Math Challenge - May 2020

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  • #26
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I was looking for separated solutions too, just in the form of a product instead of a sum (no joy). But wouldn't such a possibility (sum) be included in my power series solution though? I used double power series thinking that the only restriction I imposed was that I wouldn't be able to identify solutions not having a power series expansion
Yes, they should be included. The trick is to use the conditions such that the series break down early. We have something like ##f'+g''=0## and if the variables are separated, we can simply integrate.
 
  • #28
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inductive limit topology:)
Would you mind to define the open sets explicitly?
 
  • #29
benorin
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6 b) Answer derived using separable sum for each ##u_i## I got is
$$u(x_1,x_2,x_3,t)=(x_1^2+2t)(x_2^2+2t)x_3$$
and I checked by hand that this indeed solves both the initial conditions and the heat equation. @fresh_42 your ansatz made it really, really easy.
 
  • #30
benorin
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I have a Q regarding the alternate solution of #3: is ##\int_{-\infty}^{+\infty}\dfrac{| x\sin(\alpha x) |}{1+x^2}\,dx## (absolutely) convergent? Because I don't think it is, tell you why after some quick work. (Btw, using the upper bound of ##1## for sine yields a divergent integral).
I was attempting the differentiation under the integral sign solution:
Let $$I(\alpha ) :=\int_{-\infty}^{+\infty}\dfrac{\cos(\alpha x)}{1+x^2}\,dx \quad (\alpha \geq 0)$$
Then $$I(\alpha )-\dfrac{d^2I}{d\alpha ^2}=\int_{-\infty}^{+\infty}\dfrac{\cos (\alpha x)}{1+x^2}\,dx+\int_{-\infty}^{+\infty}\dfrac{x^2\cos(\alpha x)}{1+x^2}\,dx = \int_{-\infty}^{+\infty}\cos (\alpha x)\, dx$$
where the latter integral is obviously divergent. But @cbarker1 's solution stated that
$$I(\alpha )=\pi e^{-\alpha}\Rightarrow I(\alpha )-\dfrac{d^2I}{d\alpha ^2}=0$$
so I'm thinking that the second application of differentiation under the integral sign was not justified? The first application was clearly justified as ##I(\alpha )## is absolutely convergent. Also note that attempted DE's with the first derivative lead to some not so easy integrals, ugh.
 
  • #31
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I solved it by foot, only using symmetry properties, substitutions and integration by parts, and Fubini. My result is the same.
 
  • #32
mathwonk
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Question on #7: did you want a non - zero quotient module?
 
  • #34
benorin
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1. Let ##1<p<4## and ##f\in L^p((1,\infty))## with the Lebesgue measure ##\lambda##. We define ##g\, : \,(1,\infty)\longrightarrow \mathbb{R}## by
$$
g(x)=\dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\,d\lambda(t).
$$
Show that there exists a constant ##C=C(p)## which depends on ##p## but not on ##f## such that ##\|g\|_2 \leq C\cdot \|f\|_p## so ##g\in L^2((1,\infty)).## (FR)

Work: Let ##\tfrac{1}{p}+\tfrac{1}{q}=1## and note that ##1<p<4\Rightarrow \tfrac{4}{3}<q<\infty##.
$$\begin{gathered} \| g \| _2^2 =\int_1^\infty\left| \dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\, d\lambda(t)\right| ^2\, d\lambda(x) \\ \leq \int_1^\infty \dfrac{1}{x^2} \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ \end{gathered} $$
by Holder's inequality, also note that ##\forall x\in (1,\infty ),\quad \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\leq \| f \| _p^2## hence it follows that
$$\begin{gathered} \| g \| _2^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \tfrac{q}{4-q}\left( 10^\tfrac{4-q}{4}-1\right) \| f \| _p^2 \int_1^\infty x^{\tfrac{2}{q}-3}\, d\lambda(x) \\ \end{gathered} \\ =\tfrac{q^2}{2(4-q)(1-q)}\left( 1-10^\tfrac{4-q}{4}\right) \| f \| _p^2 $$

and the desired result comes upon taking square roots of both sides a remembering that ##q=q(p)## (q may be thought of as a function of ##p##).
Edit: post contains an error!
 
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  • #35
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I think you made an integration error, but I'm not sure since you haven't shown what you've done. And what if ##q=4##?
 
