Math Challenge - May 2019

[fresh_42]

The question described above is written on pages 74-77 of this book. The reverse is questionable.

O.k. but we don't have to fall back on rationals here. Epsilontic and continuity will do.

 

[Periwinkle]

O.k. but we don't have to fall back on rationals here. Epsilontic and continuity will do.

$\xi(x) = (\min\{x,x_0\}+\max\{x,x_0\})/2$

That's fair. There is no need for exact equality.

 

[fresh_42]

$\xi(x) = (\min\{x,x_0\}+\max\{x,x_0\})/2$

I can't see why this avoids AC. Formally we did this:
$$
\Lambda(x):=\left\{ \xi \in (\min\{x,x_0\},\max\{x,x_0\}) \, : \, \dfrac{f(x)-f(x_0)}{x-x_0}=f\,'(\xi)\right\}
$$
The mean value theorem guarantees us that all $\Lambda(x)\neq \emptyset$, but we need more: namely a function $$\xi \, : \, [a,b]-\{x_0\}\longrightarrow \bigcup_{x\in [a,b]-\{x_0\}} \Lambda(x)$$
Narrowing the interval doesn't change the argument.

 

[Periwinkle]

I can't see why this avoids AC. Formally we did this:
$$
\Lambda(x):=\left\{ \xi \in (\min\{x,x_0\},\max\{x,x_0\}) \, : \, \dfrac{f(x)-f(x_0)}{x-x_0}=f\,'(\xi)\right\}
$$
The mean value theorem guarantees us that all $\Lambda(x)\neq \emptyset$, but we need more: namely a function $$\xi \, : \, [a,b]-\{x_0\}\longrightarrow \bigcup_{x\in [a,b]-\{x_0\}} \Lambda(x)$$
Narrowing the interval doesn't change the argument.

Exist the limit $c:=\lim_{x \to x_0}f\,'(x)\,.$

Therefore, for all $\epsilon$, there is a $\delta$ that if $|x-x_0| \lt \delta$, then $|f\,'(x)-c| \lt \epsilon$.

Based on mean value theorem $\dfrac{f(x)-f(x_0)}{x-x_0}\,$ equal to one of $f\,'(\xi)$, where $\left| \xi-x_0 \right| \lt \delta$.

Therefore without choosing $\xi$ we know $\left| \dfrac{f(x)-f(x_0)}{x-x_0}\ -c \right| \lt \epsilon$ if $|x-x_0| \lt \delta$.

 

[fresh_42]

We have chosen $\xi$, but only one element from one non-empty set $\Lambda(x)$ and use $|\xi -x_0|<|x-x_0|< \delta\,.$

 

[Couchyam]

Spoiler: (Problem 9)

Let $n:=i_\ell i_{\ell-1}\dots i_0$ be the decimal representation of $n\in\mathbb N$ (that is, $n=\sum_{j=0}^\ell i_j 10^j$.) Since $n\geq 10^\ell$, $\frac{1}{n}\leq 10^{-\ell}$. Now, for a fixed $\ell$, there are no more than $9^{\ell+1}$ numbers between $10^\ell$ and $10^{\ell+1}$ whose decimal expansions do not contain the digit $9$ (this is because each number in this range has a unique decimal expansion with at most $\ell+1$ digits, and the set of length $\ell+1$ strings composed of the digits $\{0,...,8\}$ has size $9^{\ell+1}$). Hence,
\begin{align*}
\sum_{n=1}^\infty \frac{\epsilon_n}{n}=\sum_{\ell=0}^\infty \bigg(\sum_{10^\ell\leq n<10^{\ell+1}}\frac{\epsilon_n}{n}\bigg)<\sum_{\ell=0}^\infty \frac{9^{\ell+1}}{10^\ell}=90<\infty
\end{align*}
(Heuristically, the convergence of the series is related to the Cantor-set-like support of the coefficients $\frac{\epsilon_n}{n}$, in the sense that as $n$ grows, the gaps between regions where $\epsilon_n$ is nonzero expand proportionally.)

