Challenge Math Challenge - June 2020

[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?

 

[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.