initialized Fourier_2 and Fourier_A1. Now I'm working on Fourier_A1, not 2.

Author: govin08

_posts/2024-10-05-Fourier_1.md

@@ -1434,7 +1434,7 @@ $$
 The first integrand has always periodic antiderivative and thus the first term always vanishes.
 The second integrand does too if $m\ne n$, while if $m=n$, the integrand is $1$ and the second term equals to 1
 
-(4) Similarly, the left hand side in this case equals 0.to
+(4) Similarly, the left hand side in this case equals to
 
 $$
 -\frac1{2\pi}\int_{-\pi}^\pi\cos(n+m)x\,dx

_posts/2024-10-25-Fourier_2.md

@@ -0,0 +1,30 @@
+---
+layout: single
+title: "(2) Basic Properties of Fourier Series"
+categories: mathematics
+tags: [analysis, Fourier analysis, Fourier Series]
+use_math: true
+publish: false
+author_profile: false
+toc: true
+---
+
+1장에서는 푸리에 급수에 대해 맛만 봤다면 2장에서는 이에 대해 엄밀하게 논의한다.
+2장은 다음과 같은 순서로 되어있다.
+아무래도 1장보다는 양이 많아보인다.
+
+1. Examples and formulation of the problem
+2. Uniqueness of Fourier series
+3. Convolutions
+4. Good kernels
+5. Cesàro and Abel summability : applications to Fourier series
+6. Exercises
+7. Problems
+
+이번에도 Problems는 보지 않을 예정이다.
+대신 7절을 Appendix로 채우고, 전체적인 논의에 필요하지만 본문에는 없는 부분을 적을 것이다.
+
+# Chapter 2. Basic Properties of Fourier Series
+
+# 1. Examples and formulation of the problem
+

