Fourier_A1 : preamble, lemma 1, preposition 2-(b). Added ordered list of parentheses and lower case roman in _base.scss

Author: govin08

_posts/2024-10-25-Fourier_A1.md

@@ -65,6 +65,8 @@ It is clear that
 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>
+
+<a href="#" class="btn btn--primary">proof</a>
 To sketch the proof of it, let $P=(a,b)$ and $P'=(a,c,b)$ are two partition of $[a,b]$.
 Then,
 
@@ -83,7 +85,7 @@ Similarly, $\mathcal L(f,P')\ge\mathcal L(f,P)$ holds for the same $P$ and $P'$.
 
 Let
 
-$$U=\inf_P\mathcal U(f,P),\quad L=\sup_P\mathcal L(f,P).$$
+$$U=\inf_P\,\mathcal U(f,P),\quad L=\sup_P\,\mathcal L(f,P).$$
 
 Such $U$ and $L$ always exist since $f$ is bounded.
 Moreover,
@@ -95,6 +97,7 @@ $f$ is integrable if and only if $U=L$.
 
 </div>
 
+<a href="#" class="btn btn--primary">proof</a>
 Suppose the former.
 It is trivial that $U\ge L$.
 Fix $\epsilon\gt0$.
@@ -107,7 +110,7 @@ Then
 $$
 \begin{aligned}
 U-L
-&=\inf_P\mathcal U(f,P)-\sup_P\mathcal L(f,P)\\
+&=\inf_P\,\mathcal U(f,P)-\sup_P\,\mathcal L(f,P)\\
 &\le\mathcal U(f,P)-\mathcal L(f,P)\\
 &\lt\epsilon
 \end{aligned}
@@ -149,6 +152,7 @@ $$\int_a^bf(x)\,dx=U=L$$
 
 </div>
 
+For brevity, we somtimes write $\int_a^bf\,dx$ or $\int f$ instead.
 For complex functions, we have the following definitions
 
 <div class="notice--success" markdown="1">
@@ -169,6 +173,7 @@ if $f$ is continuous on $[a,b]$, then $f$ is integrable.
 
 </div>
 
+<a href="#" class="btn btn--primary">proof</a>
 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
@@ -203,8 +208,250 @@ It follows that $u$ and $v$ are both integrable and that $f$ is integrable.
 ## A1.1 Basic properties
 
 <div class="notice--info" markdown="1">
-<br><br>
+책에는 Proposition 1.1에 주요 성질들이 적혀있고 여기에 $f$와 $g$가 적분가능하면 $f+g$와 $cf$가 적분가능하다는 게 나오고 있고, 그 이후에 Lemma 1.2로 연속함수의 합성에 의한 효과가 언급되고 그 직후에 $fg$, $|f|$도 적분가능하다는 사실이 언급되고 있다.
+그런데 $fg$, $|f|$가 적분가능하다는 사실은 $f+g$, $cf$가 적분가능하다는 사실만큼이나 중요하다고 생각되고 lemma의 증명에 proposition의 결과가 필요해보이지 않아서 lemma를 앞으로 빼고 proposition을 뒤로 넣었다.
+Rudin의 책에도 이 순서로 되어 있다.
+그리고 lemma, proposition 등의 넘버링은 원래 1.1, 1.2 등으로 되어 있었지만 이 포스트에서는 Appendix 1의 내용만 쓸 것이므로 1, 2 등으로 썼다.
+
+lemma 1.1의 Stein 증명이 잘 이해가 안가서 Rudin의 증명을 더 많이 봤는데, 기본적으로 두 증명이 거의 같다.
+notation은 두 증명을 적당히 조합해서 썼다.
+증명은, $\phi\circ f$에 대한 upper-lower limit의 차가 $\epsilon$의 배수보다 적어짐을 보이기 위하는 것이 목적이다.
+$\phi$의 uniform continuity로부터 $\delta$를, $f$의 적분가능성으로부터 $P$를 얻어내고, 각 subinterval에 대하여 $f$의 최댓값 $M_j$과 최솟값 $m_j$의 차이가 $\delta$와 비교했을 때 큰지($j\in A$) 작은지($j\in B$)로 나누어 생각한다.
+만약 $j\in A$이면 $\phi\circ f$의 최댓값 $M^\ast_j$과 최솟값 $m^\ast_j$의 차이가 $\epsilon$의 상수배보다 적어져서 쉽게 해결된다.
+만약 $j\in B$이면 $M_j-m_j\ge\delta$라는 사실로부터 $\sum_{j\in B}(x_j-x_{j-1})$가 $\epsilon$의 상수배보다 적다는 것을 $P$의 construction으로부터 얻어내어 해결한다.
+둘을 조합하면 증명이 완성된다.
 </div>
 
