Challenge Math Challenge - June 2020

[benorin]

@benorin Your answer is correct, but you can simplify: What is $2\lambda(2)$?

$2\lambda (2) =\zeta (2) +\hat{\zeta} (2) = \tfrac{\pi ^2}{6}+\tfrac{\pi ^2}{12} = \tfrac{\pi ^2}{4}$
where $\hat{\zeta}$ is the alternating Zeta function, but I'm not going to prove the value of those sums. They are known :)

 

[Fred Wright]

# 8

Spoiler

Part a.)
Factor the logarithm:
$$

\log(1-2\alpha \cos(x) +\alpha^2)= \log[(\alpha - e^{ix})(\alpha - e^{-ix})]\\

=2\log(\alpha) + \log(1-\frac{ e^{ix}}{\alpha}) + \log(1-\frac{ e^{-ix}}{\alpha})\\
$$
The Integral becomes:
$$
I=\int_0^{\pi}\log(1-2\alpha \cos(x) +\alpha^2)dx= I_1+I_2+I_3\\

I_1=2\log(\alpha)\int_0^{\pi}dx=2\pi \log(\alpha)\\

I_2=\int_0^{\pi} \log(1-\frac{ e^{ix}}{\alpha})dx\\
$$
Make a substitution in $I_2$:
$$
u=\frac{ e^{ix}}{\alpha}\\

du=iudx\\

I_2=i\int_{\frac{-1}{\alpha}}^{\frac{1}{\alpha}}\frac{\log(1-u)}{u}du\\
$$
Make a substitution in $I_3$:$$
I_3=\int_0^{\pi} \log(1-\frac{ e^{-ix}}{\alpha})dx\\

u=\frac{ e^{-ix}}{\alpha}

du=-iudx\\
$$
and find,$$
I_3=-I_2\\
$$
and thus,for $\alpha \geq 1$,$$
I=2\pi \log(\alpha)\\
$$
for $\alpha \leq -1$ take the principal branch of the logarithm,$$
I=2\pi(\log(|\alpha |)+ i\pi)\\
$$
Part b.)
Extend the the Taylor series for $\log(1-x)$ to the complex plane,$$
\log(1-x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(-x)^k}{k}=-\sum_{k=1}^{\infty}\frac{x^k}{k}\\

\log(1-z)=-\sum_{k=1}^{\infty}\frac{z^k}{k}\\
$$
This series converges for $|z| \leq 1$.
From part a.),$$
I=2\pi \log(\alpha) + \int_0^{\pi} \log(1-\frac{ e^{ix}}{\alpha})dx + \int_0^{\pi} \log(1-\frac{ e^{-ix}}{\alpha})dx\\

=2\pi \log(\alpha) -\sum_{k=1}^{\infty}\ \int_0^{\pi}\frac{e^{ikx}+ e^{-ikx} }{k\alpha^k}=2\pi \log(\alpha)-2\sum_{k=1}^{\infty} \int_0^{\pi} \frac{\cos(kx)}{k\alpha^k}dx\\

=2\pi \log(\alpha)-2\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2\alpha^k}|_0^{\pi}=2\pi \log(\alpha)\\
$$
We recover the result of Part a.) because $2\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^2\alpha^k}|_0^{\pi}=0$.

 

[Halc]

@Halc The condition "will fall down" means that the rope is not supported by a nail. You could take the rope to be a loop, if that's helpful.

My solution (the latter part) works in that case.

 

[etotheipi]

@Halc The condition "will fall down" means that the rope is not supported by a nail. You could take the rope to be a loop, if that's helpful.

The question sort of reminds me of the Borromean rings... I'm sure the answer is some clever figure-of-eight type arrangement of the loop around the pins.

 

[Fred Wright]

#7
If all the off diagonal elements are positive and the sum of the entries in each row is negative implies that all the diagonal entries are negative. This implies that no column is a multiple of another column and therefore all the columns are linearly independent. The invertible matrix theorem states that if all the columns are linearly independent then the matrix is invertible.