 

[SpinFlop]

Attempt at Problem 4

Part a) follows directly from the fundamental property of self adjoint operators.
$$ (\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})$$
where $\hat T$ is a self-adjoint linear operator and ${\mathbf a},{\mathbf b}$ are two vectors in a given vector space.
In the following proof ${\mathbf a},{\mathbf b}$ will be taken as eigenvectors of $\hat T$ with eigenvalues:
$$\hat T {\mathbf a} = \alpha{\mathbf a}\\
\hat T {\mathbf b} = \beta{\mathbf b}$$
First show that eigenvalues of a self-adjoint operator are real:
$$
(\hat T{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\hat T{\mathbf a})\\
(\alpha{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\alpha{\mathbf a})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf a}) = \alpha({\mathbf a}\cdot{\mathbf a})\\
\alpha^{*} = \alpha
$$
Thus $\alpha$ is real.
Now show that eigenvectors of $\hat T$ with distinct eigenvalues are orthogonal
$$
(\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})\\
(\alpha{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\beta{\mathbf b})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})\\
\alpha({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})
$$
where the last line follows from the fact that the eigenvalues must be real.
Since the eigenvectors ${\mathbf a},{\mathbf b}$ have distinct eigenvalues this means $\alpha\neq\beta$ and so necessarily ${\mathbf a}\cdot{\mathbf b} = 0$.
Thus, by definition ${\mathbf a},{\mathbf b}$ are orthogonal.

Part b)
If I understand correctly, for a given function $f(t)$ in the Hilbert space $\mathcal {H} =L_{2}([0,1])$ the linear operator $T_g$ simply multiplies $f(t)$ by the function $g(t)$:
$$T_{g}(f)(t):=g(t)f(t)$$
If so, then the eigenvalue problem requires
$$g(t)f(t) = \lambda_{g}f(t)$$
The trivial possibility is that the function $g(t)$ is a constant and so we get $\lambda_{g} = m = M$
If $g(t)$ is not a constant, then I am not so sure I know how to proceed, the only path forward that I see is to define our eigenfunctions as delta functions, in which case our eigenvalue spectrum is continuous over the interval $[M,m]$.

 

[fresh_42]

Attempt at Problem 4

Part a) follows directly from the fundamental property of self adjoint operators.
$$ (\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})$$
where $\hat T$ is a self-adjoint linear operator and ${\mathbf a},{\mathbf b}$ are two vectors in a given vector space.
In the following proof ${\mathbf a},{\mathbf b}$ will be taken as eigenvectors of $\hat T$ with eigenvalues:
$$\hat T {\mathbf a} = \alpha{\mathbf a}\\
\hat T {\mathbf b} = \beta{\mathbf b}$$
First show that eigenvalues of a self-adjoint operator are real:
$$
(\hat T{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\hat T{\mathbf a})\\
(\alpha{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\alpha{\mathbf a})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf a}) = \alpha({\mathbf a}\cdot{\mathbf a})\\
\alpha^{*} = \alpha
$$
Thus $\alpha$ is real.
Now show that eigenvectors of $\hat T$ with distinct eigenvalues are orthogonal
$$
(\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})\\
(\alpha{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\beta{\mathbf b})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})\\
\alpha({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})
$$
where the last line follows from the fact that the eigenvalues must be real.
Since the eigenvectors ${\mathbf a},{\mathbf b}$ have distinct eigenvalues this means $\alpha\neq\beta$ and so necessarily ${\mathbf a}\cdot{\mathbf b} = 0$.
Thus, by definition ${\mathbf a},{\mathbf b}$ are orthogonal.

