Math Challenge - July 2020

[Adesh]

If you add them, using $\sin^4{x} + \cos^4{x} \in [\frac{1}{2},1]$, you can get the inequality $$x^4 + y^4 + 2x^2 y^2 -4(x^2 + y^2) < -4$$ $$(x^2 + y^2)^2 -4(x^2 + y^2) + 4 < 0$$ $$([x^2 + y^2] -2)^2 < 0$$That would seem to imply that there are no possible values of $x^2 + y^2$ that satisfy the inequality, and that there are no solutions.

I think, pal, you missed the equality. The first expression should be

$$x^4 + y^4 + 2x^2 y^2 -4(x^2 + y^2) \leq -4$$
because 1 is allowed. Hence, your last expression becomes
$$([x^2 + y^2] -2)^2 \leq 0$$
since we want only real solutions, therefore equality is something that we want (a square cannot be less than zero if it's components are less than 0). So, we have
$$
[(x^2 +y^2)-2]^2= 0$$
$$
x^2 + y^2 = 2 $$
With this value we can put it into our original given equations to get the real solutions.

(Pal I'm not alluding anything to our discussion in that thread, we are friends and I respect that).

 

[etotheipi]

I think, pal, you missed the equality. The first expression should be

You're right, see #83

 

[Adesh]

You're right, see #83

Sorry missed that. Actaully, there were so many posts and even the tab “there are more posts to display” but I didn’t click on it

 

[mathwonk]

SPOILER Remarks/hints:
Problem #1 taught me that i did not really know what the weak topology is, i.e. what the basic open sets are. nice problem.

Problem #9 taught me that you can reverse the inequality and still get the same conclusion, even wthout assuming the map is onto, (essentially same proof). Another instructive problem.

Problem #2, I don't know much about, but I would start by finding their lie algebras. (Since, as you probably know, the lie algebra is the tangent space to the lie group at the identity, and transversaity is related to relative position of tangent spaces.)

 

[Infrared]

@mathwonk Yes, computing their Lie algebras is the right way to go. And in general that should be enough, because if $G,G'\subset GL_n$ are Lie groups, and $g\in G\cap G'$, then $G$ and $G'$ intersect transversely at $g$ iff they do at $1$ (since $T_gG=g_* \left(T_1G\right)$ and similarly for $G'$).

 

[wrobel]

you can reverse the inequality and still get the same conclusion, even wthout assuming the map is onto, (essentially same proof). Another instructive problem.

Sure, to remember the Poincare recurrence theorem is also instructive in this concern

 

[mathwonk]

I'm afraid I don't know that theorem of Poincare'. Ok I googled it, yes, nice connection.
Also one gets from the reverse result that every (necessarily injective) isometry of a compact metric space into itself is also surjective, a possibly useful fact, analogous to results about linear maps of finite dimensional vector spaces, and algebraic maps of irreducible projective varieties, and of course functions on finite sets, another illustration of the principle that "compact" generalizes "finite".And hopefully the lie algebra result gives what we want, since then SU(n) has a nice lie algebra structure compatible with the other two!
Possible spoiler comments on #4.
And in looking at continuous functions on the interval, I was able to find a sequence of constant norm that converges to zero weakly, (using the known structure of the dual of C[0,1] due to Riesz in 1909, as functions of bounded variation), but so far that only illustrates problem #1, apparently not whether C[0,1] is a dual. I thought maybe I could use the theorem that the unit ball in a dual is weak star compact, but have not seen how to do that yet. I.e. even if the ball were not weakly compact, which is not at all clear either, it seems it might still be weak-star compact. Some people say the key is to look at "extreme points", i.e. maybe convexity is involved? (i.e. Krein Milman as well as Alaoglu.) Haven't written anything down yet but if the picture in my head is close, among positive valued functions in the unit ball in C[0,1], there seems to be only one extreme point, hence altogether only two? Just tossing out guesses.

 

[wrobel]

theorem that the unit ball in a dual is weak star compact,

Krein Milman

the key words have been pronounced :)

 

[nuuskur]

Spoiler: Problem 1

Since $\dim V \leq \dim V'$ always holds, the space $V'$ must be infinite dimensional. Assume for a contradiction $\sigma (V,V')$ is normable. The dual space of normed space is a Banach space, so it is enough to show that the vector space $V'$ admits a countable basis, which is impossible for Banach spaces.