 

[Infrared]

@StoneTemplePython I actually hadn't heard of Gerschgorin Discs and instead had a method in mind with just row reduction, but your solution looks good too.

@benorin Yes, you've solved it fully now. In case you weren't sure how to get those values, from $\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}=\pi^2/6$, you get $\sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{1}{4}\zeta(2)=\pi^2/24$, and then subtracting these two sums gives the sum of the reciprocals of the odd squares.

@Halc I think you're misunderstanding the problem. If you're treating the rope as a loop, then just letting it rest on top of the nails doesn't count as a valid arrangement, since it could just fall off. To rephrase it with this language: find a way of arranging a loop of rope around two nails in a wall such that if any part of the rope is pulled down, the rope will stay on the wall, but this will no longer be the case if either nail is removed.

@etotheipi Yes, and there's a general solution for $n$ nails (removing any nail will cause the rope to fall). This was just a fun puzzle a friend showed me in undergrad that I thought was fun enough to share here.

 

[Infrared]

This implies that no column is a multiple of another column and therefore all the columns are linearly independent

This isn't true. For example, none of the vectors $\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\1\\0\end{bmatrix}$ are a multiple of another, and yet they are still linearly dependent. Also for $n=2$, having the diagonal terms negative and off-diagonal terms positive is consistent with the columns being linearly dependent.

 

[Halc]

#11

To rephrase it with this language: find a way of arranging a loop of rope around two nails in a wall such that if any part of the rope is pulled down, the rope will stay on the wall, but this will no longer be the case if either nail is removed.

OK, I figured it was something like that.
Cheat answer: Bend the two nails downward but with the heads touching, forming a closed loop trapping the rope.

The question sort of reminds me of the Borromean rings... I'm sure the answer is some clever figure-of-eight type arrangement of the loop around the pins.

Good description. You should get credit for it then, or at least my like.

Spoiler

 

[Infrared]

@Halc That picture can work, but could you re-draw it so that it's clear which strands are over/under in all of the crossings?

And, if anyone has a solution for the general problem with $n$ nails, feel free to post!

 

[Halc]

@Halc That picture can work, but could you re-draw it so that it's clear which strands are over/under in all of the crossings?

Also, apologies for the free-hand drawing using paint, crude, but adequate I guess.

Spoiler

 

[Delta2]

Corrected arithmetic error from previous post, $\boxed{\boxed{\text{corrections are double boxed}}}$.
I assume the cone has a base because it's oriented. Let $S_1$ denote the upper cone-shaped surface given by $z=g(x,y):=2-\sqrt{x^2+y^2},\quad ( 0\leq x^2+y^2\leq 2^2)$. Also denote the $x,y,$ and $z$ components of $\vec{F}$ by $P,Q,$ and $R$, respectively. Then

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\ = \int_{0}^{2\pi}\int_0^2\left(\tfrac{\boxed{\boxed{-2}}r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\boxed{\boxed{-2}}\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\boxed{\tfrac{8\pi}{5}}}\\ \end{gathered} $$

I don't understand why u have -2 inside the box while it should have been +2. Then after calculating the integrals as in the last line(where -2 I put +2) above I get $$2\frac{\pi}{4}\frac{32}{5}+2\pi\frac{12}{5}=40\frac{\pi}{5}=8\pi$$

Another way to verify this result:The divergence of the vector field F can be easily found and its 0. Hence by the divergence theorem, the flux through any closed surface is zero.

I believe you calculated correctly the flux through the circular disk at the base of the cone as$-8\pi$, this means that the flux through the surface $S_1$ must be $8\pi$ so they add together and give result 0, because the surface $S_1$ and the circular disk at the base form a closed surface.

 

[fresh_42]

# 8for $\alpha \leq -1$ take the principal branch of the logarithm,$$
I=2\pi(\log(|\alpha |)+ i\pi)$$

It is a real integral with a real value! If you use complex analysis, you should eliminate it again at the end.