Part b)
If I understand correctly, for a given function $f(t)$ in the Hilbert space $\mathcal {H} =L_{2}([0,1])$ the linear operator $T_g$ simply multiplies $f(t)$ by the function $g(t)$:
$$T_{g}(f)(t):=g(t)f(t)$$
If so, then the eigenvalue problem requires
$$g(t)f(t) = \lambda_{g}f(t)$$
The trivial possibility is that the function $g(t)$ is a constant and so we get $\lambda_{g} = m = M$
If $g(t)$ is not a constant, then I am not so sure I know how to proceed, the only path forward that I see is to define our eigenfunctions as delta functions, in which case our eigenvalue spectrum is continuous over the interval $[M,m]$.

What you wrote is correct so far. But the question was to determine the spectrum, i.e. the complement of the resolvent set, not the point spectrum of eigenvalues.

 

[Periwinkle]

Spoiler: Question 8

$$ (a(x),(b(x)) =\int_{-\pi /2}^{\pi/2} (11\sin(x) + 8\cos(x))\cdot (4\sin(x) + 13\cos(x)) \, dx =

\\ \int_{-\pi /2}^{\pi/2} (44\sin^2(x) + 175 \cos(x) \sin(x) + 104 \cos^2(x)) \, dx =
44 \pi /2 +104 \pi/2 = 148 \pi /2. $$

$$ \left|a \right| ^2 = (a(x),(a(x)) =
\\ \int_{-\pi /2}^{\pi/2} (121\sin^2(x) + 176 \cos(x) \sin(x) + 64 \cos^2(x)) \, dx = 121 \pi /2 +64 \pi/2 = 185 \pi /2.$$

$$ \left|b \right| ^2 = (b(x),(b(x)) =
\\ \int_{-\pi /2}^{\pi/2} (16\sin^2(x) + 104 \cos(x) \sin(x) + 169 \cos^2(x)) \, dx = 16 \pi /2 + 169 \pi/2 = 185 \pi /2.$$

$$ \cos(\phi) = \frac {(a(x),(b(x))} {\left|a \right| \left|b \right| } = \frac {148 \pi /2} { 185 \pi /2 }= 0.8. $$

$$ \phi = 0.6435. $$

 

[fresh_42]

Spoiler: Question 8

$$ (a(x),(b(x)) =\int_{-\pi /2}^{\pi/2} (11\sin(x) + 8\cos(x))\cdot (4\sin(x) + 13\cos(x)) \, dx =

\\ \int_{-\pi /2}^{\pi/2} (44\sin^2(x) + 175 \cos(x) \sin(x) + 104 \cos^2(x)) \, dx =
44 \pi /2 +104 \pi/2 = 148 \pi /2. $$

$$ \left|a \right| ^2 = (a(x),(a(x)) =
\\ \int_{-\pi /2}^{\pi/2} (121\sin^2(x) + 176 \cos(x) \sin(x) + 64 \cos^2(x)) \, dx = 121 \pi /2 +64 \pi/2 = 185 \pi /2.$$

$$ \left|b \right| ^2 = (b(x),(b(x)) =
\\ \int_{-\pi /2}^{\pi/2} (16\sin^2(x) + 104 \cos(x) \sin(x) + 169 \cos^2(x)) \, dx = 16 \pi /2 + 169 \pi/2 = 185 \pi /2.$$

$$ \cos(\phi) = \frac {(a(x),(b(x))} {\left|a \right| \left|b \right| } = \frac {148 \pi /2} { 185 \pi /2 }= 0.8. $$

$$ \phi = 0.6435. $$

That will take me awhile, since one of us has made a mistake and I have to figure out who and where.

Correction: We were both right. I forgot a square root in the denominator.

