] is a Borel set, it can also be empty. Then [itex]C(f)^c = D(f)[/itex] is a Borel set. Let's chase epsilons from the definition of continuity. [itex]f[/itex] is continuous at [itex]x\in X[/itex] if and only if
[tex]
\forall n\in\mathbb N,\ \exists \delta >0,\ \forall z\in X,\quad z\in B(x,\delta) \Rightarrow f(z) \in B\left (f(x), n^{-1}\right ).
[/tex]
In other words we have [itex]\{x\}= \bigcap _{n\in\mathbb N} \{U\subseteq X \mid x\in U,\ U\text{ is open and }f(U) \subseteq B(f(x), n^{-1}) \}[/itex]. Do this for all points of continuity, then we have the [itex]G_\delta[/itex] set
[tex]
\bigcap _{n\in\mathbb N} \bigcup \left\{ U\subseteq X \mid U\text{ is open, }\ \sup_{u,v\in U} d_Y(f(u),f(v)) \leq n^{-1} \right \} = C(f).
[/tex]
The equality should be clear.
 
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  • #56
Infrared
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That means any [itex]u\neq 0[/itex] can be written uniquely as [itex]u=g(z)[/itex].
Why?
 
  • #57
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Why?
Oops, that's bad expression by me. I really had in mind that [itex]u[/itex] is identified uniquely with [itex]g(z)[/itex]. But now [itex]e^u = e^v \Rightarrow e^{g(z)} = e^{g(w)}[/itex] might break. Thanks for noticing. The homemorphism part still works, but usable if [itex]g[/itex] is an inclusion as @Math_QED said.

I'll stick to my rifle at #32 for now. Yet I'm not convinced continuity is a required assumption. I get a feeling [itex]e^{g(z)} =z[/itex] forces [itex]g[/itex] to be continuous and in this case, holomorphic.
 
  • #58
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Yet I'm not convinced continuity is a required assumption. I get a feeling [itex]e^{g(z)} =z[/itex] forces [itex]g[/itex] to be continuous and in this case, holomorphic
It doesn't. ##g(z)=\ln|z|+\arg(z)## is a counterexample
 
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  • #59
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It doesn't. ##g(z)=\ln|z|+\arg(z)## is a counterexample
Of course *smacks forehead*. I had a thought maybe [itex]|z|\leq |e^z|[/itex], but that's not true in [itex]\mathbb C[/itex].
 
  • #60
PeroK
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Hmm, I need a little more details here. Explain why the discontinuities arise. I find your argument a bit too handwavy.
We have a function ##\beta (\theta)## for ##0 \le \theta < 2\pi## with ##\cos \beta (\theta) = \cos \theta## and ##\sin \beta (\theta) = \sin \theta##. We know that ##\beta(0) = 2\pi n## for some ##n##. As adding a constant does not effect the continuity of a function we can, wlog, take ##\beta(0) = 0##.

The technical point outstanding is that if ##\beta## is continuous, then ##\beta(\theta) = \theta##.

In general, we have ##\beta(\theta) = \theta + 2\pi n(\theta)## with ##n(0) = 0##, as above.

The only continuous functions of the form ##2\pi n(\theta)## are constant functions. I'll spare everyone an epsilon-delta proof of that. Therefore, we must have ##g(z) = i\theta = i\arg z## for ## 0 \le \theta < 2\pi##.

The second technical point is that ##\arg(z)## is discontinuous on the unit circle. I'll quote that as a known result.

In any case, that proves that ##g## cannot be continuous.
 
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  • #61
11. Let ##a < b < c < d## be real numbers. Sort ##x = ab + cd, y = bc + ad, z = ac + bd## and prove it.
The sorted order is ##y < z < x##, which is proved as follows by first proving 2 pairwise inequalities.

$$
y - z = (bc + ad) - (ac + bd) = c(b-a) + d(a-b) = (c-d)(b-a) < 0
\Rightarrow (y - z) < 0 \Rightarrow y < z
$$
where ##(c-d)(b-a) < 0## follows from the fact that ##c < d## and ##b > a## as per the given conditions.

$$
x - z = (ab + cd) - (ac + bd) = a(b-c) + d(c-b) = (d-a)(c-b) > 0
\Rightarrow (x - z) > 0 \Rightarrow x > z
$$
where ##(d-a)(c-b) > 0## follows from the fact that ##d > a## and ##c > b## as per the given conditions.

Combining the 2 inequalities gives the ordering inequality ##x > z > y##.
 
  • #62
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Problem 8.
I first state the Bohr-Mollerup theorem,
Let ##f:(0, \infty ) \rightarrow \mathbb R^+## be a function satisfying:
(i) ##f(x+1)=xf(x)##.
(ii) f is a log-convex function.
(iii) ##f(1)=1##.
##f(x)=\Gamma (x)## on it's domain.

##\Gamma## is meromorphic. The identity theorem states: If two meromorphic functions in ##\mathbb C## agree on a set with a limit point in ##\mathbb C##, then they agree everywhere in ##\mathbb C##. In particular, two meromorphic functions that agree on ##(0, \infty )## agree everywhere on ##\mathbb C##.