Part b.)
Extend the the Taylor series for $\log(1-x)$ to the complex plane,$$
\log(1-x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{(-x)^k}{k}=-\sum_{k=1}^{\infty}\frac{x^k}{k}\\

\log(1-z)=-\sum_{k=1}^{\infty}\frac{z^k}{k}\\
$$
This series converges for $|z| \leq 1$.
...

Yes, but it is $|\alpha|$ in the solution to be exact.

Btw., at all engineers, physicists and all who deal with real problems out there:
https://www.amazon.com/s?k=Gradshteyn,+Ryzhik
is a must have!

 

[benorin]

I don't understand why u have -2 inside the box while it should have been +2. Then after calculating the integrals as in the last line(where -2 I put +2) above I get $$2\frac{\pi}{4}\frac{32}{5}+2\pi\frac{12}{5}=40\frac{\pi}{5}=8\pi$$

Another way to verify this result:The divergence of the vector field F can be easily found and its 0. Hence by the divergence theorem, the flux through any closed surface is zero.

I believe you calculated correctly the flux through the circular disk at the base of the cone as$-8\pi$, this means that the flux through the surface $S_1$ must be $8\pi$ so they add together and give result 0, because the surface $S_1$ and the circular disk at the base form a closed surface.

You’re correct of course, I must have missed either the minus from the formula or from the power rule yesterday. Thanks. It’s amazing what sleep does. Lol

 

[benorin]

@benorin Yes, you've solved it fully now. In case you weren't sure how to get those values, from $\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}=\pi^2/6$, you get $\sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{1}{4}\zeta(2)=\pi^2/24$, and then subtracting these two sums gives the sum of the reciprocals of the odd squares.

Iirc I did these in Fourier analysis way back when... but that is easier way. I was thinking the trick to analytically continue the zeta fcn to Re[z]>0 when I did that. Thanks.

 

[benorin]

It can be done with real analysis and elementary methods. However, part one is rather tricky, whereas part two only requires to find the series expansion which fits best. Here is a hint for part a):
$$\displaystyle{\int_0^{\pi/2} \log(\sin(x))\,dx=-\frac{\pi}{2}\log(2)}$$

#8) The case of $| \alpha |=1$. Recall $I(\alpha )$ is define to be the integral in question. Then
$$I(+1)= \int_{0}^\pi \log (2-2\cos (x)) \, dx$$
Make the substitution $u=\pi -x$ (I’m skipping some elementary details here) to get
$$I(+1)= \int_{0}^\pi \log (2+2\cos (x)) \, dx=I(-1)$$
Let $\beta := I(\pm 1)$ so that

$$\begin{gathered} 2\beta = \int_0^\pi \log\left[ (2-2\cos (x))(2+2\cos (x))\right]\, dx \\ = \int_0^\pi \log (4\sin ^2 (x))\,dx \\ = 2\pi\log 2+4\left( -\tfrac{\pi}{2}\log 2\right) =0\\ \end{gathered} $$

Where I have used some trig identities and log properties and symmetry over $\left[ 0,\pi \right]$ of the absolute value of $\sin x$ in the last equality. Apologies for skipping steps but I’m on my phone typing this and it’s a pain but I’m sure fresh_42 can follow me well enough. If not, I will type it up in full on my Mac later.

 

[zinq]

Question 3.

Spoiler

Let B = {e_n | n ∈ Z+} be an orthonormal basis for the separable infinite-dimensional Hilbert space H
of square-summable sequences of real numbers.

Let O = CC({±e_n | n ∈ Z+}) be the closed convex hull of ±B. Then O may be described as the Hilbert orthoplex (the generalization to Hilbert space of the regular octahedron). The set O has empty interior.* Now let

X = R ⋅ O = {t⋅y | t ∈ R+, y ∈ O }

Then X is a convex subset of H that cannot be all of H because it contains no interior. And X is contained in no half-space because it contains all the axes R⋅e_n, n ∈ Z+.
_____
* For, the distance of the origin to the set {y ∈ H | Σ_n y_n = 0} ⊂ O, where y = (y_n), is zero. For instance, consider the sequence {z_n} where z_n = e₁/n - (e₂ + ... + e_n²+1)/n², for which ||z_n|| = √2/n.

