실해석학 HW3
A. 과제 수령 및 확인 리포트
확정 문항 리스트
- Chapter 5: 1, 2, 3, 6, 11, 12, 22, 25, 26, 27
- Chapter 6: 1, 2, 3, 4, 6, 7, 8, 11, 12, 19
총 문항 수: \(10+10=\boxed{20}\)
풀이 순서: \(\text{Ch.5 전부 완료}\implies \text{Ch.6 시작}\)
근거 교재: Chapter 5 Differentiation, Chapter 6 Riemann-Stieltjes Integral.
B. 문항별 상세 솔루션
Ch.5 - Problem 1
문제 원문 및 번역
Original. Let \(f\) be defined for all real \(x\), and suppose that \(\vert f(x)-f(y) \vert \le (x-y)^2\) for all real \(x\) and \(y\). Prove that \(f\) is constant.
번역. 모든 실수 \(x\)에 대해 \(f\)가 정의되어 있고, 모든 실수 \(x,y\)에 대해 \(\vert f(x)-f(y) \vert \le (x-y)^2\) 라고 하자. \(f\)가 상수함수임을 증명하여라.
요청 사항
\[\forall x,y\in\mathbb R,\quad f(x)=f(y)\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad x<y,\quad n\in\mathbb N,\quad x_k=x+\frac{k}{n}(y-x)\quad (k=0,\dots,n)\] \[\text{Step 1:}\quad x_0=x,\quad x_n=y,\quad x_k-x_{k-1}=\frac{y-x}{n}\]\(\text{Step 2:}\quad \vert f(y)-f(x) \vert = \left\vert\sum_{k=1}^{n}{f(x_k)-f(x_{k-1})}\right\vert \le \sum_{k=1}^{n}\vert f(x_k)-f(x_{k-1}) \vert\) \(\because \text{삼각부등식}\)
\(\text{Step 3:}\quad \vert f(x_k)-f(x_{k-1}) \vert \le (x_k-x_{k-1})^2 = \left(\frac{y-x}{n}\right)^2\) \(\because \text{문제 가정}\)
\[\implies \vert f(y)-f(x) \vert \le \sum_{k=1}^{n}\left(\frac{y-x}{n}\right)^2 = n\cdot \frac{(y-x)^2}{n^2} = \frac{(y-x)^2}{n}\] \[\text{Step 4:}\quad 0\le \vert f(y)-f(x) \vert \le \frac{(y-x)^2}{n} \quad \forall n\in\mathbb N\] \[n\to\infty \implies \frac{(y-x)^2}{n}\to 0 \implies \vert f(y)-f(x) \vert =0\] \[\therefore f(y)=f(x)\] \[x,y\in\mathbb R \text{ arbitrary} \implies f \text{ is constant.}\]최종 답안
\[\boxed{\boldsymbol{Ans.}\quad f(x)=C\ \text{for all }x\in\mathbb R.}\]Ch.5 - Problem 2
문제 원문 및 번역
Original. Suppose \(f'(x)>0\) in \((a,b)\). Prove that \(f\) is strictly increasing in \((a,b)\), and let \(g\) be its inverse function. Prove that \(g\) is differentiable, and that \(g'(f(x))=\frac{1}{f'(x)}\qquad (a<x<b).\)
번역. \((a,b)\)에서 \(f'(x)>0\)라고 하자. \(f\)가 \((a,b)\)에서 엄격히 증가함을 증명하고, \(g\)를 \(f\)의 역함수라 하자. \(g\)가 미분가능하며 \(g'(f(x))=\frac{1}{f'(x)}\qquad (a<x<b)\) 임을 증명하여라.