+<div class="notice--success" markdown="1">
+<b> Lemma 1 </b>
+
+Suppose that $f:[a,b]\to\mathbb R$ is integrable and that $\phi:\mathbb R\to\mathbb R$ is continuous.
+Then, $\phi\circ f:[a,b]\to\mathbb R$ is integrable.
+</div>
+
+<!-- ![lemma_f]({{site.url}}\images\2024-10-25-Fourier_A1\lemma_f.png){: .img-50-center} -->
+
+<div style="display: flex; justify-content: center; gap: 10px;">
+    <img src="{{site.url}}/images/2024-10-25-Fourier_A1/lemma_f.png" style="width: 40%;" />
+    <img src="{{site.url}}/images/2024-10-25-Fourier_A1/lemma_phi.png" style="width: 40%;" />
+</div>
+
+<a href="#" class="btn btn--primary">proof</a>
+Since $f$ is bounded, $m\le f\le M$ on $[a,b]$ for some real $m$ and $M$.
+Since $\phi$ is continuous, its range on $[m,M]$ is bounded ; $\lvert\phi\rvert\le K$ on $[m,M]$ for some $K\ge0$.
+Fix $\epsilon\gt0$.
+Since $\phi$ is uniformly continuous on $[m,M]$, there exists $\delta\gt0$ such that $\delta\lt\epsilon$ and
+
+$$
+m\le s,t\le M,\quad |s-t|\lt\delta\quad\Longrightarrow\quad\lvert\phi(s)-\phi(t)\rvert\lt\epsilon.
+\tag{1-1}
+$$
+
+Since $f$ is integrable, there exists a partition $P=(x_0,x_1,\cdots,x_N)$ of $[a,b]$ so that
+
+$$
+\mathcal U(f, P)-\mathcal L(f, P)\lt\delta^2.
+\tag{1-2}
+$$
+
+For each $j$, denote
+
+$$
+\begin{aligned}
+M_j&=\sup_{x_{j-1}\le x\le x_j}f(x)&m_j&=\inf_{x_{j-1}\le x\le x_j}f(x)\\
+M^\ast_j&=\sup_{x_{j-1}\le x\le x_j}\phi\circ f(x)&m^\ast_j&=\inf_{x_{j-1}\le x\le x_j}\phi\circ f(x).
+\end{aligned}
+$$
+
+These values exist since $f$ is bounded and $\phi$ is continuous.
+Then, by (1-2),
+
+$$
+\sum_{j=1}^N(M_j-m_j)(x_j-x_{j-1})\lt\delta^2
+\tag{1-3}
+$$
+
+Let $A=\\{j:M_j-m_j\lt\delta\\}$ and $B=\\{j:M_j-m_j\ge\delta\\}$.
+If $j\in A$, then $M^\ast_j-m^\ast_j\le\epsilon$ by (1-1).
+If $j\in B$, then we can use (1-3) to get
+
+$$
+\sum_{j\in B}(x_j-x_{j-1})\lt\delta
+$$
+
+and we also have $\lvert M^\ast_j-m^\ast_j\rvert\le 2K$.
+Therefore,
+
+$$
+\begin{aligned}
+\mathcal U(\phi\circ f, P)-\mathcal L(\phi\circ f, P)
+&=\sum_{j=1}^N(M^\ast_j-m^\ast_j)(x_j-x_{j-1})\\
+&=\sum_{j\in A}(M^\ast_j-m^\ast_j)(x_j-x_{j-1})+\sum_{j\in B}(M^\ast_j-m^\ast_j)(x_j-x_{j-1})\\
+&\le\epsilon\sum_{j\in A}(x_j-x_{j-1})+2K\sum_{j\in B}(x_j-x_{j-1})\\