  • #36
benorin
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I think you made an integration error, but I'm not sure since you haven't shown what you've done. And what if ##q=4##?
Sorry it was late and I wanted to post it before bed... I'll show my work this time :)

1. Let ##1<p<4## and ##f\in L^p((1,\infty))## with the Lebesgue measure ##\lambda##. We define ##g\, : \,(1,\infty)\longrightarrow \mathbb{R}## by
$$
g(x)=\dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\,d\lambda(t).
$$
Show that there exists a constant ##C=C(p)## which depends on ##p## but not on ##f## such that ##\|g\|_2 \leq C\cdot \|f\|_p## so ##g\in L^2((1,\infty)).## (FR)

Work: Let ##\tfrac{1}{p}+\tfrac{1}{q}=1## and note that ##1<p<4\Rightarrow \tfrac{4}{3}<q<\infty##.
$$\begin{gathered} \| g \| _2^2 =\int_1^\infty\left| \dfrac{1}{x}\int_x^{10x}\dfrac{f(t)}{t^{1/4}}\, d\lambda(t)\right| ^2\, d\lambda(x) \\ \leq \int_1^\infty \dfrac{1}{x^2} \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ \end{gathered} $$
by Holder's inequality, also note that ##\forall x\in (1,\infty ),\quad \left\{ \int_x^{10x} | f | ^p\, d\lambda(t)\right\} ^{\tfrac{2}{p}}\leq \| f \| _p^2## hence it follows that
$$\begin{gathered} \| g \| _2^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\left[ \tfrac{t^{1-\tfrac{q}{4}}}{1-\tfrac{q}{4}}\right|_{t=x}^{10x} \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ =\tfrac{4}{4-q} \| f \| _p^2 \int_1^\infty\left\{ \dfrac{1}{x^2} \left[ (10x)^{\tfrac{4-q}{4}}-x^{\tfrac{4-q}{4}}\right] ^{\tfrac{2}{q}} \right\} \, d\lambda(x) \\ =\tfrac{4}{4-q}\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}} \| f \| _p^2 \int_1^\infty x^{\tfrac{4-q}{4}\cdot \tfrac{2}{q}-2} \, d\lambda(x) \\ =\tfrac{4}{4-q}\cdot \tfrac{2q}{4-3q}\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}} \| f \| _p^2\underbrace{ \left[ x^{\tfrac{4-3q}{2q}}\right| _{x=1}^\infty}_{=-1} \\ =\tfrac{8q}{(4-q)(3q-4)}\underbrace{\left( 10^{\tfrac{4-q}{4}}-1\right)^{\tfrac{2}{q}}}_{\rightarrow 0 \text{ as }q\rightarrow 4} \| f \| _p^2 \\ \end{gathered} $$

Edit: I should've said something about this indeterminate form ##\tfrac{0}{0}## and l'Hospital's rule here, but @fresh_42 doesn't need the explanation of how to compute the limit, he knows.

and the desired result comes upon taking square roots of both sides a remembering that ##q=q(p)## (q may be thought of as a function of ##p##).
 
  • #37
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I didn't integrate the ##x^{-2}## part, so my factors are a bit different. Anyway, your solution does not work for ##q=4##.
 
  • #39
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Did you read the Edit comment? May need to refresh the page
C'mon, a little logarithm would have done, no L'Hôpital, limits or other hints needed. The constant ##\sqrt{\log 10}## does it if I'm right.
 
  • #40
benorin
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1) the ##q=4\Rightarrow p=\tfrac{1}{4}## case:
From this part:
$$\begin{gathered} \| g \| _{2}^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-1}\, d\lambda(t) \right\} ^{\tfrac{1}{2}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \int_1^\infty \dfrac{1}{x^2} \left\{\left[ \log (t) \right|_{t=x}^{10x} \right\} ^{\tfrac{1}{2}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \log ^{\tfrac{1}{2}} 10\int_1^\infty \dfrac{1}{x^2} \, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \log ^{\tfrac{1}{2}} 10 \\ \end{gathered} $$
 
  • #41
mathwonk
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Hint, possible spoiler, for problem #7: It is surely by far the easiest problem of all. Thus I presume it is the definitions that are scary.

A commutative ring is (noetherian if it has no infinite ascending sequence of ideals, and is) artinian if it has no infinite descending sequence of ideals. E.g. the integers are noetherian, since the only ideals strictly containing say the ideal (15) are the ideals (3) and (5), but not artinian, since the ideal (3) contains strictly the ideals (9), (27), (81),.....,(3^n),.....

Similar definitions apply to modules with submodules in place of ideals, i.e. an artinian module has no infinite sequence of strictly descending submodules.

Any ideal in a ring is a module for that ring, as is the ring itself. Now basic theory of quotients says the submodules of M/N are exactly those submodules of M that contain N. This should suggest some examples, even just sticking to rings and ideals.