요청 사항
\(f'(x)>0 \implies f \text{ strictly increasing}\) \(g=f^{-1} \implies g'(f(x))=\frac1{f'(x)}\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad a<x_1<x_2<b\]\(\text{Step 1:}\quad f(x_2)-f(x_1)=(x_2-x_1)f'(c) \quad \exists c\in(x_1,x_2)\) \(\because \text{교재 Thm 5.10 Mean Value Theorem}\)
\[\text{Step 2:}\quad x_2-x_1>0,\quad f'(c)>0\] \[\implies f(x_2)-f(x_1)>0 \implies f(x_2)>f(x_1)\]\(\therefore f \text{ strictly increasing on }(a,b)\) \(\because \text{교재 Thm 5.11(a): } f'\ge0\implies f \text{ increasing; here }f'>0\implies \text{strict}\)
\[\text{Step 3:}\quad f \text{ strictly increasing} \implies f \text{ one-to-one} \implies g=f^{-1}\text{ exists on }f((a,b))\] \[\text{Def:}\quad y=f(x),\quad s=f(t),\quad t=g(s),\quad x=g(y)\] \[\text{Step 4:}\quad \frac{g(s)-g(y)}{s-y} = \frac{t-x}{f(t)-f(x)} = \left(\frac{f(t)-f(x)}{t-x}\right)^{-1}\]\(\text{Step 5:}\quad s\to y \iff f(t)\to f(x) \implies t\to x\) \(\because f \text{ strictly increasing + differentiability }\implies f \text{ continuous} \quad \because \text{교재 Thm 5.2}\)
\(\text{Step 6:}\quad \lim_{s\to y}\frac{g(s)-g(y)}{s-y} = \left( \lim_{t\to x}\frac{f(t)-f(x)}{t-x} \right)^{-1} = \frac1{f'(x)}\) \(\because f'(x)\neq0,\quad \text{교재 Def 5.1 derivative}\)
\[\therefore g'(y)=\frac1{f'(x)}\] \[y=f(x) \implies g'(f(x))=\frac1{f'(x)}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f \text{ is strictly increasing on }(a,b),\qquad g'(f(x))=\frac1{f'(x)}. }\]Ch.5 - Problem 3
문제 원문 및 번역
Original. Suppose \(g\) is a real function on \(\mathbb R^1\), with bounded derivative, say \(\vert g' \vert \le M\). Fix \(\varepsilon>0\), and define \(f(x)=x+\varepsilon g(x).\) Prove that \(f\) is one-to-one if \(\varepsilon\) is small enough. An admissible set of values of \(\varepsilon\) can be determined which depends only on \(M\).
번역. \(g\)가 \(\mathbb R^1\) 위의 실함수이고, 도함수가 유계이며 \(\vert g' \vert \le M\)라고 하자. \(\varepsilon>0\)을 고정하고 \(f(x)=x+\varepsilon g(x)\) 로 정의한다. \(\varepsilon\)이 충분히 작으면 \(f\)가 일대일임을 증명하여라. 허용 가능한 \(\varepsilon\)의 범위는 \(M\)에만 의존하도록 정할 수 있다.
요청 사항
\(\vert g' \vert \le M,\quad f(x)=x+\varepsilon g(x)\) \(\exists \varepsilon\text{-range depending only on }M \implies f \text{ one-to-one}\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad x<y,\quad M\ge0,\quad \vert g'(t) \vert \le M\quad \forall t\in\mathbb R\]\(\text{Step 1:}\quad g(y)-g(x)=(y-x)g'(c) \quad \exists c\in(x,y)\) \(\because \text{교재 Thm 5.10 Mean Value Theorem}\)
\[\implies \vert g(y)-g(x) \vert \le M\vert y-x \vert = M(y-x)\] \[\text{Step 2:}\quad f(y)-f(x) = (y-x)+\varepsilon(g(y)-g(x))\] \[\text{Step 3:}\quad g(y)-g(x)\ge -\vert g(y)-g(x) \vert \ge -M(y-x)\] \[\implies f(y)-f(x) \ge (y-x)-\varepsilon M(y-x) = (1-\varepsilon M)(y-x)\] \[\text{Step 4:}\quad 0<\varepsilon<\frac1M \quad (M>0) \implies 1-\varepsilon M>0\] \[\implies f(y)-f(x)>0\] \[\therefore x<y\implies f(x)<f(y) \implies f \text{ strictly increasing}\] \[\therefore f \text{ one-to-one}\]\(M=0 \implies g'=0 \implies g \text{ constant} \quad \because \text{교재 Thm 5.11(b)}\) \(\implies f(x)=x+\varepsilon C \implies f \text{ one-to-one for every }\varepsilon>0\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad M>0:\ 0<\varepsilon<\frac1M \implies f \text{ is one-to-one.} }\] \[\boxed{ \boldsymbol{Ans.}\quad M=0:\ \text{every }\varepsilon>0\text{ is admissible.} }\]Ch.5 - Problem 6
문제 원문 및 번역
Original. Suppose (a) \(f\) is continuous for \(x\ge 0\), (b) \(f'(x)\) exists for \(x>0\), (c) \(f'(0)=0\), (d) \(f'\) is monotonically increasing.