 

[Adesh]

In question 13 does “ $n\in \mathbb{N_0}$” mean that $n=0$ is allowed ?

 

[Infrared]

@Adesh Yes, but $n=0$ does not satisfy the condition "there are no integers $a,b\in\mathbb{Z}$ with $3a^3+b^3=n^3.$

 

[Adesh]

@Adesh Yes, but $n=0$ does not satisfy the condition "there are no integers $a,b\in\mathbb{Z}$ with $3a^3+b^3=n^3.$

For $n=0$ both $a$ and $b$ cannot be an integer to satisfy the equation. One of them can be an integer but not both.

 

[Infrared]

Why not $a=b=0$?

 

[Adesh]

Why not $a=b=0$?

Oh yeah! I didn’t take that into account, thanks.

 

[wrobel]

In my opinion the solution of probl. 3 by zinq is not completely formal but is it completely correct and nice. I would consider probl. 3 solved by zinq
I also think that the Baire category theorem is needed to turn zinq's argument into a formal proof

 

[strangerep]

2. Let $A$ and $B$ be complex $n\times n$ matrices such that $AB-BA$ is a linear combination of $A$ and $B$. Show that $A$ and $B$ must have a common eigenvector. (IR)

Thank you to @Infrared for posing this question. It made me discover some holes in my understanding of even the simpler case where $AB=BA$. (E.g., certain conditions on the intersection of the nullspaces of $A$ and $B$ must be met.)

Anyway, I'll start with a non-general partial solution to get things moving...

For this initial simple case, assume we are given $$ AB - BA = \alpha A + \beta B ~,~~~~~ (1)$$ where the scalar coefficients $\alpha,\beta$ are non-zero.

Let $|a\rangle$ be an eigenvector of $A$ with eigenvalue $a$, i.e., $A |a\rangle = a |a\rangle$. Then $|a\rangle$ is also an eigenvector of $(A+k)$, with eigenvalue $a+k$.

Now, (1) implies $$0 ~=~ AB|a\rangle - BA|a\rangle - \alpha A |a\rangle - \beta B|a\rangle ~=~ AB|a\rangle - aB |a\rangle - a \alpha|a\rangle - \beta B|a\rangle~.~~~~~ (2)$$We need to recast (2) in the form: $$0 ~=~ A(B+v) |a\rangle ~-~ w (B+v) |a\rangle ~=~ AB|a\rangle + va|a\rangle - w (B+v) |a\rangle ~,~~~~~ (3)$$ because that means $(B+v)|a\rangle$ is an eigenvector of $A$, with eigenvalue $w$.

Rearranging (3) and comparing with (2), we find that we need $w=a+\beta$ and $a\alpha = v (w-a)$, which implies $v = a\alpha/\beta$.

Therefore, $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$ with eigenvalue $(a+\beta)$.

[Edit: I have struck out the following paragraph. Anyone following the proof above should now go straight to the later post in this thread by PeroK, and later by me, which complete the answer.]
The proof then continues (in the simple case of distinct $A$-eigenvectors) by the standard technique of realizing that the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector, resulting (after a shift of the eigenvalue) in an eigenvector of $B$.

A more general solution would require a condition on the intersection of the nullspaces of $A$ and $B$, among other things (IIUC). As an illustration of what can go wrong in the simpler case of $AB=BA$, consider $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} ~~~\mbox{and}~~ B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ~,$$ for which $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is an eigenvector of $B\;$ but not $A$ (since that vector lies in the nullspace of $A$).

I'll leave it at that for now, and ask the question: how far did you [ @Infrared ] envisage that the answer should go? I'm guessing you'll require total completeness to qualify as a proper answer?