How do you find my solution:

We define $f(x)=\sin(x)-6\cos(x)\; , \;g(x)=6\sin(x)+\cos(x)$ and observe, that $\{\,f,g\,\}$ is a orthogonal basis for a two dimensional subspace of $L^2\left( \left[ -\frac{\pi}{2},+\frac{\pi}{2} \right] \right)$ with $\gamma :=|f|=|g|=\sqrt{\dfrac{37 \pi}{2}}$. As we are interested in an angle, we won't have to bother the length of our coordinate vectors, i.e. we do not need to normalize them. Now we have $a=-f+2g\, , \,b=-2f+g$ and
\begin{align*}
\cos \varphi &= \cos (\sphericalangle (a,b))\\
&= \cos(\sphericalangle (-f+2g,-2f+g))\\
&= \dfrac{\langle -f+2g,-2f+g \rangle}{|-f+2g|\cdot |-2f+g|}\\
&= 2 \; \dfrac{\langle f,f\rangle + \langle g,g \rangle}{\sqrt{\left( |f|^2+4|g|^2\right)} \cdot \sqrt{\left( 4|f|^2+|g|^2 \right)}}\\
&= 2\; \dfrac{\gamma^2+\gamma^2}{\sqrt{5\gamma^2 \cdot 5\gamma^2}}\\
&= \dfrac{4}{5}
\end{align*}
and $\varphi \approx 36.87° \approx 0.2 \pi$

 

[Periwinkle]

That will take me awhile, since one of us has made a mistake and I have to figure out who and where.

Correction: We were both right. I forgot a square root in the denominator.

How do you find my solution:

We define $f(x)=\sin(x)-6\cos(x)\; , \;g(x)=6\sin(x)+\cos(x)$ and observe, that $\{\,f,g\,\}$ is a orthogonal basis for a two dimensional subspace of $L^2\left( \left[ -\frac{\pi}{2},+\frac{\pi}{2} \right] \right)$ with $\gamma :=|f|=|g|=\sqrt{\dfrac{37 \pi}{2}}$. As we are interested in an angle, we won't have to bother the length of our coordinate vectors, i.e. we do not need to normalize them. Now we have $a=-f+2g\, , \,b=-2f+g$ and
\begin{align*}
\cos \varphi &= \cos (\sphericalangle (a,b))\\
&= \cos(\sphericalangle (-f+2g,-2f+g))\\
&= \dfrac{\langle -f+2g,-2f+g \rangle}{|-f+2g|\cdot |-2f+g|}\\
&= 2 \; \dfrac{\langle f,f\rangle + \langle g,g \rangle}{\sqrt{\left( |f|^2+4|g|^2\right)} \cdot \sqrt{\left( 4|f|^2+|g|^2 \right)}}\\
&= 2\; \dfrac{\gamma^2+\gamma^2}{\sqrt{5\gamma^2 \cdot 5\gamma^2}}\\
&= \dfrac{4}{5}
\end{align*}
and $\varphi \approx 36.87° \approx 0.2 \pi$

I was only the inner product of the three-dimensional Euclidean space before my eyes, which has the same meaning in Hilbert space.

 

[Periwinkle]

Question 4

​

Self-adjoint linear operator's eigenvalues are real ​

$$ \lambda (x,x) = (\lambda x, x) = (Ax,x) = (x, Ax) = (x,\lambda x,) = \bar \lambda (x,x) $$ However, in the Euclidean space $(x,x) =0$ follows $x = 0$, so $\lambda = \bar \lambda$.

The eigenvectors belonging to different eigenvalues are orthogonal ​

$$ (Ax, Ay) = (AAx, y) = (\lambda _1 Ax, y) = \lambda_1^2(x,y) $$ $$ (Ax, Ay) = (x, AAy) = (x, \lambda _2 Ay) = \lambda_2^2(x,y) $$ Because $\lambda_1$ and $\lambda_2$ is different, so $(x,y)$ is equal to $0$.