Put \(g(x)=\frac{f(x)}{x}\qquad (x>0)\) and prove that \(g\) is monotonically increasing.
번역. 다음을 가정하자. (a) \(f\)는 \(x\ge0\)에서 연속이다. (b) \(x>0\)에서 \(f'(x)\)가 존재한다. (c) \(f'(0)=0\). (d) \(f'\)는 단조증가한다.
\(g(x)=\frac{f(x)}{x}\qquad (x>0)\) 로 두고, \(g\)가 단조증가함을 증명하여라.
요청 사항
\[f'\uparrow,\quad f'(0)=0 \implies g(x)=\frac{f(x)}x \text{ monotonically increasing on }(0,\infty)\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad 0<x<y,\quad g(x)=\frac{f(x)}x\]\(\text{Step 1:}\quad f(x)-f(0)=x f'(c_x) \quad \exists c_x\in(0,x)\) \(\because \text{교재 Thm 5.10 Mean Value Theorem}\)
\[\implies \frac{f(x)-f(0)}x=f'(c_x)\] \[\text{Step 2:}\quad f'\text{ monotonically increasing},\quad 0<c_x<x \implies f'(c_x)\le f'(x)\] \[\implies \frac{f(x)-f(0)}x\le f'(x)\]\(\text{Step 3:}\quad \text{문제의 }f'(0)=0\text{ 은 우도함수 의미}\) \(f'(0)=\lim_{t\to0+}\frac{f(t)-f(0)}t=0\) \(\because \text{교재 Def 5.1 derivative at endpoint: one-sided derivative}\)
\(\text{Step 4:}\quad g(x)=\frac{f(x)}x\) \(\text{단, }g'(x)=\frac{xf'(x)-f(x)}{x^2}\) \(\because \text{교재 Thm 5.3(c) quotient rule}\)
\[\text{Step 5:}\quad xf'(x)-f(x) = x f'(x)-[f(x)-f(0)]-f(0)\]여기서 문제의 표준 형태는 보통 \(f(0)=0\)을 포함한다. 만약 \(f(0)\ne0\), 결론은 일반적으로 거짓이다. 예: \(f(x)=1\Rightarrow f'=0\), 그러나 \(g(x)=1/x\) 감소.
따라서 결론 성립에는 필요한 조건: \(f(0)=0\)
\[\text{Step 6:}\quad f(0)=0 \implies xf'(x)-f(x) = x f'(x)-[f(x)-f(0)] \ge0\] \[\implies g'(x)=\frac{xf'(x)-f(x)}{x^2}\ge0\]\(\therefore g \text{ monotonically increasing}\) \(\because \text{교재 Thm 5.11(a): }g'\ge0\implies g\text{ increasing}\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad \text{문제에 }f(0)=0\text{ 가 포함되어야 하며, 그때 }g(x)=\frac{f(x)}x \text{ 는 단조증가한다.} }\] \[\boxed{ \boldsymbol{Note.}\quad f(0)=0\text{ 없이는 반례 }f(x)=1,\ g(x)=1/x. }\]Ch.5 - Problem 11
문제 원문 및 번역
Original. Suppose \(f\) is defined in a neighborhood of \(x\), and suppose \(f''(x)\) exists. Show that \(\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x).\) Show by an example that the limit may exist even if \(f''(x)\) does not. Hint: Use Theorem 5.13.
번역. \(f\)가 \(x\)의 근방에서 정의되어 있고 \(f''(x)\)가 존재한다고 하자. 다음을 보여라. \(\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x).\) 또한 이 극한이 존재하더라도 \(f''(x)\)가 존재하지 않을 수 있음을 예로 보여라. 힌트: 정리 5.13을 사용하라.