 

[zinq]

I want to elaborate on wrobel's point: If A = r e^iθ (r > 0) and B = x+iy are two complex numbers, then the definition of A^B is the set

A^B = {exp(B log(A))} as log(A) ranges over all its values. In other words

A^B = {exp((ln(r) + i(θ+2nπ)) ⋅ (x + iy) | n ∈ Z), i.e.,

A^B = {exp(x ln(r) - y (θ+2nπ) + i (y ln(r) + x (θ+2nπ))) | n ∈ Z}

where ln(r) = $\int_{1}^{r} t^{-1} ~dt$ as usual.

 

[nuuskur]

Spoiler: Problem 1

By linearity of T, it suffices to show T is continuous at 0. Take (x_n) \subseteq H_1 such that x_n \xrightarrow[n\to\infty]{}0\in H_1. Let z\in H_2, then
<br /> \langle z,Tx_n \rangle = \langle Sz, x_n \rangle \xrightarrow[n\to\infty]{}0 \in \mathbb K.<br />
As z is arbitrary it implies Tx_n\xrightarrow[n\to\infty]{}0 \in H_2.

 

[member 587159]

Spoiler: Problem 1

By linearity of T, it suffices to show T is continuous at 0. Take (x_n) \subseteq H_1 such that x_n \xrightarrow[n\to\infty]{}0\in H_1. Let z\in H_1, then
<br /> \langle z,Tx_n \rangle = \langle Sz, x_n \rangle \xrightarrow[n\to\infty]{}0 \in \mathbb K.<br />
As z is arbitrary it implies Tx_n\xrightarrow[n\to\infty]{}0 \in H_2.

I think you mean $z\in H_2$.

Please justify the step

$$\forall z \in H_2: \langle z, Tx_n\rangle\to 0\implies T x_n\to 0$$

PS: Glad to see you back here!

 

[Adesh]

The answer to question number 13 is ${\Large 6}$. But right now I don’t have an analytic proof (actually margins of my paper are too small to contain the proof ) . Will a graphical proof be counted? I can proof that there exist no integral solution to $$3a^3 +b^3 =6$$ by means of graph.

 

[member 587159]

The answer to question number 13 is ${\Large 6}$. But right now I don’t have an analytic proof (actually margins of my paper are too small to contain the proof ) . Will a graphical proof be counted? I can proof that there exist no integral solution to $$3a^3 +b^3 =6$$ by means of graph.

You can try to prove it using modular arithmetic. To get you started, modulo $3$ your equation becomes $b^3=0$ from which it follows that $b$ must he a multiple of $3$.

 

[fresh_42]

... from which it follows that $b$ must he a multiple of $3$.

I will probably ask: why does that follow?

That's a quirk of mine. I fight for the correct definition which is the reason here like Don Quichote fought his windmill.

 

[Adesh]

You can try to prove it using modular arithmetic. To get you started, modulo $3$ your equation becomes $b^3=0$ from which it follows that $b$ must he a multiple of $3$.

Actually, I don’t know Modular Arithemtic and I need to study it. I will do it and come with a proof.

 

[member 587159]

I will probably ask: why does that follow?

That's a quirk of mine. I fight for the correct definition which is the reason here like Don Quichote fought his windmill.

$b^3 = 0 \implies b = 0$ since $\mathbb{Z}/3\mathbb{Z}$ has no zero divisors ($3$ is prime), but I was just giving a hint :)

 

[fresh_42]

$b^3 = 0 \implies b = 0$ since $\mathbb{Z}/3\mathbb{Z}$ has no zero divisors ($3$ is prime), but I was just giving a hint :)s

I wasn't criticizing you at all. Au contraire! I liked your hint. I wanted @Adesh to read my answer and think about it ... and maybe learn the difference between prime and irreducible.

 

[member 587159]

I wasn't criticizing you at all. I wanted @Adesh to read my answer and think about it ... and maybe learn the difference between prime and irreducible.

