Challenge Math Challenge - June 2020

[nuuskur]

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

This is what the closed graph theorem tells us. We can assume without loss of generality that the sequence of images also converges.

 

[member 587159]

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

The closed graph theorem allows us to assume that. That's why it is a useful theorem. You must show $y_n \to y$ implies $Ty_n \to Ty$ to deduce continuity but the theorem says you can already assume that $(Ty_n)_n$ converges, which makes life easier.

 

[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}.

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 = \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}.

The second appproach is how I solved it as well! However, some care has to be taken because if $X$ is infinite dimensional you wrote $\infty-\infty$.

 

[PeroK]

This is what the closed graph theorem tells us. We can assume without loss of generality that the sequence of images also converges.

The closed graph theorem allows us to assume that. That's why it is a useful theorem. You must show $y_n \to y$ implies $Ty_n \to Ty$ to deduce continuity but the theorem says you can already assume that $(Ty_n)_n$ converges, which makes life easier.

For any linear operator?

 

[nuuskur]

For any linear operator?

Yes. If the domain and codomain are complete spaces, then the proposition holds. In problem 1, it is applicable.

 

[member 587159]

For any linear operator?

For any linear operator between $F$-spaces (Banach spaces are $F$-spaces).

 

[wrobel]

Probl 5 has been got by nuuskur Congratulations!
By the way: there are no unbounded functionals in finite dimensional space

 

[nuuskur]

By the way: there are no unbounded functionals in finite dimensional space

That's right, so the second approach is useless. :D I am so rusty ..

 

[member 587159]

That's right, so the second approach is useless. :D I am so rusty ..

No that approach can be made to work:

By the isomorphism theorem, together with the fact that $f \neq 0$, we have
$$X/\ker f \cong \mathbb{R}$$
Since $f$ is unbounded, we know that $\ker f$ is not closed. Hence, $\ker f \subsetneq \overline{\ker f}$.
Consider the natural surjection $T: X/\ker f \to X/ \overline{\ker f}$. If the dimension of domain and codomain is equal, we get that $T$ is an isomorphism and in particular then $\ker T = 0$. But $\ker T = \overline{\ker f} / \ker f$, so this is impossible.

We conclude that $\dim (X/\overline{\ker f}) < \dim (X/\ker f)=1$ and thus $X = \overline{\ker f}$, proving the claim.

Also works in topological vector spaces. No need for norms.

 

[PeroK]

The closed graph theorem allows us to assume that. That's why it is a useful theorem. You must show $y_n \to y$ implies $Ty_n \to Ty$ to deduce continuity but the theorem says you can already assume that $(Ty_n)_n$ converges, which makes life easier.

I've looked up the closed graph theorem. Hopefully, this is correct:

Let $X, Y$ be Banach spaces and $T$ a closed linear operator from $X$ to $Y$. Then $T$ is bounded, hence continuous.

And, the definition of a closed linear operator is one whose graph is closed. I.e. if $x_n \rightarrow x$ and $Tx_n \rightarrow y$, then $Tx = y$.

And, for Hilbert spaces, if $x_n \rightarrow x$, then $Tx_n \rightarrow y$, for some $y$?

 

[member 587159]

I've looked up the closed graph theorem. Hopefully, this is correct:

Let $X, Y$ be Banach spaces and $T$ a closed linear operator from $X$ to $Y$. Then $T$ is bounded, hence continuous.

And, the definition of a closed linear operator is one whose graph is closed. I.e. if $x_n \rightarrow x$ and $Tx_n \rightarrow y$, then $Tx = y$.

In this case, we assumed that $T$ was closed hence continuous? If all linear operators on a Hilbert space are closed, then was the adjoint property unnecessary?

Shouldn't the solution to 1) have been to prove that that particular $T$ is closed? Not assume it?

I think you are misunderstanding. The proof provided showed that $Graf(T)$ was closed. The proof did this as follows:

We must show $\overline{Graf(T)} = Graf(T)$. So fix $(y,x) \in \overline{Graf(T)}$. Then there exists a sequence $(y_n)_n$ with $(y_n, Ty_n) \to (y,x)$ and thus by definition of product topology $y_n \to y, T y_n \to x$.

Here the proof of @nuuskur begins and shows that $(y,x) = (y,Ty) \in Graf(T)$, showing the desired inclusion.