In fact not only is it easy to give examples of artinian modules occurring as such quotients (of non artinian modules), but one can show almost as easily that every artinian module occurs this way.
 
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  • #42
wrobel
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Problem 2c admits a stupid solution : ##\mathcal{T}=\{\mathbb{R}^\infty,\emptyset\}## as well
 
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  • #43
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Thus I presume it is the definitions that are scary.
However, one has to distinguish between Artian modules and Artian rings. And of course I expect a proof of all three properties.
 
  • #44
mathwonk
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fresh, could you give us your definitions of artinian modules and rings, or is it what I said?
 
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  • #45
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fresh, could you give us your definitions of these terms? What 3 properties are you referring to?
Right, it are four: ##R,M,N## not Artinian (3, ##R## as ring, ##M,N## as ##R-##modules) and ##M/N## Artinian (1). I used the DCC: every descending chain of submodules becomes stationary. I find this the most transparent one, but equivalent definitions can of course be used, too.
 
  • #46
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Spoiler for #7: I will finesse all 4 proofs.

Assume R is a non - artinian ring, that N is a non - artinian module, and that A is an artinian module. Then M = AxN is also non - artinian, since N is a submodule of M. Then M/N ≈ A is artinian.

I leave it to someone else to actually produce examples, e.g. using my hint above.
 
  • #47
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Forgive me, but I still suspect that abstract concepts like "artinian rings and modules" just make people reluctant. I know when I read the problem I was a little unnerved by the terms until I spent a while thinking about it. So I make another try at minimalizing this feeling, by pointing out that a module with only finitely many elements is definitely artinian. Maybe this will shake loose a solution.
 
  • #48
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Forgive me, but I still suspect that abstract concepts like "artinian rings and modules" just make people reluctant. I know when I read the problem I was a little unnerved by the terms until I spent a while thinking about it. So I make another try at minimalizing this feeling, by pointing out that a module with only finitely many elements is definitely artinian. Maybe this will shake loose a solution.

I'm wondering if you have a concrete example in mind?
 
  • #49
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Forgive me, but I still suspect that abstract concepts like "artinian rings and modules" just make people reluctant.
I'm afraid this is true for many algebra questions. It shouldn't be that difficult these times to look up a definition and to find examples. I remember a question about amalgams where the term had a similar effect, or Lie algebra questions which were very easy as soon as one looked up the definitions.
 
  • #50
mathwonk
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OK, The example I gave in my hint was that the ideals (3^n) in the integers, are clearly (by unique factorization) descending so not only is the ring of integers not artinian, hence also not as a Z module, but also the Z-module (i.e. ideal) (3) is also not artinian. The quotient however Z/(3), is a finite Z- module, hence is artinian.

This shows how to build lots of examples, Z/(n), for any n≥1.

Now not only is a finite module artinian, but also any module with only a finite number of submodules is artinian, so given any ring R at all, the quotient ring R/m by a maximal ideal m, is also an artinian R module, since it has only 2 ideals, hence 2 submodules. So if we take k[x1,...,xn] a polynomial ring over a field, and m any maximal ideal, e.g. (x1,...,xn), then since the sequence (x1,...,xn)^r is ? clearly descending, we have another similar example k[x1,...,xn]/m, and also k[x1,..,xn]/m^r, for r ≥ 1. Or we can also use m/m^r, r ≥ 2, since this quotient is a finite dimensional k vector space, hence has the DCC on k subspaces, hence also on k[x1,...,xn] submodules. In fact if R is any integral domain which is not a field, then R is not artinian as a ring nor as a module over itself, and no maximal ideal M is artinian over R. But the quotient R/M is a field, hence is artinian.

Similar examples exist for quotients of polynomial rings, e.g. k[x1,...,xn]/P where P is any prime ideal which is not maximal, and say k algebraically closed, is not artinian, and we can choose any maximal ideal m containing P and play the same game. Indeed if R is any integral domain which is not a field, then R is not artinian it seems (by the "Nakayama lemma", which is apparently due to Krull), and every non zero ideal is a non artinian module, but for m a maximal ideal, of course R/m is a field, hence artinian as an R module.

I guess the point to made here is that the submodules of a quotient M/N are those submodules of M that lie between N and M, i.e. the submodules "above" N in M. Since N being non artinian means there are lots of submodules "below" N, we ought to be able to find an example by taking a submodule N of M that is very big, i.e. with lots of submodules between N and (0) but very few between N and M. Then most submodules will disappear when we take the quotient M/N. Consequently if we can find a non artinian submodule E of a given M, we just choose a maximal proper submodule N of M that contains E, (which exists by "Zorn's lemma", which is apparently due to Kuratowski).

Do these scan?
 
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