+&\le\epsilon\sum_{j=1}^\infty(x_j-x_{j-1})+2K\delta\\
+&\le\epsilon\left[(b-a)+2K\right]
+\end{aligned}
+$$
+
+Since $\epsilon$ was arbitrary, $\phi\circ f$ is integrable.
+<a href="#" class="btn btn--primary">QED</a>
+
+<div class="notice--success" markdown="1">
+<b> Proposition 2 </b>
+
+Suppose that $f$ and $g$ are integrable function on $[a,b]$.
+Then
+
+<ol class="parenthesis">
+  <li>
+  $f+g$ is integrable and $\int_a^b(f+g)\,dx=\int_a^bf\,dx+\int_a^bg\,dx$.
+  </li>
+  <li>
+  If $c$ is a constant, then $cf$ is integrable and $\int_a^bcf\,dx=c\int_a^bf\,dx$.
+  </li>
+  <li>
+  If $f$ and $g$ are both real valued and if $f\le g$, then $\int_a^bcf\,dx\le\int_a^bg\,dx$.
+  </li>
+  <li>
+  If $a\le c\le b$, then $\int_a^bf\,dx=\int_a^cf\,dx+\int_c^bf\,dx$.
+  </li>
+  <li>
+  $fg$ is integrable.
+  </li>
+  <li>
+  $\lvert f\rvert$ is integrable and $\lvert\int_a^b f\,dx\rvert\le \int_a^b\lvert f\rvert\,dx$.
+  </li>
+</ol>
+
+</div>
+
+<a href="#" class="btn btn--primary">proof - a</a>
+
+Suppose first that $f$ and $g$ are real valued.
+Fix $\epsilon\gt0$.
+Since $f$ and $g$ are integrable, there exist $P_1$ and $P_2$ such that
+
+$$
+\begin{align*}
+(\mathcal U-\mathcal L)(f,P_1)&\lt\frac\epsilon2\\
+(\mathcal U-\mathcal L)(g,P_2)&\lt\frac\epsilon2
+\end{align*}
+$$
+
+Take $P=P_1\cap P_2=(x_0,x_1,\cdots,x_N)$.
+Since $P$ is finer than $P_1$ and $P_2$,
+
+$$
+\begin{align*}
+(\mathcal U-\mathcal L)(f,P)&\lt\frac\epsilon2\\
+(\mathcal U-\mathcal L)(g,P)&\lt\frac\epsilon2
+\tag{2-1}
+\end{align*}
+$$
+
+Thus,
+
+$$
+\begin{align*}
+(\mathcal U-\mathcal L)(f+g,P)
+&=\sum_{j=1}^N\left(
+    \sup_{x_{j-1}\le x\le x_j}\left(f(x)+g(x)\right)
+    -
+    \inf_{x_{j-1}\le x\le x_j}\left(f(x)+g(x)\right)
+\right)\\
+&\le\sum_{j=1}^N\left(
+    \sup_{x_{j-1}\le x\le x_j}f(x)+\sup_{x_{j-1}\le x\le x_j}g(x)
+    -
+    \inf_{x_{j-1}\le x\le x_j}f(x)-\inf_{x_{j-1}\le x\le x_j}g(x)
+\right)\\
+&=\sum_{j=1}^N\left(
+    \sup_{x_{j-1}\le x\le x_j}f(x)
+    -
+    \inf_{x_{j-1}\le x\le x_j}f(x)
+\right)
++\sum_{j=1}^N\left(
+    \sup_{x_{j-1}\le x\le x_j}g(x)
+    -
+    \inf_{x_{j-1}\le x\le x_j}g(x)
+\right)\\
+&=(\mathcal U-\mathcal L)(f,P)+(\mathcal U-\mathcal L)(g,P)\\
+&\lt\epsilon
+\end{align*}
+$$
+
+Thus $f+g$ is integrable.
 