 

[PeroK]

I think you are misunderstanding. The proof provided showed that $Graf(T)$ was closed. The proof did this as follows:

We must show $\overline{Graf(T)} = Graf(T)$. So fix $(y,x) \in \overline{Graf(T)}$. Then there exists a sequence $(y_n)_n$ with $(y_n, Ty_n) \to (y,x)$ and thus by definition of product topology $y_n \to y, T y_n \to x$.

Here the proof of @nuuskur begins and shows that $(y,x) = (y,Ty) \in Graf(T)$, showing the desired inclusion.

Sorry, I edited that post. Definitely confused!

 

[member 587159]

Sorry, I edited that post. Definitely confused!

Well, let me know if anything is still unclear! The goal is that everybody can learn from the challenges!

 

[PeroK]

Well, let me know if anything is still unclear! The goal is that everybody can learn from the challenges!

See the edited post. Thanks.

 

[member 587159]

I've looked up the closed graph theorem. Hopefully, this is correct:

Let $X, Y$ be Banach spaces and $T$ a closed linear operator from $X$ to $Y$. Then $T$ is bounded, hence continuous.

Yes, and the converse holds even without completeness (that is, if $T$ is continuous, the graph is closed).

And, the definition of a closed linear operator is one whose graph is closed. I.e. if $x_n \rightarrow x$ and $Tx_n \rightarrow y$, then $Tx = y$.

Exact.

And, for Hilbert spaces, if $x_n \rightarrow x$, then $Tx_n \rightarrow y$, for some $y$?

Not necessarily. Due to completeness, such an operator sends Cauchy sequences to Cauchy sequences and must be continuous:

https://math.stackexchange.com/ques...y-if-it-maps-cauchy-sequences-to-cauch/989611

Thus in general this can't be true.

 

[PeroK]

Yes, and the converse holds even without completeness (that is, if $T$ is continuous, the graph is closed).

Okay, I think I've got it.

By the closed graph theorem, if $T$ is closed it is continuous.

@nuuskur proved that $T$ is closed.

The pudding must be in the proof of the closed graph theorem, though!

 

[member 587159]

Okay, I think I've got it.

By the closed graph theorem, if $T$ is closed it is continuous.

@nuuskur proved that $T$ is closed.

The pudding must be in the proof of the closed graph theorem, though!

Well, it takes some work! First you need Baire's category theorem (countable intersections of open dense sets remain dense in complete metric spaces). Then as a corollary of that you prove the open mapping theorem (a continuous surjection between Banach spaces is an open map) and the closed graph theorem then becomes an easy corollary of that one, but it uses some deep results to get there.

 

[PeroK]

Well, it takes some work! First you need Baire's category theorem (countable intersections of open dense sets remain dense in complete metric spaces). Then as a corollary of that you prove the open mapping theorem (a continuous surjection between Banach spaces is an open map) and the closed graph theorem then becomes an easy corollary of that one, but it uses some deep results to get there.

I see now that's where my efforts at solving this problem were leading. I had a feeling the axiom of choice was at the root of it somewhere.

 

[member 587159]

I see now that's where my efforts at solving this problem were leading. I had a feeling the axiom of choice was at the root of it somewhere.

Yes, Baire Category theorem is equivalent with the axiom of dependent choice.

 

[nuuskur]

Alternatively, Zabreiko's lemma does also imply the theorem of closed graph for Banach spaces. Without completeness, it need not hold. For instance, the graph of \mathrm{id} : (c_{00}, \|\cdot\|_\infty) \to (c_{00}, \|\cdot\|_1) is closed, but the operator is not continuous.

 

[strangerep]

[...]
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 $B$.

A very minor improvement is that the equation above implies immediately that $v_k$ is an eigenvector of $B$ with eigenvalue $-\lambda_k\, \alpha/\beta~$ .

I'd been thinking along related lines, but my logic was a bit different. In your terminology, we have $$A v_2 ~=~ (\lambda_1 + \beta) v_2 ~.$$ If $A$ does NOT have a (nonzero) eigenvector with that eigenvalue, then the only solution is $v_2 = 0$, hence $$B v_1 = -\lambda_1 \alpha/\beta \; v_1$$ so $v_1$ is an eigenvector of $B$.

Else, we act again with $(B + \lambda_1 \alpha/\beta)$ on $v_2$, giving a $v_3$ with a different constant, and we can apply the same reasoning as in the previous paragraph. This algorithm must terminate because an $n\times n$ matrix has at most $n$ eigenvalues.