Thus condition (i) above holds in the complex plane. The two meromorphic functions, ##\Gamma (z+1)## and ##z\Gamma (z)##, that agree on ##(0, \infty )## implies ##\Gamma (z+1)=z\Gamma (z)## for all ##z \in \mathbb C##.
Wielant's theorem (https://www.jstor.org/stable/2975370) states that condition (ii) "f is a log-convex function" can be replaced by "##f(x)## is in the bounded strip ##\{ z\in \mathbb C | 1 \leq \mathfrak{ R}(z) \leq 2\}##".
Because ##F(z)## is in the bounded strip ##\{ z\in \mathbb C | 1 \leq \mathfrak{ R}(z) \leq 2\}## it satisfies the Bohr-Mollerup theorem, extended to the complex plane, up to a real constant ##F(1) = a## and thus ##F(z)=F(1)\Gamma (z)##.
 
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  • #63
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Problem 8.
I first state the Bohr-Mollerup theorem,
Let ##f:(0, \infty ) \rightarrow \mathbb R^+## be a function satisfying:
(i) ##f(x+1)=xf(x)##.
(ii) f is a log-convex function.
(iii) ##f(1)=1##.
##f(x)=\Gamma (x)## on it's domain.

##\Gamma## is meromorphic. The identity theorem states: If two meromorphic functions in ##\mathbb C## agree on a set with a limit point in ##\mathbb C##, then they agree everywhere in ##\mathbb C##. In particular, two meromorphic functions that agree on ##(0, \infty )## agree everywhere on ##\mathbb C##.

Thus condition (i) above holds in the complex plane. The two meromorphic functions, ##\Gamma (z+1)## and ##z\Gamma (z)##, that agree on ##(0, \infty )## implies ##\Gamma (z+1)=z\Gamma (z)## for all ##z \in \mathbb C##.
Wielant's theorem (https://www.jstor.org/stable/2975370) states that condition (ii) "f is a log-convex function" can be replaced by "##f(x)## is in the bounded strip ##\{ z\in \mathbb C | 1 \leq \mathfrak{ R}(z) \leq 2\}##".
Because ##F(z)## is in the bounded strip ##\{ z\in \mathbb C | 1 \leq \mathfrak{ R}(z) \leq 2\}## it satisfies the Bohr-Mollerup theorem, extended to the complex plane, up to a real constant ##F(1) = a## and thus ##F(z)=F(1)\Gamma (z)##.
Well, problem #8 IS Wielandt's theorem. You just shifted the problem to Bohr-Mollerup. This is a bit like proving AC by Zorn.

There is still another - in a way elementary - proof possible. It uses a common theorem of complex analysis.
 
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  • #64
Infrared
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Of course *smacks forehead*. I had a thought maybe [itex]|z|\leq |e^z|[/itex], but that's not true in [itex]\mathbb C[/itex].
Nor is it true in ##\mathbb{R}## (negative numbers)!
 
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  • #65
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SPOILER #9:


g would have to restrict a continuous injection from the unit circle to the line {it: all t in R}, which is impossible. I.e. g is injective since e^g is injective, and the only complex numbers z with e^z lying on the unit circle are of form z = iy with y real, so g would hve to rstrict to a continuous injection from the unit circle to the "imaginary real line" of complex numbers of form iy with y real. Now it is immediate that a continuous map from the unit circle to a copy of the real line cannot be injective, since it has a maximum say M at p and a minimum say m at q, and then both arcs joining p to q on the circle must map onto the same interval [m,M], by the intermediate value theorem, so g is not injective.

With a larger weapon, one could say that since g is locally inverse to exp, it must be smooth if it is continuous, so using the strong smooth jordan curve theorem, it maps the unit circle isomorphically onto a smooth manifold, whose interior in the complex plane serves as its boundary. Then integrating the pullback of the closed form dtheta, gives a contradiction of the sort suggested earlier using Cauchy's theorem to integrate dz/z. I.e. since the pullback is closed, the integral of d of it over the interior of the manifold is zero, but since the boundary is parametrized by g, which pulls back the form dtheta to itself, the integral is 2π, contradicting stokes theorem.

More abstractly, these maps, if they existed, would induce a group homomorphism of fundamental groups, or of 1st homology groups, whose composition would be the identity map Z-->Z, while nonetheless factoring though the zero group Z-->0-->Z, an impossibility. Or as Bott put it long ago, to prove there is no such map, all you need is "a homotopy invariant functor that does not vanish on the circle".

By covering space theory, which is essentially the same argument, since exp is a covering space of C*, via C-->C*, a map C*-->C* can only factor through exp if it induces a map on fundamental groups, whose image is zero, not the case for the identity map. The existence of g would also violate unique path lifting, since the parametrization t-->e^2πit of the circle is via a lift t-->2πit, through the exponential covering that sends the two ends points 0 and 2π to different points of C, while the existence of g would give a lift that sends them both to the same point. This is the essential content of the earlier answer that any lift via g must be discontinuous as a map on the circle.