I didn't take it as critisism, no worries.

 

[nuuskur]

Oh dear, I made a mistake. I will revise, @Math_QED (small world :) ). The implication in question is false. I think I was thinking about finite dimensions at the time I was writing the response :/

 

[member 587159]

Oh dear, I made a mistake. I will revise, @Math_QED (small world :) ). The implication in question is false. I think I was thinking about finite dimensions at the time I was writing the response :/

No worries! It was a trap carefully set up (you are not the first to walk into it)! As a hint: Do you know about the closed graph theorem in functional analysis?

 

[member 587159]

Ok, if I understand the closed graph theorem correctly, we get ..

Spoiler: Problem 1, take two

It's equivalent to show the graph of T is closed. Take (y_n,Ty_n)\in \mathrm{gr}T,\ n\in\mathbb N, such that
<br /> y_n\xrightarrow [n\to\infty]{H_1} y,\quad Ty_n \xrightarrow [n\to\infty]{H_2} x.<br />
By linearity of T we have
<br /> \|Ty-Ty_n\|^2 = \langle Ty - Ty_n, Ty-Ty_n \rangle = \langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0.<br />
Thus Ty_n \to Ty i.e Ty = x.

I think I see why the closed graph theorem is so useful here. We can assume without loss that the Ty_n converge.

$$\langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0$$

Why is this?

 

[nuuskur]

$$\langle ST(y-y_n), y-y_n \rangle \xrightarrow [n\to\infty]{} 0$$

Why is this?

I think I made a similar mistake :/ my thoughts were
<br /> \langle STz_n, z_n \rangle \leq \|ST\| \|z_n\|^2 \xrightarrow [n\to\infty]{}0,<br />
but that would make sense if ST was continuous and it needn't be :(

 

[member 587159]

I think I made a similar mistake :/ my thoughts were
<br /> \langle STz_n, z_n \rangle \leq \|ST\| \|z_n\|^2 \xrightarrow [n\to\infty]{}0,<br />
but that would make sense if ST was continuous and it needn't be :(

Yes, exactly. However, the solution I wrote down is not much longer than your attempt, so maybe try another attempt :)

 

[Infrared]

Therefore, $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$ with eigenvalue $(a+\beta)$.

I'm having a little bit of trouble following this: If I apply $A$, I get:

$A(B+ a\alpha/\beta) |a\rangle=AB|a\rangle+\frac{a\alpha}{\beta}A|a\rangle=AB|a\rangle+\frac{a^2\alpha}{\beta}|a\rangle,$

which I don't see how to simplify to $(a+\beta)|a\rangle.$ You can substitute $AB=BA+\alpha A+\beta B$, but it doesn't look like your terms cancel.

A more general solution would require a condition on the intersection of the nullspaces of $A$ and $B$, among other things (IIUC). As an illustration of what can go wrong in the simpler case of $AB=BA$, consider $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} ~~~\mbox{and}~~ B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ~,$$ for which $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is an eigenvector of $B\;$ but not $A$ (since that vector lies in the nullspace of $A$).

Actually, this is allowed! The nullspace of a matrix is the same thing as its $0$-eigenspace. Eigenvectors are not allowed to be zero, but eigenvalues are.

 

[zinq]

Here is what I'd say is a simpler answer to question 3. than the one I posted in #66:

Spoiler

Let the vector space be H = the separable infinite-dimensional Hilbert space of square-summable sequences of real numbers, and let the convex subset be R^∞, defined as the increasing union

R^∞ = $\bigcup_{n=1}^\infty$ Rⁿ

where Rⁿ = Rⁿ × {0} ⊂ Rⁿ⁺¹.

Clearly R^∞ is convex. R^∞ contains only points with finitely many nonzero components, so it is not all of H.