 

[Infrared]

Nice work to @PeroK and @strangerep! I think this completely covers that case that $(\alpha,\beta)\neq (0,0)$. Since the case $AB=BA$ is well-known, I'll count this as a solution.

 

[strangerep]

[...] I'll count this as a solution [to Q2].

Phew. Now I need to make some progress on Samalkhaiat's challenge #001 (since it looks like nobody else is having a go?). But I need to review some relativistic elasticity theory first.

 

[Haorong Wu]

Thank you to @Infrared for posing this question.

Hi, strangerep. Great solutions. But I still have a problem.

I do not understand this statement

the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

 

[PeroK]

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

It was shown that:
$$A(B+ a\alpha/\beta) |a\rangle = (a + \beta)(B+ a\alpha/\beta) |a\rangle$$
And that is the condition for $(B+ a\alpha/\beta) |a\rangle$ to be an eigenvector of $A$ with eigenvalue $a + \beta$.

It may be even clearer if we let $v = (B+ a\alpha/\beta) |a\rangle$, then it was shown that:
$$Av = (a + \beta)v$$
And we see that $v$ is indeed an eigenvector of $A$.

 

[Haorong Wu]

It was shown that:
$$A(B+ a\alpha/\beta) |a\rangle = (a + \beta)(B+ a\alpha/\beta) |a\rangle$$
And that is the condition for $(B+ a\alpha/\beta) |a\rangle$ to be an eigenvector of $A$ with eigenvalue $a + \beta$.

It may be even clearer if we let $v = (B+ a\alpha/\beta) |a\rangle$, then it was shown that:
$$Av = (a + \beta)v$$
And we see that $v$ is indeed an eigenvector of $A$.

Thanks, PeroK.

I can see $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$. But, in order to let $\left | a \right >$ be an eigenvector of $B$, $(B+ a\alpha/\beta) |a\rangle$ must equal to $\lambda \left | a \right >$. Otherwise, $(B+ a\alpha/\beta) |a\rangle =\left | b \right >$ where $\left |b \right > \neq \left |a \right >$ would not give something like $B \left | a \right > = k \left | a \right >$.

 

[PeroK]

Thanks, PeroK.

I can see $(B+ a\alpha/\beta) |a\rangle$ is an eigenvector of $A$. But, in order to let $\left | a \right >$ be an eigenvector of $B$, $(B+ a\alpha/\beta) |a\rangle$ must equal to $\lambda \left | a \right >$. Otherwise, $(B+ a\alpha/\beta) |a\rangle =\left | b \right >$ where $\left |b \right > \neq \left |a \right >$ would not give something like $B \left | a \right > = k \left | a \right >$.

True. It's an eigenvector of $A$. It's not necessarily an eigenvector of $B$.

 

[Adesh]

Summary:: Normed, Banach and Hilbert spaces, topology, geometry, linear algebra, integration, flux.
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

Which is the smallest natural number n∈N0 such that there are no integers a,b∈Z with 3a3+b3=n?

Spoiler

So, we have $3a^3=n−b^3$, that is $n−b^3\equiv 0\ mod 3$. We know
$$b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$

$$b\equiv 1 \mod 3 \implies −b^3 \equiv −1~~~~~~~~~~ (2)$$

$$b\equiv 2\ mod3⟹b^3≡8⟹b^3≡2⟹−b^3≡−2 ~~~~~~~` (3)$$

And we know
$$n\equiv 0\ mod 3 ~~~~~~~~~~~~~ (i)$$

$$n\equiv 1\mod 3 ~~~~~~~~~~~~~~ (ii)$$

$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)$$
So, the condition $n−b^3\equiv 0 \mod 3$ can be fulfilled by any $n$ among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).
For $n=0,1,2,3,4,5$ we can easily find solutions. My claim is that the equation $3a^3+b^3=6$ doesn't have any integral solutions.
${\large PROOF}$
$$b^3 = 6-3a^3$$

$$\implies b^3 \equiv 0 \mod 3$$
@fresh_42 schon seit, $b^3\equiv 0$ ist teilbar durch 3, das heißt$$\frac{b \times b \times b}{3} = M $$