요청 사항
\[f''(x)\text{ exists} \implies \lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)\] \[\exists f:\quad \lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\text{ exists},\quad f''(x)\text{ DNE}\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad A(h)=f(x+h)+f(x-h)-2f(x)\] \[\text{Step 1:}\quad A(0)=0,\quad h^2\to0\]\(\text{Step 2:}\quad \lim_{h\to0}\frac{A(h)}{h^2} = \lim_{h\to0}\frac{A'(h)}{2h}\) \(\because \text{교재 Thm 5.13 L'Hospital, }0/0\text{ form}\)
\[\text{Step 3:}\quad A'(h)=f'(x+h)-f'(x-h)\] \[\implies \lim_{h\to0}\frac{A'(h)}{2h} = \lim_{h\to0} \frac{f'(x+h)-f'(x-h)}{2h}\]\(\text{Step 4:}\quad f''(x)\text{ exists} \implies \lim_{h\to0}\frac{f'(x+h)-f'(x)}{h}=f''(x)\) \(\because \text{교재 Def 5.14 higher derivative}\)
\[\lim_{h\to0} \frac{f'(x)-f'(x-h)}{h} = f''(x)\] \[\text{Step 5:}\quad \frac{f'(x+h)-f'(x-h)}{2h} = \frac12 \left[ \frac{f'(x+h)-f'(x)}h + \frac{f'(x)-f'(x-h)}h \right]\] \[\implies \lim_{h\to0} \frac{f'(x+h)-f'(x-h)}{2h} = \frac12[f''(x)+f''(x)] = f''(x)\] \[\therefore \lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)\] \[\text{Counterexample}\] \[\text{Def:}\quad x=0,\quad f(t)=\vert t \vert\]\(\text{Step 6:}\quad \frac{f(h)+f(-h)-2f(0)}{h^2} = \frac{\vert h \vert +\vert -h \vert -0}{h^2} = \frac{2\vert h \vert}{h^2} = \frac2{\vert h \vert} \to\infty\) \(\text{not finite; unsuitable}\)
더 적합한 예:
\[\text{Def:}\quad f(t)=t\vert t \vert,\quad x=0\] \[\text{Step 7:}\quad f(h)=h\vert h \vert,\quad f(-h)=(-h)\vert -h \vert=-h\vert h \vert\] \[\implies f(h)+f(-h)-2f(0) = h\vert h \vert-h\vert h \vert-0 = 0\] \[\therefore \frac{f(h)+f(-h)-2f(0)}{h^2}=0 \implies \lim_{h\to0}\frac{f(h)+f(-h)-2f(0)}{h^2}=0\] \[\text{Step 8:}\quad f'(t)= \begin{cases} 2t,&t>0,\\ -2t,&t<0, \end{cases} \quad f'(0)=\lim_{t\to0}\frac{t\vert t \vert}{t}=\vert t \vert\to0\] \[\text{Step 9:}\quad f''(0)=\lim_{t\to0}\frac{f'(t)-f'(0)}t\]\(t>0:\quad \frac{2t}{t}=2\) \(t<0:\quad \frac{-2t}{t}=-2\)
\[\implies \lim_{t\to0}\frac{f'(t)-f'(0)}t\text{ DNE}\] \[\therefore f''(0)\text{ does not exist}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f''(x)\text{ exists} \implies \lim_{h\to0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} = f''(x). }\] \[\boxed{ \boldsymbol{Example.}\quad f(t)=t\vert t \vert,\ x=0: \quad \lim_{h\to0}\frac{f(h)+f(-h)-2f(0)}{h^2}=0, \quad f''(0)\text{ DNE}. }\]Ch.5 - Problem 12
문제 원문 및 번역
Original. If \(f(x)=\vert x \vert^3\), compute \(f'(x), f''(x)\) for all real \(x\), and show that \(f^{(3)}(0)\) does not exist.
번역. \(f(x)=\vert x \vert^3\)일 때, 모든 실수 \(x\)에 대해 \(f'(x), f''(x)\)를 계산하고, \(f^{(3)}(0)\)이 존재하지 않음을 보여라.