The spectrum of the $T_g$ operator is the interval $[m,M]$​

The regular values of the operator $T_g$ are the $\lambda$ numbers for which the $(T_g-\lambda I)^{-1}$ operator has a value in the whole space. The other values of $\lambda$ are the spectrum of the operator. The inverse operator is defined by the following formula: $$ (T_g-\lambda I)^{-1} f(t) = \frac 1 {g(t) -\lambda} f(t). $$ It must be proved that for each number $\lambda$ in interval $[m,M]$, there is an $s(t)$element of space $L^2([0,1])$, that is not mapped to an element of space $L^2([0,1])\,$ by the above inverse operator. It follows from the continuity of the $g(t)$ function that if $\lambda$ is a point in the $[m, M]$ interval, then $g (t) = \lambda$ on some $t_{\lambda}$. Also, provided that $\lambda$ is different from $m$, there is in the $[0,1]$ interval a $t_1, t_2, \dots, t_i, \dots$ monotone growing sequence that if $t_i \leq t \lt t_{\lambda}$, then $$\frac 1 {|g(t) - \lambda|} \gt i.$$ If $\lambda = m$, there is a similar monotone descending sequence. The appropriate $s(t)\,$element of $L^2([0,1])\,$ is constructed as follows.

The value of the function $s(t)$ on the $[t_i, t_{i+1})$ interval is $$ \frac 1 {i\sqrt {t_{i+1} - t_i}}. $$ At points outside all $(t_i, t_{i+1})$ intervals $s(t)$ is $0$ everywhere. The integral of the $s^2 (t)$ function is equal $$ \sum_{i=1}^\infty {(t_{i+1} - t_i)} {\left( \frac 1 {i \sqrt {t_{i+1} - t_i}} \right)^2} = \sum_{n=1}^\infty \frac 1 {i^2}.$$ However, for each $(t_i, t_{i+1})$ interval, the $$S(t) = \frac 1 {g(t) -\lambda} s(t) $$ function is greater than $i s(t)$, therefore, therefore $S^2(t)$ integral is divergent, thus the transformed function cannot belong to space $L^2([0,1])\,$.

At points outside $[m,M]$ interval $\lambda$ values are regular. The values of $$\left|\frac 1 {g(t) -\lambda}\right|$$ in this case have an $K\,$upper bound. However, if there is an integral of $s^2(t)$, then there is also an integral of $K^2s^2(t)$, so the smaller $$\frac 1 {(g(t) -\lambda)^2} s^2(t)$$ also has an integral. So in this case, the $(T_g-\lambda I)^{-1}$ operator has a value in the whole space.

 

[fresh_42]

Question 4

​

Self-adjoint linear operator's eigenvalues are real ​

$$ \lambda (x,x) = (\lambda x, x) = (Ax,x) = (x, Ax) = (x,\lambda x,) = \bar \lambda (x,x) $$ However, in the Euclidean space $(x,x) =0$ follows $x = 0$, so $\lambda = \bar \lambda$.

The eigenvectors belonging to different eigenvalues are orthogonal ​

$$ (Ax, Ay) = (AAx, y) = (\lambda _1 Ax, y) = \lambda_1^2(x,y) $$ $$ (Ax, Ay) = (x, AAy) = (x, \lambda _2 Ay) = \lambda_2^2(x,y) $$ Because $\lambda_1$ and $\lambda_2$ is different, so $(x,y)$ is equal to $0$.

The spectrum of the $T_g$ operator is the interval $[m,M]$​

...

Correct. It could be said a bit shorter if we don't specify a potential inverse:

From the boundaries of $g$ we get that $m,M$ are a lower, resp. upper bound of $T_g\,.$ Hence $\sigma(T_g) \subseteq [m,M]$. According to the mean value theorem for continuous functions we know, that $g$ takes every value in $[m,M]$ at least once, i.e for every $\mu \in [m,M]$ there is a real number $t_\mu \in [0,1]$ such that $g(t_\mu)=\mu\,.$ Thus $$T_g(f)(t_\mu)=g(t_\mu)f(t_\mu)=\mu\cdot f(t_\mu)$$
and $T-\mu$ isn't bounded invertible, hence $\mu \in \sigma(T_g)$ and $\sigma(T_g)=[m,M]\,.$

 

[Periwinkle]

I noticed my own mistake. Correctly: $$ |S(t)| = \left| \frac 1 {g(t) -\lambda} s(t) \right|$$ function is greater than $i |s(t)|.$

Tomorrow I will consider the above solution.