Deshalb, 3 muss teilen b and hence $b=3k$ (I wrote German only for fun purpose not to offend you sir).
Now, substituting this value of b in our original equation we have :
$$3a^3 + 27k^3 = 6 $$

$$9k^3=2−a^3$$

$$⟹2−a^3\equiv 0 \mod 9$$

Let's check if that's possible (in every congruence that follows have modulus 9)

$$a\equiv 0⟹−a^3\equiv 0$$

$$a\equiv 1⟹−a^3\equiv −1$$

$$a \equiv 2 \implies a^3 \equiv 8

\implies -a^3 \equiv -8$$
$$a \equiv 3 \implies a^3 \equiv 27

\implies -a^3 \equiv 0$$

$$ a \equiv 4 \implies a^3 \equiv 64

\implies -a^3 \equiv -1$$

$$a\equiv 5 \implies a^3 \equiv 125

\implies -a^3 \equiv -8 $$

$$a \equiv 6 \implies a^3 \equiv 216

\implies -a^3 \equiv 0$$

$$ a\equiv 7 \implies a^3 \equiv 343

\implies -a^3 \equiv -1$$

$$a \equiv 8 \implies a^3 \equiv 512

\implies-a^3 \equiv -8$$

That is to say, we have only three possibilities for cubic residues:

$$−a^3\equiv 0$$

$$−a^3\equiv−1$$

$$−a^3\equiv−8 $$

And $2\equiv 2\mod 9$. So, only possibilities:

$$2−a^3\equiv 2\mod 9$$

$$2−a^3\equiv 1\ mod 9$$

$$2−a^3\equiv−6\mod 9$$

Which means $2−a^3\equiv 0 \mod 9$ is not possible, hence there is no integral solution to $a^3+9k^3=6$, which in turn means that
$$3a^3 +b^3 =6$$ have no integral solution
${\tiny Maths} ~{\large QED}$

 

[fresh_42]

Spoiler

So, we have 3a3=n−b3, that is n−b3≡0mod3. We know
$$
b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$
b≡1mod3⟹−b3≡−1 (2)
b≡2mod3⟹b3≡8⟹b3≡2⟹−b3≡−2 (3)
And we know

n≡0mod3 (i)
n≡1mod3 (ii)
$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)
$$
So, the condition n−b3≡0mod3 can be fulfilled by any n among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).

For n=0,1,2,3,4,5 we can easily find solutions. My claim is that the equation 3a3+b3=6 doesn't have any integral solutions.

PROOF:
$$
b^3 = 6-3a^3 \
\implies b^3 \equiv 0 \mod 3
$$
@fresh_42 schon seit, b3−0 ist teilbar durch 3, das heißt
$$
\frac{b \times b \times b}{3} = M $$
Deshalb, 3 muss teilen b and hence b=3k (I wrote German only for fun purpose not to offend you sir).

Now, substituting this value of b in our original equation we have :
$$
3a^3 + 27k^3 = 6 $$
9k3=2−a3
⟹2−a3≡0mod9
Let's check if that's possible
a≡0⟹−a3≡0
a≡1⟹−a3≡−1
$$a \equiv 2 \implies a^3 \equiv 8 \implies
a^3 \equiv 2 \implies -a^3 \equiv -2$$

$$a \equiv 3 \implies a^3 \equiv 27 \implies
-a^3 \equiv 0$$
$$ a \equiv 4 \implies a^3 \equiv 64 \implies
-a^3 \equiv -1$$
$$a\equiv 5 \implies a^3 \equiv 125 \implies
-a^3 \equiv -8 $$
$$a \equiv 6 \implies a^3 \equiv 216 \implies
-a^3 \equiv 0$$
$$ a\equiv 7 \implies a^3 \equiv 343 \implies
-a^3 \equiv -1$$
$$a \equiv 8 \implies a^3 \equiv 512 \implies
-a^3 \equiv -8$$
That is to say, we have only three possibilities for cubic residues:
−a3≡0
−a3≡−1
−a3≡−8
And 2≡2mod9. So, only possibilities:
2−a3≡2mod9
2−a3≡1mod9
2−a3≡−6mod9
Which means 2−a3≡0mod9 is not possible, hence there is no integral solution to a3+9k3=6, which in turn means that
$$
3a^3 +b^3 =6$$ have no integral solution.

##{\tiny Maths_} ~{\large QED}

Can you either edit this properly, or say in words what you mean. I have difficulties to follow your reasoning.
See https://www.physicsforums.com/help/latexhelp/

 

[Adesh]

Can you either edit this properly, or say in words what you mean. I have difficulties to follow your reasoning.
See https://www.physicsforums.com/help/latexhelp/

I really don't know what happened in actuality, but I have repaired everything.