요청 사항
\(f(x)=\vert x \vert^3 \implies f'(x),\ f''(x)\text{ 계산}\) \(f^{(3)}(0)\text{ DNE 증명}\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad f(x)=\vert x \vert^3 = \begin{cases} x^3,&x\ge0,\\ -x^3,&x<0 \end{cases}\]\(\text{Step 1:}\quad x>0:\quad f'(x)=3x^2,\quad f''(x)=6x\) \(\because \text{교재 Ex 5.4: }(x^n)'=nx^{n-1}\)
\(\text{Step 2:}\quad x<0:\quad f(x)=-x^3 \implies f'(x)=-3x^2,\quad f''(x)=-6x\) \(\because \text{교재 Thm 5.3 linearity/product rules}\)
\[\text{Step 3:}\quad f'(0)=\lim_{t\to0}\frac{\vert t \vert^3-0}{t}\]\(t>0:\quad \frac{t^3}{t}=t^2\to0\) \(t<0:\quad \frac{(-t)^3}{t}=\frac{-t^3}{t}=-t^2\to0\)
\[\therefore f'(0)=0\] \[\text{Step 4:}\quad f''(0)=\lim_{t\to0}\frac{f'(t)-f'(0)}t\]\(t>0:\quad \frac{3t^2}{t}=3t\to0\) \(t<0:\quad \frac{-3t^2}{t}=-3t\to0\)
\[\therefore f''(0)=0\] \[\text{Step 5:}\quad f''(x)= \begin{cases} 6x,&x>0,\\ 0,&x=0,\\ -6x,&x<0 \end{cases} = 6\vert x \vert\]\(\text{Step 6:}\quad f^{(3)}(0) = \lim_{t\to0}\frac{f''(t)-f''(0)}t = \lim_{t\to0}\frac{6\vert t \vert}{t}\) \(\because \text{교재 Def 5.14 higher derivative}\)
\(t\to0+:\quad \frac{6\vert t \vert}{t}=6\) \(t\to0-:\quad \frac{6\vert t \vert}{t}=-6\)
\[6\ne -6 \implies f^{(3)}(0)\text{ DNE}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f'(x)= \begin{cases} 3x^2,&x>0,\\ 0,&x=0,\\ -3x^2,&x<0, \end{cases} }\] \[\boxed{ \boldsymbol{Ans.}\quad f''(x)= \begin{cases} 6x,&x>0,\\ 0,&x=0,\\ -6x,&x<0, \end{cases} =6\vert x \vert. }\] \[\boxed{ \boldsymbol{Ans.}\quad f^{(3)}(0)\text{ does not exist.} }\]Ch.5 - Problem 22
문제 원문 및 번역
Original. Suppose \(f\) is a real function on \((-\infty,\infty)\). Call \(x\) a fixed point of \(f\) if \(f(x)=x\).
(a) If \(f\) is differentiable and \(f'(t)\ne1\) for every real \(t\), prove that \(f\) has at most one fixed point.
(b) Show that the function \(f\) defined by \(f(t)=t+(1+e^t)^{-1}\) has no fixed point, although \(0<f'(t)<1\) for all real \(t\).
(c) However, if there is a constant \(A<1\) such that \(\vert f'(t) \vert \le A\) for all real \(t\), prove that a fixed point \(x\) of \(f\) exists, and that \(x=\lim x_n\), where \(x_1\) is an arbitrary real number and \(x_{n+1}=f(x_n)\qquad(n=1,2,3,\dots).\)
번역. \(f\)가 \((-\infty,\infty)\) 위의 실함수라고 하자. \(f(x)=x\)이면 \(x\)를 \(f\)의 고정점이라 부른다.
(a) \(f\)가 미분가능하고 모든 실수 \(t\)에 대해 \(f'(t)\ne1\)이면, \(f\)는 많아야 하나의 고정점만 가짐을 증명하여라.
(b) \(f(t)=t+(1+e^t)^{-1}\) 로 정의된 함수는 모든 실수 \(t\)에 대해 \(0<f'(t)<1\)임에도 고정점이 없음을 보여라.
(c) 반면, 어떤 상수 \(A<1\)가 존재하여 모든 실수 \(t\)에 대해 \(\vert f'(t) \vert \le A\)이면, \(f\)의 고정점 \(x\)가 존재하며, 임의의 실수 \(x_1\)에서 시작하여 \(x_{n+1}=f(x_n)\) 로 정의한 수열에 대해 \(x=\lim x_n\)임을 증명하여라.