Let $\|\cdot\|$ be the inducing norm. I will attempt to justify that we may assume a countable neighborhood basis of $0$ which consists of kernels of functionals in $V'$. Consider a nh basis of $0$, for instance $T_n = \{x\in V \mid \|x\| < 1/n\},\quad n\in\mathbb N.$
Fix $n\in\mathbb N$. By Hahn-Banach (the point separation corollary), take $\varphi _n \in V'$ such that $\overline{T_n}\subseteq \mathrm{Ker}\varphi _n$. Then the sequence of kernels constitute a nh basis of $0$. In fact,
$$\left\{\bigcap _{k=1}^n\varphi _k^{-1}(B(0,\varepsilon)) \mid n\in\mathbb N, \varepsilon >0\right\}$$
is a nh basis of $0$ w.r.t $\sigma (V,V')$. Take $\varphi \in V'$. For every $\varepsilon >0$ we have $N_\varepsilon \in\mathbb N$ s.t
$$\bigcap _{k=1} ^{N_\varepsilon} \varphi _{k}^{-1}(B(0,\varepsilon)) \subseteq \varphi ^{-1}(B(0,\varepsilon))$$
Put $N := \min _{\varepsilon }N_{\varepsilon}$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in\mathrm{Ker}\varphi$, thus $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$.

I think Hahn Banach is overkill and I also think a similar argument passes assuming $\sigma (V,V')$ is metrisable. I can't put my finger on it atm. It's likely something stupidly simple.

 

[member 587159]

Spoiler: Problem 1

Since $\dim V \leq \dim V'$ always holds, the space $V'$ must be infinite dimensional. Assume for a contradiction $\sigma (V,V')$ is normable. The dual space of normed space is a Banach space, so it is enough to show that $V'$ admits a countable basis, which is impossible for Banach spaces.

Let $\|\cdot\|$ be the inducing norm. I will attempt to justify that we may assume a countable nh basis of $0$ which consists of kernels of functionals in $V'$. Consider a neighborhood basis of $0$, for instance $T_n = \{x\in V \mid \|x\| < 1/n\},\quad n\in\mathbb N.$
Fix $n\in\mathbb N$. By Hahn-Banach (the point separation corollary), take $\varphi _n \in V'$ such that $\overline{T_n}\subseteq \mathrm{Ker}\varphi _n$. Then the sequence of kernels constitute a nh basis of $0$. In fact,
$$\mathcal N := \left\{\bigcap _{k=1}^n\varphi _k^{-1}(B(0,\varepsilon)) \mid n\in\mathbb N, \varepsilon >0\right\}$$
is a nh basis of $0$ w.r.t $\sigma (V,V')$. Take $\varphi \in V'$. For every $\varepsilon >0$ we have $N_\varepsilon \in\mathbb N$ s.t
$$\bigcap _{k=1} ^{N_\varepsilon} \varphi _{k}^{-1}(B(0,\varepsilon)) \subseteq \varphi ^{-1}(B(0,\varepsilon))$$
Put $N := \min _{\varepsilon }N_{\varepsilon}$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _n$ implies $x\in\mathrm{Ker}\varphi$, thus $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$.

I think Hahn Banach is overkill and I also think a similar argument passes assuming $\sigma (V,V')$ is metrisable. I can't put my finger on it atm. It's likely something stupidly simple.

I didn't read yet in detail but what do you mean with "basis"? A topological basis? In that case $l^2(\Bbb{N})$ is separable and thus has a countable basis, yet has infinite dimension. So I don't quite see how you would get a contradiction.

 

[nuuskur]

I didn't read yet in detail but what do you mean with "basis"? A topological basis? In that case $l^2(\Bbb{N})$ is separable and thus has a countable basis, yet has infinite dimension. So I don't quite see how you would get a contradiction.

Now, you're confusing me. The $\varphi_n$ make a countable basis of the vector space $V'$, which is impossible for Banach spaces. Iirc it's a consequence of Baire category theorem.. now that I think about it, maybe P1 more generally can also be shown with BCT.

I edited the post to emphasise it's a basis in the sense of vector spaces.