 

[fresh_42]

Which part? I write everything in Latex but I really don't know what has happened, I'm surprised.

Everything before

So, we have 3a3=n−b3, that is n−b3≡0mod3. We know
$$
b \equiv 0 \mod 3 \implies -b^3\equiv 0 ~~~~~~~(1) $$
b≡1mod3⟹−b3≡−1 (2)
b≡2mod3⟹b3≡8⟹b3≡2⟹−b3≡−2 (3)
And we know

n≡0mod3 (i)
n≡1mod3 (ii)
$$n\equiv 2 \mod 3 ~~~~~~~~~~~~(iii)
$$
So, the condition n−b3≡0mod3 can be fulfilled by any n among (i), (ii) and (iii) (if b is allowed to be any integer). Because of this we have to move for inspection (trial and error method).

For n=0,1,2,3,4,5 we can easily find solutions. My claim is that the equation 3a3+b3=6 doesn't have any integral solutions.

here. But this isn't needed, so let's go further.

PROOF:
$$
b^3 = 6-3a^3 \
\implies b^3 \equiv 0 \mod 3
$$
@fresh_42 schon seit, b3−0 ist teilbar durch 3, das heißt
$$
\frac{b \times b \times b}{3} = M $$
Deshalb, 3 muss teilen b and hence b=3k (I wrote German only for fun purpose not to offend you sir).

Now, substituting this value of b in our original equation we have :
$$
3a^3 + 27k^3 = 6 $$
9k3=2−a3
⟹2−a3≡0mod9
Let's check if that's possible

Beside that the LaTeX code is broken, I also don't know what your modules are. You switch between modulo $3,2,9$ so how am I supposed to know which one you mean?
The correct notation is
$$
a \equiv b \mod n \Longleftrightarrow n\,|\,(a-b) \Longleftrightarrow a\equiv b (n)
$$

a≡0⟹−a3≡0
a≡1⟹−a3≡−1

So you need to write $a\equiv 0 (9)$ and $a\equiv 1 (9)$ if nine is the module you mean. However, $a\equiv 1 (n)$ implies $a^3\equiv 1 (n)$ for all $n\in \mathbb{N}$. This equals $-1$ only in case $n=2$. How should I know that you are considering the remainders of division by two?

$$a \equiv 2 \implies a^3 \equiv 8 \implies
a^3 \equiv 2 \implies -a^3 \equiv -2$$

$$a \equiv 3 \implies a^3 \equiv 27 \implies
-a^3 \equiv 0$$
$$ a \equiv 4 \implies a^3 \equiv 64 \implies
-a^3 \equiv -1$$
$$a\equiv 5 \implies a^3 \equiv 125 \implies
-a^3 \equiv -8 $$
$$a \equiv 6 \implies a^3 \equiv 216 \implies
-a^3 \equiv 0$$
$$ a\equiv 7 \implies a^3 \equiv 343 \implies
-a^3 \equiv -1$$
$$a \equiv 8 \implies a^3 \equiv 512 \implies
-a^3 \equiv -8$$
That is to say, we have only three possibilities for cubic residues:
−a3≡0
−a3≡−1
−a3≡−8
And 2≡2mod9. So, only possibilities:
2−a3≡2mod9
2−a3≡1mod9
2−a3≡−6mod9
Which means 2−a3≡0mod9 is not possible, hence there is no integral solution to a3+9k3=6, which in turn means that
$$
3a^3 +b^3 =6$$ have no integral solution.

##{\tiny Maths_} ~{\large QED}

 

[Adesh]

So you need to write a≡0(9) and a≡1(9) if nine is the module you mean. However, a≡1(n) implies a3≡1(n) for all n∈N. This equals −1 only in case n=2. How should I know that you are considering the remainders of division by two?

(Please tell me why when I quote some post the latex code doesn't get quoted rather the mathematical symbols get copied).

Well, for getting the minus sign I'm using this fact
$$a \equiv b \mod n $$
$$-1 \equiv -1 \mod n$$

Now, multiplying corresponding sides of congruences (multiplication preserves the congruency), so we have
$$
-a \equiv - b \mod n
$$

 

[fresh_42]

(Please tell me why when I quote some post the latex code doesn't get quoted rather the mathematical symbols get copied).