Any half-space D of H is of the form D = {x ∈ H | u ⋅ x ≤ c} where u is a fixed unit vector in H and c is a fixed real number. Given this half-space D, we show that R^∞ is not contained in it:

Let B = {e_n | n ∈ Z+} denote an orthonormal basis of H, and define the set C via C = ±B = {±e_n | n ∈ Z+}. Clearly an element of C making the smallest possible angle with the unit vector u is < 90º from u. Call this element e_u. Then the line

R ⋅ e_u

is entirely contained in R^∞ but is not entirely contained in the half-space D. Since D was arbitrary, this shows R^∞ is not contained in any half-space.

 

[strangerep]

I'm having a little bit of trouble following this: If I apply $A$, I get:

$A(B+ a\alpha/\beta) |a\rangle=AB|a\rangle+\frac{a\alpha}{\beta}A|a\rangle=AB|a\rangle+\frac{a^2\alpha}{\beta}|a\rangle,$

which I don't see how to simplify to $(a+\beta)|a\rangle.$ You can substitute $AB=BA+\alpha A+\beta B$, but it doesn't look like your terms cancel.

I get: $$\begin{array}{l}
A(B+ a\alpha/\beta) |a\rangle - (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (BA + \alpha A + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (aB + a \alpha + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle\\~~
~=~ a \alpha |a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta) a\alpha/\beta|a\rangle\\
~~ ~=~ 0 ~.
\end{array}
$$

 

[nuuskur]

I can't live with myself if I don't solve this problem. This is becoming personal.

Spoiler: Problem 1, the destroyer of souls

By theorem of closed graph, assume
<br /> y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.<br />
Was it really this simple all along ?!
<br /> \begin{align*}<br /> \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &amp;= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\<br /> &amp;= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0.<br /> \end{align*}<br />
Thus Ty=x and T is continuous.

 

[Adesh]

I can't live with myself if I don't solve this problem. This is becoming personal.

Spoiler: Problem 1, the destroyer of souls

By theorem of closed graph, assume
<br /> y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.<br />
Was it really this simple all along ?!
<br /> \begin{align*}<br /> \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &amp;= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\<br /> &amp;= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0.<br /> \end{align*}<br />
Thus Ty=x and T is continuous.

I appreciate and share that feeling.

 

[member 587159]

I can't live with myself if I don't solve this problem. This is becoming personal.

Spoiler: Problem 1, the destroyer of souls

By theorem of closed graph, assume
<br /> y_n \xrightarrow[n\to\infty]{H_1}y\qquad Ty_n \xrightarrow[n\to\infty]{H_2}x.<br />
Was it really this simple all along ?!
<br /> \begin{align*}<br /> \|x-Ty\|^2 = \langle x-Ty,x-Ty \rangle &amp;= \langle x-Ty, x \rangle - \langle x-Ty, Ty \rangle \\<br /> &amp;= \lim\langle x-Ty, Ty_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \lim \langle S(x-Ty), y_n \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle S(x-Ty), y \rangle - \langle x-Ty,Ty \rangle \\<br /> &amp;= \langle x-Ty,Ty \rangle - \langle x-Ty,Ty \rangle = 0.<br /> \end{align*}<br />
Thus Ty=x and T is continuous.

Well done!

 

[nuuskur]

Spoiler: Problem 5

Since f is unbounded, pick x_n\in X such that |f(x_n)|\geq n\|x_n\|,n\in\mathbb N. Without loss of generality, assume \|x_n\| \equiv 1 so we have |f(x_n)| \geq n. Fix x\in X. Define
<br /> y_n := x - \frac{f(x)}{f(x_n)}x_n, n\in\mathbb N.<br />
One readily verifies the y_n\in\mathrm{Ker}f. We also see that \|y_n-x\| = \left\lvert\frac{f(x)}{f(x_n)} \right\rvert \xrightarrow[n\to\infty]{}0. Thus y_n\to x and x\in\overline{\mathrm{Ker} f}.

