Linear Algebra

Problem 1

Prove that if $T\in \mathcal{L}(V)$ is normal, then
$$\mbox{null}T^k=\mbox{null}T \mbox{ and } \mbox{range}T^k=\mbox{range}T$$

Solution:
easy to check that $\mbox{null}T\subset\mbox{null}T^k$
to prove an inclusion in the other direction, suppose that $v\in\mbox{null}T^k$
$T^kv=T(T^{k-1}v)=0\implies T^{k-1}v\in\mbox{null}T \mbox{ and } T^{k-1}v\in\mbox{range}T^{k-1}\subset\mbox{range}T$
but $V=\mbox{null}T\oplus(\mbox{null}T)^{\perp}=\mbox{null}T\oplus\mbox{range}T^*=\mbox{null}T\oplus\mbox{range}T$
the third equality holds because $T$ is normal
$T^{k-1}v\in \mbox{null}T\cap\mbox{range}T\implies T^{k-1}v=0$, inductively, we have $Tv=0$
This shows that $\mbox{null}T^{k}\subset\mbox{null}T$, completing the proof that $\mbox{null}T=\mbox{null}T^k$

Now we are going to show that $\mbox{range}T^k=\mbox{range}T$
$\mbox{range}T=\mbox{range}T^*=(\mbox{null}T)^{\perp}=(\mbox{null}T^k)^{\perp}=\mbox{range}(T^k)^*=\mbox{range}T^k$
first equality holds because T is normal, third equality is come from the previous proof, the last equality holds because $T^k$ is normal

Lebesgue Integral

Problem 1

Lebesgue Measure

Problem 1

Let \(A_n\) be any sequence of subsets of \(\mathbb{R}\)
(i) if \(A_n\uparrow A\) then \(\lambda^*(A_n)\uparrow \lambda^*(A)\)
(ii)\(\lambda^*(\liminf A_n)\leq\liminf\limits_{n\to\infty} \lambda^*(A_n)\), where \(\liminf A_n=\{x:x\in A_n \mbox{ultimately}\}\)

solution:
(i)
Let  \(F_n=\bigcup_{i=1}^{n}E_i\), \(F=\cup E_n,A=\cup A_n\), where \(E_n\) is measurable such that \(A_n\subset E_n\), and \(E_n-A_n\) is negligible (it is easy to prove such \(E_n\) exists), clearly \(A\subset F\),
\(F_n\uparrow F\), therefore \(\lambda^*(F_n)\uparrow \lambda^*(F)\)
\(\begin{align}
\lambda^*(A_n)\leq\lambda^*(F_n)&\leq\lambda^*(A_n)+\lambda^*(F_n-A_n)\\
&\leq\lambda^*(A_n)+\sum_{i=1}^{n}\lambda^*(E_i-A_i)\\
&= \lambda^*(A_n)
\end{align}\)
\(\lambda^*(F_n)=\lambda^*(A_n)\leq \lambda^*(A)\leq \lambda^*(F)\), and since \(F_n\uparrow F\), we have \(\lambda^*(A_n)\uparrow \lambda^*(A)\)
(ii)
Let \(B_n=\bigcap_{k\geq n}A_k\), then \(B_n\uparrow B=\liminf A_n\)
\(\lambda^*(B_n)\leq \lambda^*(A_n)\Rightarrow \liminf\limits_{n\to\infty} \lambda^*(B_n)=\lambda^*(\liminf A_n)\leq \liminf\limits_{n\to\infty} \lambda^*(A_n)\)
The equality follows from (i)

Problem 2

For \(A\subset [a,b]\subset \mathbb{R}\), \(\lambda^*(A)\) is also called the exterior measure of A in \([a,b]\) and is denoted \(\lambda_e(A)\), the interior measure of A in \([a,b]\), denoted \(\lambda_i(A)\), is defined by the formula
\[\lambda_i(A)=(b-a)-\lambda_e([a,b]-A)\]in general, \(\lambda_i(A)\leq\lambda_e(A)\) ; the set A is measurable if and only if \(\lambda_i(A)=\lambda_e(A)\)