This is a bug when you use "quote".

Well, for getting the minus sign I'm using this fact
$$a \equiv b \mod n $$
$$-1 \equiv -1 \mod n$$

Now, multiplying corresponding sides of congruences (multiplication preserves the congruency), so we have
$$
-a \equiv - b \mod n
$$

Sorry, I've overseen the minus sign. But this doesn't change the fact that you need to note the module with every $\equiv$ sign.

 

[Adesh]

This is a bug when you use "quote".

Sorry, I've overseen the minus sign. But this doesn't change the fact that you need to note the module with every $\equiv$ sign.

I have made corresponding repairs in my solution, sorry for disturbance caused.

 

[fresh_42]

Let's check if that's possible (in every congruence that follows have modulus 9)

$$a\equiv 0⟹−a^3\equiv 0$$

$$a\equiv 1⟹−a^3\equiv −1$$

$$a \equiv 2 \implies a^3 \equiv 8

\implies a^3 \equiv 2 \implies -a^3 \equiv -2$$

If $a^3\equiv 8 (9)$ then $a^3 \not\equiv 2 (9)$.

Hint: From $2=a^3+9k^3$ continue by considering the equation modulo $3$. This is shorter and easier.

 

[Adesh]

If $a^3\equiv 8 (9)$ then $a^3 \not\equiv 2 (9)$.

Hint: From $2=a^3+9k^3$ continue by considering the equation modulo $3$. This is shorter and easier.

I'm editing my post every time you're pointing out an error, is that okay or should I consider reposting?

I will surely solve it by your hint also.

 

[fresh_42]

I'm editing my post every time you're pointing out an error, is that okay or should I consider reposting?

I will surely solve it my your hint also.

Repost it. That's better.
You were here:

Assume there is a solution $6=3a^3+b^3.$
Then $0\equiv b^3 \,(3)$ and $b=3k$, i.e. $6=3a^3+27k^3$ or $2=a^3+9k^3$.
This means $a^3\equiv 2\,(3)$.

Now go ahead and examine how $a$ must look like so that is possible.

 

[Adesh]

Now go ahead and examine how a must look like so that is possible.

I get $a^3 = 3m +2$.

 

[fresh_42]

I get $a^3 = 3m +2$.

Yes, but which $a$ have this shape? $a$ cannot be divisible by three, can it have the remainder $1$ by division by $3$?

 

[Adesh]

Yes, but which $a$ have this shape? $a$ cannot be divisible by three, can it have the remainder $1$ by division by $3$?

$$a^3 \equiv 2 \mod 3$$
$$ 8\equiv 2 \mod 3$$
By Symmetry we have
$$ a^3 \equiv 2^3 \mod 3$$
$$ a \equiv 2 \mod 3$$
That is to say, $a = 3k +2$.

 

[fresh_42]

$$a^3 \equiv 2 \mod 3$$
$$ 8\equiv 2 \mod 3$$
By Symmetry transitivity we have
$$ a^3 \equiv 2^3 \mod 3$$
$$ a \equiv 2 \mod 3$$

Why is this? E.g. $5^3 \equiv 2^3 \mod 13$ but $5\not\equiv 2 \mod 13$.
You cannot take the root in general. Why is it true here?

That is to say, $a = 3k +2$.

And now you can go back: Take $a^3=(3m+2)^3= \ldots$ and consider it modulo $9$.
Btw., you shouldn't take the same letter ($k$) for two different numbers. We already used $k$.

 

[strangerep]

I do not understand this statement

the vector $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector

Why $(B+ a\alpha/\beta) |a\rangle$ must be a multiple of an $A$-eigenvector?

The paragraph in which I wrote the above was misleading at best, so I will go back a put a strike through it.

I suggest you ignore that paragraph and proceed directly to the 2nd part of the solution given first by PeroK, and later by me in a different form.

 

[Haorong Wu]

The paragraph in which I wrote the above was misleading at best, so I will go back a put a strike through it.

I suggest you ignore that paragraph and proceed directly to the 2nd part of the solution given first by PeroK, and later by me in a different form.

Thanks, strangerep.

 

[Adesh]

Why is this? E.g. $5^3 \equiv 2^3 \mod 13$ but $5\not\equiv 2 \mod 13$.
You cannot take the root in general. Why is it true here?