 

[member 587159]

Now, you're confusing me. The $\varphi_n$ make a countable basis of the vector space $V'$, which is impossible for Banach spaces. Iirc it's a consequence of Baire category theorem.. now that I think about it, maybe P1 more generally can also be shown with BCT.

I edited the post to emphasise it's a basis in the sense of vector spaces.

Ah yes, I see now. Your idea was to show that $V^*$ admits a countable Hamel basis by showing that $V^*$ is spanned by countably many functionals. Interesting! I will look at the details soon and get back to you.

 

[member 587159]

Spoiler: Problem 1

Since $\dim V \leq \dim V'$ always holds, the space $V'$ must be infinite dimensional. Assume for a contradiction $\sigma (V,V')$ is normable. The dual space of normed space is a Banach space, so it is enough to show that the vector space $V'$ admits a countable basis, which is impossible for Banach spaces.

Let $\|\cdot\|$ be the inducing norm. I will attempt to justify that we may assume a countable neighborhood basis of $0$ which consists of kernels of functionals in $V'$. Consider a nh basis of $0$, for instance $T_n = \{x\in V \mid \|x\| < 1/n\},\quad n\in\mathbb N.$
Fix $n\in\mathbb N$. By Hahn-Banach (the point separation corollary), take $\varphi _n \in V'$ such that $\overline{T_n}\subseteq \mathrm{Ker}\varphi _n$. Then the sequence of kernels constitute a nh basis of $0$. In fact,
$$\left\{\bigcap _{k=1}^n\varphi _k^{-1}(B(0,\varepsilon)) \mid n\in\mathbb N, \varepsilon >0\right\}$$
is a nh basis of $0$ w.r.t $\sigma (V,V')$. Take $\varphi \in V'$. For every $\varepsilon >0$ we have $N_\varepsilon \in\mathbb N$ s.t
$$\bigcap _{k=1} ^{N_\varepsilon} \varphi _{k}^{-1}(B(0,\varepsilon)) \subseteq \varphi ^{-1}(B(0,\varepsilon))$$
Put $N := \min _{\varepsilon }N_{\varepsilon}$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in\mathrm{Ker}\varphi$, thus $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$.

I think Hahn Banach is overkill and I also think a similar argument passes assuming $\sigma (V,V')$ is metrisable. I can't put my finger on it atm. It's likely something stupidly simple.

I'm not sure how you apply Hahn-Banach, but even if its use is justified there seems to be a problem:

You have $T_n \subseteq \ker(\varphi_n)$ and consequently $V=\operatorname{span}(T_n) \subseteq \ker(\varphi_n)$ so we have $\varphi_n = 0$ for all $n$, so it is impossible that these functionals span the dual space.

Somewhere you must have made a mistake.

 

[nuuskur]

Oops, I'm an idiot..
Will revise, apologies for wasting your time.

 

[member 587159]

Will revise, apologies for wasting your time.

No time wasted! I learned something from it :)

 

[mathwonk]

SPOILER-HINT for #5:

Take any two points a,b on an integral curve for the gradient flow of f. Then their distance apart along that curve seems to be f(b)-f(a), but is greater along any other path joining them. That seems to do it, at least based on an intuitive meaning of "geodesic".

 

[Infrared]

@mathwonk Yes, that's exactly the idea (and in fact shows that the integral curves are minimizing geodesics)

 

[mathwonk]

Spoiler/Hints for #9:

given any two points a,b in X and f:X-->X a surjective map with d(f(x),f(y)) ≤ d(x,y) for all x,y in X, choose a0 = f(a), b0 = f(b), a1 = a, b1 = b, a2: f(a2) = a1, b2: f(b2) = b1, a3: f(a3) = a2, b3: f(b3) = b2,...
an+1: f(an+1) = an, bn+1: f(bn+1) = bn,...

Show d(a0,b0) ≤ d(an,bn) all n. and that d(a0,an,) ≤ d(a1,an+1) all n.

Show that a0 is an accumulation point of the sequence an, n ≥ 1, and that the pair (a0,b0) is an accumulation point of the sequence (an,bn), n≥1. Conclude that d(a,b) ≤ d(f(a),f(b)).

 

[mathwonk]

Spoiler/hint: for #5:

Let c be a unit speed path, from time s to time t, joining two points of an integral curve for the gradient flow of f. Compare the integrals of <c',c'>, and <gradf, c'> from s to t, and think about what they tell you.