_posts/2024-10-25-Fourier_A1.md

@@ -0,0 +1,210 @@
+---
+layout: single
+title: "(A1) Riemann integral"
+categories: mathematics
+tags: [analysis, Riemann integral, Riemann]
+use_math: true
+publish: false
+author_profile: false
+toc: true
+---
+
+Stein 책에는 리만적분에 대한 내용이 Appendix의 첫 장으로 되어 있다.
+본문의 2장은 리만적분을 가정하고 진행하고 있으므로 이에 대해 정리해보았다.
+
+가장 먼저는, 리만적분에 관한 기본적인 사실들을 인터넷으로 찾아보다가, nontrivial한 사실인 적분가능한 두 함수의 곱이 적분가능하다는 사실을 증명해보려고 하다가, baby rudin의 6장을 다시 읽어보게 되었다.
+baby rudin은 Stieltjies 적분까지 포함한 논의를 하고 있어서 좀 복잡하지만, 그래도 찬찬히 읽어보니 필요한 내용을 다 정리할 수 있겠다 싶었다.
+그런데 Stein의 2장을 계속 읽다보니 해당 책의 Appendix의 1장 (이상 Stein A1)에 해당 내용이 정리되어있다는 걸 보고 거길 읽어보았다.
+Stein A1이 훨씬 간결하게, 꼭 필요한 내용들만 있는데다가, 기본적으로 Rudin 책의 내용과 거의 비슷하기 때문에, Stein A1을 정리하려고 한다.
+
+<!-- <div class="notice--info" markdown="1"> -->
+<!-- <div class="notice--success" markdown="1"> -->
+<!-- <a href="#" class="btn btn--primary">proof</a> -->
+<!-- <a href="#" class="btn btn--primary">QED</a> -->
+
+# Appendix 1. Riemann Integral
+
+<div class="notice--info" markdown="1">
+<br><br>
+</div>
+
+Let $f:[a,b]\to\mathbb R$ be bounded.
+A *partition* of $[a,b]$ is a sequence $(x_0,x_1,\cdots,x_{N-1},x_N)$ where
+
+$$a=x_0\lt x_1\lt\cdots\lt x_{N-1}\lt x_N=b.$$
+
+The upper sum $\mathcal U(f,P)$ and lower sum $\mathcal L(f,P)$ of $f$ with respect to $P$ are
+
+$$
+\begin{aligned}
+\mathcal U(f,P)
+&=\sum_{j=1}^N\sup\{f(x):x_{j-1}\le x\le x_j\}(x_j-x_{j-1})\\
+\mathcal L(f,P)
+&=\sum_{j=1}^N\inf\{f(x):x_{j-1}\le x\le x_j\}(x_j-x_{j-1})\\
+\end{aligned}
+$$
+
+All the supremums and infimums exist since $f$ is bounded.
+It is immediate that $\mathcal U(f,P)\ge\mathcal L(f,P)$.
+
+<div class="notice--success" markdown="1">
+<b> Definition </b>
+
+$f$ is said to be *(Riemann) integrable* if for every $\epsilon\gt0$, there exists $P$ such that
+
+$$(\mathcal U-\mathcal L)(f,P)\lt\epsilon.$$
+
+</div>
+
+If another partition $P'$ of $[a,b]$ is a subset of $P$ where we conceive $P$ and $P'$ as sets, then $P'$ is called a *refinement* of $P$.
+It is clear that
+
+<div class="notice--success" markdown="1">
+<b> Proposition </b>
+
+if $P'$ is a refinement of $P$, then $\mathcal U(f,P')\le\mathcal U(f,P)$ and $\mathcal L(f,P')\ge\mathcal L(f,P)$.
+
+</div>
+To sketch the proof of it, let $P=(a,b)$ and $P'=(a,c,b)$ are two partition of $[a,b]$.
+Then,
+
+$$
+\begin{aligned}
+\mathcal U(f,P')
+&=\sup\{f(x):a\le x\le c\}(c-a) + \sup\{f(x):c\le x\le b\}(b-c)\\
+&\le\sup\{f(x):a\le x\le b\}(c-a) + \sup\{f(x):a\le x\le b\}(b-c)\\
+&=\sup\{f(x):a\le x\le b\}(b-a)\\
+&=\mathcal U(f,P)
+\end{aligned}
+$$
+
+Similarly, $\mathcal L(f,P')\ge\mathcal L(f,P)$ holds for the same $P$ and $P'$.
+<a href="#" class="btn btn--primary">QED</a>
+
+Let
+
+$$U=\inf_P\mathcal U(f,P),\quad L=\sup_P\mathcal L(f,P).$$
+
+Such $U$ and $L$ always exist since $f$ is bounded.
+Moreover,
+
+<div class="notice--success" markdown="1">
+<b> Proposition </b>
+
+$f$ is integrable if and only if $U=L$.
+
+</div>
+
+Suppose the former.
+It is trivial that $U\ge L$.
+Fix $\epsilon\gt0$.
+Since $f$ is integrable, there exists a partition $P$ such that
+
+$$\mathcal U(f,P)-\mathcal L(f,P)\lt\epsilon$$
+
+Then
+
+$$
+\begin{aligned}
+U-L
+&=\inf_P\mathcal U(f,P)-\sup_P\mathcal L(f,P)\\
+&\le\mathcal U(f,P)-\mathcal L(f,P)\\
+&\lt\epsilon
+\end{aligned}
+$$
+
+Since $\epsilon$ was arbitrary, $U-L\le0$.
+
+Suppose the latter and fix $\epsilon\gt0$.
+There exist $P_1$ and $P_2$ such that
+
+$$
+\begin{aligned}
+\mathcal U(f,P_1) - U\lt\frac\epsilon2\\
+L - \mathcal L(f,P_2)\lt\frac\epsilon2
+\end{aligned}
+$$
+
+Let $P$ be the common refinement of $P_1$ and $P_2$, the enumeration of the intersection of $P_1$ and $P_2$ as sets.
+Then
+
+$$
+\begin{aligned}
+\mathcal U(f,P) - \mathcal L(f,P)
+&\le\mathcal U(f,P_1) - \mathcal L(f,P_2)\\
+&\lt U-L+\epsilon\\
+&=\epsilon.
+\end{aligned}
+$$
+
+Thus $f$ is integrable as desired.
+<a href="#" class="btn btn--primary">QED</a>
+
+<div class="notice--success" markdown="1">
+<b> Definition </b>
+
+If $f$ is integrable so that $U=L$, define
+
+$$\int_a^bf(x)\,dx=U=L$$
+
+</div>
+
+For complex functions, we have the following definitions
+
+<div class="notice--success" markdown="1">
+<b> Definition </b>
+
+$f:[a,b]\to\mathbb C$ is such that $f=u+iv$, then $f$ is integrable on $[a,b]$ if both of $u$ and $v$ are integrable on $[a,b]$ and
+
+$$\int_a^bf(x)\,dx=\int_a^bu(x)\,dx+i\int_a^bv(x)\,dx$$
+
+</div>
+
+It is a fundamental fact that
+
+<div class="notice--success" markdown="1">
+<b> Proposition </b>
+
+if $f$ is continuous on $[a,b]$, then $f$ is integrable.
+
+</div>
+
+Suppose that $f$ is real valued.
+Since $[a,b]$ is compact (closed and bounded), $f$ is uniformly continuous on this interval.
+That is, for each $\epsilon\gt0$, there exists $\delta\gt0$ such that
+
+$$|x-y|\lt\delta\quad\Longrightarrow\quad |f(x)-f(y)|\lt\frac\epsilon{b-a}.$$
+
+Choose $n$ so that $\frac{b-a}n\lt\delta$ and let
+
+$$P=\left(a, a+\frac{b-a}n, a+2\frac{b-a}n, \cdots, a+(n-1)\frac{b-a}n,b\right).$$
+
+Then
+
+$$
+\begin{aligned}
+(\mathcal U - \mathcal L)(f, P)
+&=\sum_{j=1}^N\left(
+    \sup\{f(x):x_{j-1}\le x\le x_j\}
+    -
+    \inf\{f(x):x_{j-1}\le x\le x_j\}
+    \right)\frac{b-a}n\\
+&=\sum_{j=1}^N\epsilon\frac{b-a}n\\
+&=\frac\epsilon{b-a}\cdot(b-a)=\epsilon.
+\end{aligned}
+$$
+
+Suppose now that $f$ is complex valued so that $f=u+iv$.
+Since $f$ is continuous, it is trivial that $u$ and $v$ are also continuous.
+It follows that $u$ and $v$ are both integrable and that $f$ is integrable.
+<a href="#" class="btn btn--primary">QED</a>
+
+
+## A1.1 Basic properties
+
+<div class="notice--info" markdown="1">
+<br><br>
+</div>
+
+
+