And now you can go back: Take $a^3=(3m+2)^3= \ldots$ and consider it modulo $9$.
Btw., you shouldn't take the same letter ($k$) for two different numbers. We already used $k$.

Okay Herr,
$$ a^3 \equiv 2^3 \mod 3 ~~~~~(1) $$
$$a \equiv 2 \mod 3 ~~~~~~~ (2)$$
(1) is given and (2) is possible, knowing that if (2) is given we can get (1), therefore (I’m not taking cubic root, I just reasoned it out, maybe not very nicely)
$$ a \equiv 2 \mod 3$$
So, for an integer $t$ we have $a = 3t +2$. Plugging this value in the Eqaution
$$ a^3 = 2-9m^3 $$
We get
$$ 27 t^3 +8 + 54t^2 +36t = 2-9m^3$$
$$27t^3 +6 +54t^2 +36t =-9m^3$$
Well, that means
$$ 27t^3 + 54t^2 +36t +6 \equiv 0 \mod 9$$
$$ 27t^3 +54t^2 +36t \equiv -6 \mod 9$$
But we know that $27t^3 +54t^2 +36t \equiv 0 \mod 9$, hence the contradiction and therefore the equation doesn’t have any integral solution.

 

[fresh_42]

Okay Herr,
$$ a^3 \equiv 2^3 \mod 3 ~~~~~(1) $$
$$a \equiv 2 \mod 3 ~~~~~~~ (2)$$
(1) is given and (2) is possible, knowing that if (2) is given we can get (1), therefore (I’m not taking cubic root, I just reasoned it out, maybe not very nicely)

This is not logical. The fact that $a\equiv 2\mod 3$ is possible, does not mean that it is true. You have to consider the possible cases! There are only three possible cases modulo $3$, the remainders $0,1,2$. Hence you can check them:
Case $1$: if $a\equiv 0 \mod 3 \text{ then } a^3 \equiv ... \mod 3$
Case $2$: if $a\equiv 1 \mod 3 \text{ then } a^3 \equiv ... \mod 3$
Case $3$: if $a\equiv 2 \mod 3 \text{ then } a^3 \equiv ... \mod 3$

The rest is ok, although a bit lengthy. From $27t^3+ 8+54t^2+36t=2-9m^3$ you can directly conclude $8\equiv 2 \mod 9$ which is the desired contradiction.

 

[nuuskur]

Some thoughts, but potentially a dead-end. Don't know how to make use of completeness of X.

Spoiler: Problem 9, incomplete

Let A:X\to Y be linear. Assume
<br /> B:Y&#039;\to X&#039;,\quad g\mapsto g\circ A<br />
is well-defined. Take x_n\in X,n\in\mathbb N, such that x_n\to 0. Fix g\in Y&#039;, then
<br /> (Bg)(x_n) = g(Ax_n),\quad n\in\mathbb N.<br />
Since g\circ A is bounded, we have |g(Ax_n)| \leq M_g\|x_n\|\to 0. Thus
<br /> \forall g\in Y&#039;,\quad \lim g(Ax_n) = 0.<br />
If \lim Ax_n exists, then we can switch \lim and g. Since g is arbitrary, it must hold that \lim Ax_n = 0. Thus, A would be continuous at 0. Equivalently, A would be bounded.

..but why would \lim Ax_n have to exist?

 

[wrobel]

Is it time to give a hint?

 

[wrobel]

It is time I guess.

THEOREM. A subset of a normed space is bounded iff it is weakly bounded.

Consider problems 1, 9 in this light

 

[nuuskur]

I hope I have the definition right. A\subseteq X of a normed space is weakly bounded if \sup _{x\in A} |x^*(x)|&lt;\infty for every x^* \in X^*.

Spoiler: Problem 9, gibberish..

So, we could show \{Ax \mid x\in X\}\subseteq Y is bounded. Fix g\in Y&#039;, then
<br /> \sup _{x\in X} |g(Ax)| &lt; \infty<br />
because by assumption Y&#039;\to X&#039;,\ g\mapsto g\circ A, is well-defined i.e g\circ A is bounded.

But this is not right. \|Ax\| \leq K doesn't imply \|Ax\| \leq L\|x\|.

I am short-circuiting. Something somewhere is false. My functional analysis is a pile of garbage..

 

[strangerep]

My functional analysis is a pile of garbage..

Heh, I once heard a description of FA as being a "pile of abstract nonsense".