요청 사항
\[(a)\quad f'(t)\ne1\ \forall t \implies \#\{x:f(x)=x\}\le1\] \[(b)\quad f(t)=t+\frac1{1+e^t} \implies \text{no fixed point},\quad 0<f'(t)<1\] \[(c)\quad \vert f' \vert \le A<1 \implies \exists !x:f(x)=x,\quad x_n\to x\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad F(t)=f(t)-t\](a)
\[\text{Step 1:}\quad F'(t)=f'(t)-1\ne0\] \[\text{Step 2:}\quad \text{Assume }x<y,\quad F(x)=F(y)=0\]\(\implies \exists c\in(x,y):\quad F(y)-F(x)=(y-x)F'(c)\) \(\because \text{교재 Thm 5.10 Mean Value Theorem}\)
\[0=(y-x)F'(c),\quad y-x>0 \implies F'(c)=0\] \[\Rightarrow \Leftarrow \quad F'(c)=f'(c)-1\ne0\] \[\therefore \text{at most one fixed point}\](b)
\[\text{Def:}\quad f(t)=t+\frac1{1+e^t}\] \[\text{Step 1:}\quad f(t)-t=\frac1{1+e^t}>0 \quad \forall t\in\mathbb R\] \[\implies f(t)\ne t \quad \forall t\] \[\therefore \text{no fixed point}\]\(\text{Step 2:}\quad f'(t) = 1-\frac{e^t}{(1+e^t)^2}\) \(\because \text{교재 Thm 5.3(c) quotient rule + Thm 5.5 chain rule}\)
\[\text{Step 3:}\quad 0<\frac{e^t}{(1+e^t)^2}\] \[(1+e^t)^2-e^t = 1+e^t+e^{2t}>0\] \[\implies \frac{e^t}{(1+e^t)^2}<1\] \[\therefore 0<1-\frac{e^t}{(1+e^t)^2}<1\] \[\therefore 0<f'(t)<1\](c)
\[\text{Def:}\quad \vert f'(t) \vert \le A<1\]\(\text{Step 1:}\quad \vert f(y)-f(x) \vert \le A\vert y-x \vert \quad \forall x,y\) \(\because \text{교재 Thm 5.10 MVT applied to }f\)
\[\text{Step 2:}\quad x_{n+1}=f(x_n)\] \[\implies \vert x_{n+1}-x_n \vert = \vert f(x_n)-f(x_{n-1}) \vert \le A\vert x_n-x_{n-1} \vert\] \[\implies \vert x_{n+1}-x_n \vert \le A^{n-1}\vert x_2-x_1 \vert\] \[\text{Step 3:}\quad m>n: \quad \vert x_m-x_n \vert \le \sum_{k=n}^{m-1}\vert x_{k+1}-x_k \vert \le \sum_{k=n}^{m-1}A^{k-1}\vert x_2-x_1 \vert\] \[\le \frac{A^{n-1}}{1-A}\vert x_2-x_1 \vert \to0\]\(\therefore \{x_n\}\text{ is Cauchy} \implies \exists x\in\mathbb R:\ x_n\to x\) \(\because \mathbb R\text{ complete}\)
\(\text{Step 4:}\quad f \text{ differentiable}\implies f\text{ continuous}\) \(\because \text{교재 Thm 5.2}\)
\[x_{n+1}=f(x_n),\quad x_n\to x \implies x=\lim x_{n+1}=\lim f(x_n)=f(x)\] \[\therefore x\text{ fixed point}\] \[\text{Step 5:}\quad \text{uniqueness: }x,y\text{ fixed}\] \[\vert x-y \vert = \vert f(x)-f(y) \vert \le A\vert x-y \vert\] \[(1-A)\vert x-y \vert \le0,\quad 1-A>0 \implies x=y\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad (a)\ \text{at most one fixed point.} }\] \[\boxed{ \boldsymbol{Ans.}\quad (b)\ f(t)>t\ \forall t\implies \text{no fixed point},\quad 0<f'(t)<1. }\] \[\boxed{ \boldsymbol{Ans.}\quad (c)\ \vert f' \vert \le A<1\implies \exists !x=f(x),\quad x_n\to x. }\]Ch.5 - Problem 25
문제 원문 및 번역
Original. Suppose \(f\) is twice differentiable on \([a,b]\), \(f(a)<0\), \(f(b)>0\), \(f'(x)\ge \delta>0\), and \(0\le f''(x)\le M\) for all \(x\in[a,b]\). Let \(\xi\) be the unique point in \((a,b)\) at which \(f(\xi)=0\). Complete the details in the following outline of Newton’s method for computing \(\xi\).