solution:
\(\Rightarrow\) is easy
\(\Leftarrow\) We use \(\lambda^*\) instead of \(\lambda_e\) denote \(I=[a,b]\), let \(J\) be any bounded interval \begin{align*} \lambda^*(I)&=\lambda^*(I\cap A)+\lambda^*(I\cap A')\\ &=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A'\cap J') \\
&=\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)+\lambda^*(I\cap A\cap J')+\lambda^*(I\cap A'\cap J')\\
&\geq \lambda^*(I\cap J)+\lambda^*(I\cap J')\\
&=\lambda^*(I)
\end{align*}
Therefore we have \(\lambda^*(I\cap A\cap J)+\lambda^*(I\cap A'\cap J)=\lambda^*(I\cap J)\), and it is easy to see that \(\lambda^*(A\cap J)+\lambda^*(A'\cap J)=\lambda^*(J)\) Suppose \(S\) is a subset of \(\mathbb{R}\) for every \(\varepsilon>0\), there exists \(U,S\subset U=\bigcup J_n\) and \(\sum \lambda^*(J_n)<\lambda^*(S)+\varepsilon\)
\begin{align*}
\lambda^*(A\cap S)+\lambda^*(A'\cap S)&\leq\lambda^*(A\cap U)+\lambda^*(A'\cap U)\\
&\leq \sum\lambda^*(A\cap J_n)+\sum\lambda^*(A'\cap J_n)\\ &=\sum(\lambda^*(A\cap J_n)+\lambda^*(A'\cap J_n))\\
&=\sum\lambda^*(J_n)\\
&<\lambda^*(S)+\varepsilon
\end{align*}
Since \(\varepsilon\) is arbitrary, we can conclude that A is measurable 

Problem 3

If A is a Lebesgue-measurable subset of \(\mathbb{R}\) such that \(\lambda(A)>0\), then the difference set \(D=\{a-b:a,c\in A\}\) is a neighborhood of 0, that is, \((-\delta,\delta)\subset D\) for some \(\delta>0\).

Sequences of Functions

Problem 1
Let \(\{g_n\}\) be a sequence of real valued functions such that \(g_{n+1}(x)\leq g_n(x)\) for each \(x\) in T and for every \(n=1,2,...\). If \(\{g_n\}\) is uniformly bounded on T and if \(\sum f_n(x)\) converges uniformly on T, then \(\sum f_n(x)g_n(x)\) also converges uniformly on T.

Problem 2
Let \({a_n}\) be a decreasing sequence of positive terms. Prove that the series \( \sum a_n \sin nx\) converges uniformly on \(\mathbb{R}\) if, and only if, \(na_n\to 0\) as \(n\to\infty\)

Only if direction is easy so we omit the proof here, we only prove the if direction
Suppose \(na_n\to 0\) as \(n\to\infty\), let \(B_n(x)=\sum\limits_{k=1}^{n}\sin kx\) \(\sum\limits_{k=1}^{n}a_k \sin kx=B_n a_{n+1}-\sum\limits_{k=1}^{n}B_k (a_{k+1}-a_k)\) \begin{align*} \bigg|\sum\limits_{k=n+1}^{m}a_k \sin kx\bigg|&=\bigg| B_n a_{n+1}-B_m a_{m+1}+\sum\limits_{k=n+1}^{m}B_k (a_{k+1}-a_k)\bigg|\\ &=\bigg| B_n a_{n+1}-B_m a_{m+1}+ B(a_{m+1}-a_{n+1}) \bigg|,\;\;\ \\ &\mbox{where}\;min\{B_{n+1},B_{n+2},...,B_{m}\}\leq B \leq max\{B_{n+1},B_{n+2},...,B_{m}\}\\ &\leq \bigg|(B_n-B) a_{n+1} \bigg|+ \bigg|(B-B_m) a_{m+1} \bigg|\\ &<2Na_{n+1}+2Ma_{m+1}\\
&<\varepsilon
\end{align*}

problem 1 can be proved in a similar way

Problem 3
Given a power series \(\sum_{n=0}^{\infty} a_n z^n\) whose coefficients are related by an equation of the form \[a_n+Aa_{n-1}+Ba_{n-2}=0\] (n=2,3,...)
Show that for any \(x\) for which ther series converges, its sum is \[\frac{a_0+(a_1+Aa_0)x}{1+Ax+Bx^2}\]

Problem 4
Show that the binomial series \((1+x)^{\alpha}=\sum_{n=0}^{\infty} {\alpha \choose k} x^n\) exhibits the following behavior at the points \(x=\pm 1\)
a) If \(x=-1\), the series converges for \(\alpha\geq 0\) and diverges for \(\alpha<0\)
b) If \(x=1\), the series diverges for \(\alpha \leq -1\), converges conditionally for \(\alpha\) in the inteval \(-1<\alpha<0\), and converges absolutely for \(\alpha\geq0\)