+With the same $P$, we can use (2-1) again to get
+
+$$
+\mathcal U(f,P)\le\mathcal U(f,P)-\mathcal L(f,P)+\int f\,dx\lt\int f\,dx+\frac\epsilon2.
+$$
 
+Similarly,
+
+$$
+\mathcal U(g,P)\lt\int g\,dx+\frac\epsilon2.
+$$
+
+By the subadditivity of the supremum,
+
+$$
+\begin{aligned}
+\int f+g
+&\le\mathcal U(f+g,P)\\
+&\le\mathcal U(f,P)+\mathcal U(g,P)\\
+&\lt\int f+\int g+\epsilon.
+\end{aligned}
+$$
+
+Since $\epsilon$ was arbitrary, $\int f+g\le\int f+\int g.$
+Similarly, $\int f+g\ge\int f+\int g$ and thus
+
+$$
+\int f+g=\int f+\int g.
+$$
+
+Now suppos that $f$ and $g$ are complex valued so that $f=f_R+if_I$ and $g=g_R+ig_I$.
+If $f$ and $g$ are both integrable, $f_R$, $f_I$, $g_R$ and $g_I$ are all integrable.
+It follows that $f_R+g_R$ and $f_I+g_I$ are both integrable and that
+
+$$f+g=(f_R+g_R)+i(f_I+g_I)$$
+
+is integrable.
+Moreover
+
+$$
+\begin{align*}
+\int f+g
+&=\int (f_R+g_R)+i(f_I+g_I)\\
+&=\int (f_R+g_R)+i\int(f_I+g_I)\\
+&=\int f_R+i\int f_I+\int g_R+i\int g_I\\
+&=\int f+\int g
+\end{align*}
+$$
+
+<a href="#" class="btn btn--primary">proof - b</a>
+
+$$
+\begin{aligned}
+\int_a^bcf(x)\,dx
+&=\inf_P\,\mathcal U(cf,P)\\
+&=\inf_P\,\sum_{j=1}^N\sup_{x_{j-1}\le x\le x_j}cf(x)(x_j-x_{j-1})\\
+&=c\inf_P\,\sum_{j=1}^N\sup_{x_{j-1}\le x\le x_j}f(x)(x_j-x_{j-1})\\
+&=c\inf_P\,\mathcal U(f,P)\\
+&=c\int_a^bf(x)\,dx.
+\end{aligned}
+$$
+
+<a href="#" class="btn btn--primary">proof - c</a>
+
+$$
+\begin{aligned}
+\int_a^bf(x)\,dx
+&=\inf_P\,\mathcal U(f,P)\\
+&=\inf_P\,\sum_{j=1}^N\sup_{x_{j-1}\le x\le x_j}f(x)(x_j-x_{j-1})\\
+&\le\inf_P\,\sum_{j=1}^N\sup_{x_{j-1}\le x\le x_j}g(x)(x_j-x_{j-1})\\
+&=\inf_P\,\mathcal U(g,P)\\
+&=\int_a^bg(x)\,dx
+\end{aligned}
+$$

_sass/minimal-mistakes/_base.scss

@@ -181,6 +181,23 @@ li ol {
   margin-top: 0.5em;
 }
 
+/* parenthesis list, added in 2024-10-25 */
+
+ol.parenthesis {
+  counter-reset: list;
+  list-style-type: none;
+  // margin-left: 0.5em;  /* 설정가능 */
+  padding-left: 0.5em;  /* 설정가능 */
+}
+
+ol.parenthesis > li {
+  counter-increment: list;
+}
+
+ol.parenthesis > li::before {
+  content: "(" counter(list, lower-alpha) ") ";
+}
+
 /*
    Media and embeds
    ========================================================================== */