# Math Challenge - May 2020

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Mentor
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.

inductive limit topology:)

Mentor
inductive limit topology:)
Would you mind to define the open sets explicitly?

benorin
Homework Helper
Gold Member
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.

fresh_42
benorin
Homework Helper
Gold Member
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.

Mentor
I solved it by foot, only using symmetry properties, substitutions and integration by parts, and Fubini. My result is the same.

mathwonk
Homework Helper
Question on #7: did you want a non - zero quotient module?

Mentor
Question on #7: did you want a non - zero quotient module?
That would be nice of course.

benorin
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Gold Member
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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Mentor
I think you made an integration error, but I'm not sure since you haven't shown what you've done. And what if $q=4$?

benorin
Homework Helper
Gold Member
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 $p$).

Mentor
I didn't integrate the $x^{-2}$ part, so my factors are a bit different. Anyway, your solution does not work for $q=4$.

benorin
Homework Helper
Gold Member
Did you read the Edit comment? May need to refresh the page

Mentor
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.

benorin
Homework Helper
Gold Member
1) the $q=4\Rightarrow p=\tfrac{1}{4}$ case:
From this part:
$$\begin{gathered} \| g \| _{2}^2 \leq \| f \| _p^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-\tfrac{q}{4}}\, d\lambda(t) \right\} ^{\tfrac{2}{q}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \int_1^\infty \dfrac{1}{x^2} \left\{\int_x^{10x} t^{-1}\, d\lambda(t) \right\} ^{\tfrac{1}{2}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \int_1^\infty \dfrac{1}{x^2} \left\{\left[ \log (t) \right|_{t=x}^{10x} \right\} ^{\tfrac{1}{2}}\, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \log ^{\tfrac{1}{2}} 10\int_1^\infty \dfrac{1}{x^2} \, d\lambda(x) \\ = \| f \| _{\tfrac{1}{4}}^2 \log ^{\tfrac{1}{2}} 10 \\ \end{gathered}$$

fresh_42
mathwonk
Homework Helper
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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Problem 2c admits a stupid solution : $\mathcal{T}=\{\mathbb{R}^\infty,\emptyset\}$ as well

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Mentor
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.

mathwonk
Homework Helper
fresh, could you give us your definitions of artinian modules and rings, or is it what I said?

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Mentor
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.

mathwonk
Homework Helper
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.

mathwonk
Homework Helper
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.

Math_QED