(a) Choose \(x_1\in(\xi,b)\), and define \(\{x_n\}\) by \(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.\)
(b) Prove that \(x_{n+1}<x_n\) and that \(\lim x_n=\xi.\)
(c) Use Taylor’s theorem to show that \(x_{n+1}-\xi = \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2\) for some \(t_n\in(\xi,x_n)\).
(d) If \(A=M/2\delta\), deduce that \(0\le x_{n+1}-\xi\le \frac1A[A(x_1-\xi)]^{2^n}.\)
번역. \(f\)가 \([a,b]\)에서 두 번 미분가능하고, \(f(a)<0\), \(f(b)>0\), \(f'(x)\ge\delta>0\), 그리고 모든 \(x\in[a,b]\)에 대해 \(0\le f''(x)\le M\)라고 하자. \(f(\xi)=0\)을 만족하는 유일한 점 \(\xi\in(a,b)\)가 있다고 하자. 뉴턴 방법의 세부사항을 완성하여라.
(a) \(x_1\in(\xi,b)\)를 택하고 \(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\) 로 정의한다.
(b) \(x_{n+1}<x_n\)이고 \(\lim x_n=\xi\) 임을 증명하여라.
(c) 테일러 정리를 이용하여 어떤 \(t_n\in(\xi,x_n)\)에 대해 \(x_{n+1}-\xi = \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2\) 임을 보여라.
(d) \(A=M/2\delta\)이면 \(0\le x_{n+1}-\xi\le \frac1A[A(x_1-\xi)]^{2^n}\) 를 유도하여라.
요청 사항
\[\text{Newton iteration convergence + quadratic error estimate}\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},\quad f(\xi)=0,\quad f'\ge\delta>0,\quad 0\le f''\le M\]\(\text{Step 1:}\quad f'(x)>0 \implies f\text{ strictly increasing}\) \(\because \text{교재 Thm 5.11(a)}\)
\[f(a)<0<f(b) \implies \exists !\xi\in(a,b):f(\xi)=0\](a) Geometric meaning
\[\text{Def:}\quad T_n(x)=f(x_n)+f'(x_n)(x-x_n)\] \[T_n(x_{n+1})=0 \iff x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\] \[\therefore x_{n+1} = \text{ tangent line at }(x_n,f(x_n))\text{ intersects }x\text{-axis}\](b)
\(\text{Step 2:}\quad x_n>\xi \implies f(x_n)>0\) \(\because f\uparrow,\ f(\xi)=0\)
\[\implies \frac{f(x_n)}{f'(x_n)}>0 \implies x_{n+1}<x_n\]\(\text{Step 3:}\quad f''\ge0 \implies f'\text{ increasing}\) \(\because \text{교재 Thm 5.11(a) applied to }f'\)
\(\implies f\text{ convex}\) \(\because \text{교재 Ch.5 Ex.14 criterion: }f''\ge0\iff f\text{ convex}\)
\[\text{Convex tangent inequality:}\quad f(\xi)\ge f(x_n)+f'(x_n)(\xi-x_n)\] \[0\ge f(x_n)+f'(x_n)(\xi-x_n)\] \[\implies x_n-\frac{f(x_n)}{f'(x_n)}\ge \xi\] \[\implies x_{n+1}\ge \xi\] \[\therefore \xi\le x_{n+1}<x_n\] \[\text{Step 4:}\quad \{x_n\}\downarrow,\quad x_n\ge\xi \implies x_n\to L\ge\xi\] \[\text{Step 5:}\quad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\] \[n\to\infty: \quad L=L-\frac{f(L)}{f'(L)}\] \[\because f,f'\text{ continuous; }f'(L)\ge\delta>0\] \[\implies f(L)=0 \implies L=\xi\] \[\therefore x_n\to\xi\](c)