 

[mathwonk]

Hint for #1:In the weak topology, linear functions to R must be continuous, so if f is a real valued function, the inverse image of (-e,e) must be open. To be a topology, also finite intersections of such sets must be open. Also unions of such sets must be open, and that is all; hence that means the "smallest" open sets are finite intersections of such sets. Show that these sets all contain positive dimensional linear subspaces, hence are not norm-bounded if the space is infinite dimensional. E.g. a weak nbhd of 0 contains the intersection of the kernels of a finite sequence of linear functions, i.e. the kernel of a linear map to a finite dimensional space.

 

[julian]

I'm v.tired and about to hit the hay, so I'll just give this response to #5.

Write $X^a = g^{ab} \partial_b f$ and note that

\begin{align}
0 = \partial_c (1) = \nabla_c (g^{ab} \partial_a f \partial_b f) = 2 X^b \nabla_c \partial_b f = 2 X^b \nabla_b \partial_c f = 2 X^b \nabla_b X_c
\nonumber
\end{align}

where we have used $\nabla_c \partial_b f = \partial_c \partial_b f - \Gamma^d_{cb} \partial_d f = \partial_b \partial_c f - \Gamma^d_{bc} \partial_d f = \nabla_b \partial_c f$ and $\nabla_c g^{ab} = 0$ (I'm using $\nabla_a$ to mean the covariant derivative). We have obtained

\begin{align}
X^b \nabla_b X^a = 0 .
\nonumber \\
\end{align}

(where we have used $\nabla_c g^{ab} = 0$ again). An integral curve $x^a = x^a(u)$ of a vector field is a curve such that

\begin{align}
\frac{d x^a (u)}{du} = X^a (x^b (u)) .
\nonumber \\
\end{align}

Substituting this into $X^b \nabla_b X^a = 0$ gives

\begin{align}
0 & = X^b \nabla_b X^a
\nonumber \\
& = \frac{d x^b (u)}{du} \nabla_b \left( \frac{d x^a (u)}{du} \right)
\nonumber \\
& = \frac{d x^b (u)}{du} \left\{ \frac{\partial}{\partial x^b} \left( \frac{d x^a (u)}{du} \right) + \Gamma^a_{bc} \frac{d x^c (u)}{du} \right\}
\nonumber \\
& = \frac{d^2 x^a (u)}{du^2} + \Gamma^a_{bc} \frac{d x^b (u)}{du} \frac{d x^c (u)}{du}
\nonumber
\end{align}

which is the metric geodesic equation obtained by minimizing the distance between two fixed points (with $u$ an affine parameter).

 

[member 587159]

Show that these sets all contain positive dimensional linear subspaces, hence are not norm-bounded if the space is infinite dimensional.

How would you show this?

 

[wrobel]

Regarding #5 another way to get the result is to apply the vector field straightening theorem to the vector field

$$g^{ij}\frac{\partial f}{\partial x^j}$$

 

[wrobel]

Spoiler/Hints for #9:

given any two points a,b in X and f:X-->X a surjective map with d(f(x),f(y)) ≤ d(x,y) for all x,y in X, choose a0 = f(a), b0 = f(b), a1 = a, b1 = b, a2: f(a2) = a1, b2: f(b2) = b1, a3: f(a3) = a2, b3: f(b3) = b2,...
an+1: f(an+1) = an, bn+1: f(bn+1) = bn,...

Show d(a0,b0) ≤ d(an,bn) all n. and that d(a0,an,) ≤ d(a1,an+1) all n.