Oh yes, and now I see that my hint amounted to noticing that such a map g exists, only if one also exists for the restricted diagram: S^1-->iR-->S^1.

here is another similar argument. If such a g existed, its restriction to the unit circle would factor the injection from the unit circle to C*, through C, and hence would prove that this injection is homotopic to a constant. Then Cauchy's theorem, the homotopy version, would imply that any holomorphic differential in C* would integrate to zero over the unit circle, but that violates the integral of dz/z being 2πi, as pointed out earlier.

By the way this last argument, as well as those using fundamental group and 1st homology group, prove that even if you are allowed to replace exp by any continuous map of your choice, there still is no such map g. The fact that the given map is indeed exp, allows the more elementary first argument above, which only uses the intermediate value theorem.
 
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  • #66
benorin
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@Fred Wright Nice solution for #8, I had thought of the Bohr-Mollerup theorem but didn't know how to prove log convexity.
 
  • #67
Math_QED
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We have a function ##\beta (\theta)## for ##0 \le \theta < 2\pi## with ##\cos \beta (\theta) = \cos \theta## and ##\sin \beta (\theta) = \sin \theta##. We know that ##\beta(0) = 2\pi n## for some ##n##. As adding a constant does not effect the continuity of a function we can, wlog, take ##\beta(0) = 0##.

The technical point outstanding is that if ##\beta## is continuous, then ##\beta(\theta) = \theta##.

In general, we have ##\beta(\theta) = \theta + 2\pi n(\theta)## with ##n(0) = 0##, as above.

The only continuous functions of the form ##2\pi n(\theta)## are constant functions. I'll spare everyone an epsilon-delta proof of that. Therefore, we must have ##g(z) = i\theta = i\arg z## for ## 0 \le \theta < 2\pi##.

The second technical point is that ##\arg(z)## is discontinuous on the unit circle. I'll quote that as a known result.

In any case, that proves that ##g## cannot be continuous.
This is much more readable to me than your previous post. Your approach has all the right ideas. Especially the line "The only continuous functions of the form ##2\pi n(\theta)## are constant functions." is the key to an elementary approach that does not use black magic. I consider this question solved. Well done!
 
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  • #68
benorin
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@Math_QED +1 style point (and a like) for using the phrase "black magic" in reference to mathematics.
 
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  • #69
Math_QED
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For every [itex]x\in\mathbb R[/itex] we have [itex]\mathbb P\{X\leq x\}\in \{0,1\}[/itex]. Let [itex]F[/itex] be the distribution function for [itex]X[/itex]. [itex]F[/itex] is right continuous and we have [itex]\lim _{x\to\infty} F(x) = 1[/itex] and [itex]\lim _{x\to -\infty}F(x) = 0[/itex]. This implies there exists [itex]c\in\mathbb R[/itex] such that [itex]F = I_{[c,\infty]}[/itex]. Now [itex]\mathbb P\{X<c\} = F(c-) = 0[/itex], therefore [itex]X=c[/itex] a.s.
Your first solution + the clarification in post #54 solves the question (I didn't look at the second one though)! Well done! I guess I must come up with less routine exercises since you seem to solve all of them ;)
 
  • #70
benorin
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@Infrared I wish to know... is #2 really as innocent as it looks? I was thinking IVT might be the only thing I need... on the other hand, it is bivariate (in a way) but the notion of continuity for many variables seems too strong for this problem. FYI you have the right to respond with your best "poker face."
 
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  • #71
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There is still another - in a way elementary - proof possible. It uses a common theorem of complex analysis.
My spider sense is tingling. [itex]\Gamma (z) / F(z)[/itex] is entire. Liouville's theorem? Or maybe the theorem that allows finit order entire functions without zeros be written as [itex]e^{P(z)}[/itex] for some polynomial [itex]P[/itex]. (Hadamard?)
I guess I must come up with less routine exercises since you seem to solve all of them ;)
It's not like I look at the exercise and come up with a solution. At some point I've done similar things, so it's a matter of reminding myself definitions/results and relevant techniques. As you can also see, I make mistakes, so it's not smooth sailing. Most of what I do is wrong, I have stacks of papers scribbled full of some gibberish and failed attempts at the problems in OP.
 
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  • #72
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For question 15, should it be understood that it simply means that there is no solution where ##p, q, r \in \mathbb{Q}##?
 
  • #73
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My spider sense is tingling. [itex]\Gamma (z) / F(z)[/itex] is entire. Liouville's theorem? Or maybe the theorem that allows finit order entire functions without zeros be written as [itex]e^{P(z)}[/itex] for some polynomial [itex]P[/itex]. (Hadamard?)
Liouville, but not for the quotient.
 
  • #74
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For question 15, should it be understood that it simply means that there is no solution where ##p, q, r \in \mathbb{Q}##?
Yes. There are certainly real solutions, but none with three rational numbers.
Hint: Reduce the question to an integer version of the statement.
 
  • #75
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@Math_QED +1 style point (and a like) for using the phrase "black magic" in reference to mathematics.
I think he gets that from me. I've been calling complex analysis black magic, because..well..it is :D
That said, @Math_QED , I gave a black magic approach in #32.
 
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