Spoiler: Problem 5, finite dimensional argument

We know f must be non-zero, thus its image is \mathbb K i.e \dim\mathrm{Im} f = 1. We also know the kernel is closed if and only if f is continuous, thus \mathrm{Ker}f \neq \overline{\mathrm{Ker}f}. Since X \cong \mathrm{Ker} f \oplus \mathrm{Im}f we see
<br /> \dim X - \dim \overline{\mathrm{Ker}f} &lt; \dim X - \dim \mathrm{Ker}f = 1.<br />
So it must be that X = \overline{\mathrm{Ker}f}.

 

[member 587159]

Spoiler: Problem 5

Since f is unbounded, pick x_n\in X such that |f(x_n)|\geq n\|x_n\|,n\in\mathbb N. Without loss of generality, assume \|x_n\| \equiv 1 so we have |f(x_n)| \geq n. Fix x\in X. Define
<br /> y_n := x - \frac{f(x)}{f(x_n)}x_n, n\in\mathbb N.<br />
One readily verifies the y_n\in\mathrm{Ker}f. We also see that \|y_n-x\| = \left\lvert\frac{f(x)}{f(x_n)} \right\rvert \xrightarrow[n\to\infty]{}0. Thus y_n\to x and x\in\overline{\mathrm{Ker} f}.

Why is $y_n \in \ker f$?

 

[nuuskur]

Why is $y_n \in \ker f$?

By linearity of f
<br /> f\left ( x - \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f \left ( \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f(x) = 0.<br />

 

[member 587159]

By linearity of f
<br /> f\left ( x - \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f \left ( \frac{f(x)}{f(x_n)}x_n \right ) = f(x) - f(x) = 0.<br />

Yes, obviously. Sorry I missed a factor when I read it first. The solution seems correct to me but I'm not the one moderating it.

 

[PeroK]

I get: $$\begin{array}{l}
A(B+ a\alpha/\beta) |a\rangle - (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (BA + \alpha A + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle \\~~
~=~ (aB + a \alpha + \beta B)|a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta)(B+ a\alpha/\beta)|a\rangle\\~~
~=~ a \alpha |a\rangle + a^2 \alpha/\beta |a\rangle ~-~ (a+\beta) a\alpha/\beta|a\rangle\\
~~ ~=~ 0 ~.
\end{array}
$$

I think you can finish this off by:

Spoiler

I'm going to switch notation:

If $v_1$ is an eigenvector of $A$ with eigenvalue $\lambda_1$, then:
$$v_2 = [B + \frac {\lambda_1 \alpha}{\beta}I]v_1$$ Is an eigenvector of $A$ with eigenvalue $\lambda_2 = \lambda + \beta$. And:
$$v_3 = [B + \frac {\lambda_2 \alpha}{\beta}I]v_2$$ Is an eigenvector of $A$ with eigenvalue $\lambda_3 = \lambda_2 + \beta$.

This generates an infinite sequence of distinct eigenvalues, unless for some $k$ we have:
$$v_{k+1} = [B + \frac {\lambda_k \alpha}{\beta}I]v_k = 0$$ In which case, $v_k$ is a common eigenvector of $A$ and $B + \frac {\lambda_k \alpha}{\beta}I$, hence also an eigenvector of $B$.

We also have the case when $\beta = 0$ and:
$$AB - BA = \alpha A$$ In whch case:
$$A[B + \frac{\lambda \alpha }{\lambda - 1}I]v = \lambda [B + \frac{\lambda \alpha}{\lambda - 1}I]v \ \ (\lambda \ne 1)$$
In which case, $C = [B + \frac{\lambda \alpha}{\lambda - 1}I]v$ is an eigenvector of $A$ with eigenvalue $\lambda$. Hence $C$ maps the $\lambda$ eigenspace of $A$ into itself and must have a eigenvector restricted to this space. Therefore, $C$ and hence $B$ have a common eigenvector with $A$.

Finally, if $A$ only has eigenvalues $\lambda = 1$, then $A' = 2A$ has the same eigenvectors of $A$ and the result follows as above.

 

[PeroK]

Well done!

How do we know that the limit $x = \lim Ty_n$ exists?