Show that a0 is an accumulation point of the sequence an, n ≥ 1, and that the pair (a0,b0) is an accumulation point of the sequence (an,bn), n≥1. Conclude that d(a,b) ≤ d(f(a),f(b)).

seems to be ok

 

[nuuskur]

Spoiler: Problem 1

Assume for a contradiction $\sigma (V,V')$ is metrisable. The Banach space $V'$ is infinite dimensional. Since $\sigma (V,V')$ is a metrisable locally convex topology, its neighborhood basis of zero is generated by a sequence of seminorms $p_{\varphi _n}$, where $\varphi _n\in V', n\in\mathbb N$. Wo.l.o.g this basis can be expressed as
$$S_{n} := \bigcap _{k=1}^n \varphi _k^{-1} (B(0,1)),\quad n\in\mathbb N.$$
Take $\varphi \in V'$. Fix $\varepsilon >0$. Due to continuity w.r.t $\sigma (V,V')$ we have $S_{N_\varepsilon} \subseteq \varphi ^{-1}(B(0,\varepsilon))$ for some $N_\varepsilon\in\mathbb N$. Put $N := \min _\varepsilon N_\varepsilon$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in \mathrm{Ker}\varphi$. Equivalently, $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$. Thus, $V'$ admits a countable spanning set, which is impossible. Therefore, $\sigma (V,V')$ cannot be metrisable.

 

[member 587159]

Spoiler: Problem 1

Assume for a contradiction $\sigma (V,V')$ is metrisable. The Banach space $V'$ is infinite dimensional. Since $\sigma (V,V')$ is a metrisable locally convex topology, its neighborhood basis of zero is generated by a sequence of seminorms $p_{\varphi _n}$, where $\varphi _n\in V', n\in\mathbb N$. Wo.l.o.g this basis can be expressed as
$$S_{n} := \bigcap _{k=1}^n \varphi _k^{-1} (B(0,1)),\quad n\in\mathbb N.$$
Take $\varphi \in V'$. Fix $\varepsilon >0$. Due to continuity w.r.t $\sigma (V,V')$ we have $S_{N_\varepsilon} \subseteq \varphi ^{-1}(B(0,\varepsilon))$ for some $N_\varepsilon\in\mathbb N$. Put $N := \min _\varepsilon N_\varepsilon$. Then $x\in \bigcap _{k=1}^N \mathrm{Ker}\varphi _k$ implies $x\in \mathrm{Ker}\varphi$. Equivalently, $\varphi \in \mathrm{span}(\varphi _1,\ldots, \varphi _N)$. Thus, $V'$ admits a countable spanning set, which is impossible. Therefore, $\sigma (V,V')$ cannot be metrisable.

I'm aware that that every locally convex metrizable topological vector space has topology generated by a countable family of seminorms. However, you seem to be assuming that these seminorms are induced by functionals (given a functional $\omega: V \to \Bbb{K}$, its modulus $|\omega|: V \to [0,\infty[$ gives a seminorm). Can you explain why this is true or give a reference to that result?

 

[nuuskur]

I'm aware that that every locally convex metrizable topological vector space has topology generated by a countable family of seminorms. However, you seem to be assuming that these seminorms are induced by functionals (given a functional $\omega: V \to \Bbb{K}$, its modulus $|\omega|: V \to [0,\infty[$ gives a seminorm). Can you explain why this is true or give a reference to that result?

Given the dual pair $(V,V')$, its weak topology $\sigma (V,V')$ is generated by the family of seminorms $p_f : x\mapsto |f(x)|,\ f\in V'$. If the topology is metrisable, this family may be assumed to be at most countable.

 

[Infrared]

@julian This looks right, but I think the coordinate-free approach suggested by @mathwonk is a bit cleaner (and also has the advantage of showing that the integral curves are minimizing geodesics).

@wrobel Could you sketch your argument with the vector field straightening theorem?

 

[mathwonk]

Spoiler #5:Everything follows just from looking at the integral of <gradf, c'>, where c is a unit speed curve.

summary; | integral of <gradf,c'> dt| ≤ |t-s| = length of path c, with equality iff gradf = ± c'.

 

[wrobel]

@l Could you sketch your argument with the vector field straightening theorem?

There are local coordinates such that $g^{ij}\frac{\partial f}{\partial x^j}=\delta_1^i$ thus
$$\frac{\partial f}{\partial x^j}=g_{1j};$$ and
$$|\nabla f|=1\Longrightarrow g_{11}=1\Longrightarrow\frac{\partial f}{\partial x^1}=1\Longrightarrow \Gamma_{11}^i=0.$$
Furthermore,
$$\dot x^i=\delta^i_1\Longrightarrow x^i(t)=\delta^i_1t+x^i_0.$$ This function obviously satisfies the equation
$$\ddot x^i+\Gamma_{ks}^i\dot x^k\dot x^i=0$$