Math Challenge - July 2020

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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.

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Math_QED
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2019 Award
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.

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Math_QED
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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.

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

Math_QED
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Will revise, apologies for wasting your time.
No time wasted! I learnt something from it :)

fresh_42
mathwonk
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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
Infrared
Gold Member
@mathwonk Yes, that's exactly the idea (and in fact shows that the integral curves are minimizing geodesics)

mathwonk
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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
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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.

Infrared
mathwonk
Homework Helper
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.

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Math_QED and Infrared
julian
Gold Member
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).

Math_QED
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2019 Award
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

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.

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Math_QED
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2019 Award
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?

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.

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Infrared
Gold Member
@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
Homework Helper
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'.

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Infrared
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
$$\ddot x^i+\Gamma_{ks}^i\dot x^k\dot x^i=0$$

mathwonk
Homework Helper
@Math_QED: post #110 edited to include more argument.

Math_QED
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2019 Award
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 guess the idea is correct, but it is unclear to me why you work with the ##\epsilon## and the minimum? Doesn't the following work instead?

Let ##\varphi \in V^*##. Then there exists ##n## with ##S_n = \bigcap_{k=1}^n \varphi_k^{-1}(B(0,1)) \subseteq \varphi^{-1}(B(0,1))##. This implies that ##\bigcap_{k=1}^n \ker \varphi_k \subseteq \ker \varphi##: indeed if ##\varphi_k(x)=0## for ##k=1, \dots, n##, then also ##\varphi_k(tx)=0## for all ##t\in \Bbb{R}## and thus ##tx\in S_n\subseteq \varphi^{-1}(B(0,1))##, so that ##t\varphi(x) \in B(0,1)##. Since this holds for all ##t \in \Bbb{R}##, this forces ##\varphi(x)=0##. You then can conclude like you did.

I think it is non-trivial that you can choose the basis like you did: I agree that the topology ##\sigma(V,V')## is generated by the seminorms ##\{p_f: f \in V^*\}##. But then the following two questions remain:

(1) How does metrizability imply that you can choose ##(f_n)_n## such that the family of seminorms ##\{p_{f_n}: n \geq 1\}## still generate the topology? (Provide proof or a reference).

(2) Why can we choose the basis in this particular form: i.e. as inverse images of the unit ball?

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benorin
Homework Helper
Gold Member
Summary:: Functional Analysis, Topology, Differential Geometry, Analysis, Physics
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

9. Let ##(X,d)## be a compact metric space and ##f:X\to X## be a mapping onto. Assume that ##d(f(x),f(y))\le d(x,y),\quad \forall x,y\in X.## Show that ##d(f(x),f(y))= d(x,y),\quad \forall x,y\in X.## (WR)
This is what I got so far, but my expertise here is lacking to say the least, if you've any skills here please correct what I've got (or anybody really) as Topology is really new to me, thanks.
Work: ##f:X\to X## is continuous by Munkres' Topology Theorem 21.1, pg 129 which is the ##\epsilon -\delta## definition of continuity of ##f:X\to Y## for metric spaces ##X## and ##Y## with respective metrics ##d_X## and ##d_Y##, here taking ##X=Y## and metrics ##d_X (x,y) = d_Y (x,y) :=d(x,y)## since given ##x\in X## and ##\epsilon >0, \forall n\in \mathbb{N}## choose ##\delta_n = \epsilon +\tfrac{1}{n}## such that ##d_X(x,y)<\delta _n = \epsilon +\tfrac{1}{n}\implies d_Y(f(x), f(y)) \leq d_X (x,y) < \epsilon +\tfrac{1}{n}\implies d_X(f(x),f(y)) < \epsilon## where the business with adding the ##\tfrac{1}{n}## term I'm not certain is quite right but was intended to deal with the strictness of the inequalities.

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wrobel
ork: is continuous by Munkres' Topology Theorem 21.1, pg 129 which
sure it is continuous ,moreover it is Lipschitz

I guess the idea is correct, but it is unclear to me why you work with the ##\epsilon## and the minimum? Doesn't the following work instead?

Let ##\varphi \in V^*##. Then there exists ##n## with ##S_n = \bigcap_{k=1}^n \varphi_k^{-1}(B(0,1)) \subseteq \varphi^{-1}(B(0,1))##. This implies that ##\bigcap_{k=1}^n \ker \varphi_k \subseteq \ker \varphi##: indeed if ##\varphi_k(x)=0## for ##k=1, \dots, n##, then also ##\varphi_k(tx)=0## for all ##t\in \Bbb{R}## and thus ##tx\in S_n\subseteq \varphi^{-1}(B(0,1))##, so that ##t\varphi(x) \in B(0,1)##. Since this holds for all ##t \in \Bbb{R}##, this forces ##\varphi(x)=0##. You then can conclude like you did.
It does work. I don't really understand your objection, though. Similarly, one could ask why you would work with the following if some other method is also sufficient. I think it's apples and oranges. I might be wrong.
I think it is non-trivial that you can choose the basis like you did: I agree that the topology ##\sigma(V,V')## is generated by the seminorms ##\{p_f: f \in V^*\}##. But then the following two questions remain:

(1) How does metrizability imply that you can choose ##(f_n)_n## such that the family of seminorms ##\{p_{f_n}: n \geq 1\}## still generate the topology? (Provide proof or a reference).

(2) Why can we choose the basis in this particular form: i.e. as inverse images of the unit ball?
Nh basis of zero for $\sigma (V,V')$ is given by the family
$$B_{\varepsilon, f_1,\ldots,f_n} := \left\{ x\in V \mid \max\limits_{1\leq j\leq n} |f_j(x)| \leq \varepsilon \right\},\quad \varepsilon > 0,\quad f_1,\ldots,f_n\in V',\quad n\in\mathbb N.$$
Since $V'$ is a vector space, we can instead take $\frac{1}{\varepsilon}f_j$ and obtain a basis
$$B_{1,f_1,\ldots,f_n} = \left\{ x\in V \mid \max\limits _{1\leq j\leq n} |f_j(x)| \leq 1 \right\} = \bigcap _{k=1}^n f_k^{-1}(B(0,1)),\quad f_1,\ldots,f_n\in V',\quad n\in\mathbb N.$$
The second equality is clear, I think.
Given a locally convex topology its nh basis of zero can be assumed to consist of closed absolutely convex subsets. Let the topology be metrisable, then this basis can be assumed to be countable. Suppose such a basis is $S_n,\ n\in\mathbb N$. Its generating family of seminorms can be picked as the Minkowski functionals $p_{S_n}$ of basis elements. By Hahn-Banach, there exists $\varphi _n\in V'$ s.t $|\varphi _n | \leq p_{S_n},\quad n\in\mathbb N$.

More generally, nh basis of zero generated by seminorms is given as
$$B_{\varepsilon, n} = \left\{ x\in V \mid \max\limits_{1\leq j\leq n} p_{S_j}(x) \leq \varepsilon \right\},\quad \varepsilon >0,\quad n\in\mathbb N.$$
Now, given $n\in\mathbb N$ and $\varepsilon >0$, we can find $\varepsilon _0 >0$ s.t
$$\left\{ x\in V \mid \max\limits_{1\leq j\leq n} |\varphi _n(x)| \leq \varepsilon _0 \right\} \subseteq B_{\varepsilon,n}.$$
Thus, the sequence of functionals $\varphi _n$ also generates a nh basis of zero.
I appreciate the scrutiny. Don't let me get away with handwaving.

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