\(\text{Step 6:}\quad 0=f(\xi) = f(x_n)+f'(x_n)(\xi-x_n)+\frac{f''(t_n)}2(\xi-x_n)^2\) \(\exists t_n\in(\xi,x_n)\) \(\because \text{교재 Thm 5.15 Taylor theorem with remainder}\)
\[\implies f(x_n)=f'(x_n)(x_n-\xi)-\frac{f''(t_n)}2(x_n-\xi)^2\] \[\text{Step 7:}\quad x_{n+1}-\xi = x_n-\xi-\frac{f(x_n)}{f'(x_n)}\] \[= x_n-\xi- \left[ (x_n-\xi)-\frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2 \right]\] \[= \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2\](d)
\[\text{Step 8:}\quad 0\le f''(t_n)\le M,\quad f'(x_n)\ge\delta\] \[\implies 0\le x_{n+1}-\xi \le \frac{M}{2\delta}(x_n-\xi)^2 = A(x_n-\xi)^2\] \[\text{Def:}\quad e_n=x_n-\xi\ge0\] \[e_{n+1}\le A e_n^2\] \[\text{Step 9:}\quad A e_{n+1}\le (A e_n)^2\] \[\implies A e_{n+1}\le (A e_1)^{2^n} \quad \text{by induction}\] \[\therefore e_{n+1}\le \frac1A[Ae_1]^{2^n}\] \[\therefore 0\le x_{n+1}-\xi \le \frac1A[A(x_1-\xi)]^{2^n}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad x_{n+1}<x_n,\quad \xi\le x_{n+1},\quad x_n\to\xi. }\] \[\boxed{ \boldsymbol{Ans.}\quad x_{n+1}-\xi = \frac{f''(t_n)}{2f'(x_n)}(x_n-\xi)^2, \quad t_n\in(\xi,x_n). }\] \[\boxed{ \boldsymbol{Ans.}\quad 0\le x_{n+1}-\xi \le \frac1A[A(x_1-\xi)]^{2^n}, \quad A=\frac{M}{2\delta}. }\]Ch.5 - Problem 26
문제 원문 및 번역
Original. Suppose \(f\) is differentiable on \([a,b]\), \(f(a)=0\), and there is a real number \(A\) such that \(\vert f'(x) \vert \le A\vert f(x) \vert\) on \([a,b]\). Prove that \(f(x)=0\) for all \(x\in[a,b]\).
번역. \(f\)가 \([a,b]\)에서 미분가능하고, \(f(a)=0\), 어떤 실수 \(A\)가 존재하여 \([a,b]\)에서 \(\vert f'(x) \vert \le A\vert f(x) \vert\) 라고 하자. 모든 \(x\in[a,b]\)에 대해 \(f(x)=0\)임을 증명하여라.
요청 사항
\[f(a)=0,\quad \vert f' \vert \le A\vert f \vert \implies f\equiv0\text{ on }[a,b]\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad S=\{x\in[a,b]:f=0\text{ on }[a,x]\}\] \[\text{Step 1:}\quad a\in S \implies S\ne\varnothing\] \[\text{Def:}\quad c=\sup S\]\(\text{Step 2:}\quad f\text{ differentiable}\implies f\text{ continuous}\) \(\because \text{교재 Thm 5.2}\)
\[\implies f=0\text{ on }[a,c]\] \[\text{Step 3:}\quad \text{If }c<b,\quad \text{choose }x_0\in(c,b]\text{ with }A(x_0-c)<1\] \[\text{Def:}\quad M_0=\sup_{c\le x\le x_0}\vert f(x) \vert,\quad M_1=\sup_{c\le x\le x_0}\vert f'(x) \vert\] \[\text{Step 4:}\quad \vert f'(x) \vert \le A\vert f(x) \vert \implies M_1\le A M_0\] \[\text{Step 5:}\quad f(c)=0\]\(\forall x\in[c,x_0]: \quad f(x)-f(c)=(x-c)f'(t_x) \quad \exists t_x\in(c,x)\) \(\because \text{교재 Thm 5.10 Mean Value Theorem}\)
\[\implies \vert f(x) \vert \le (x-c)M_1 \le (x_0-c)A M_0\] \[\implies M_0\le A(x_0-c)M_0\] \[A(x_0-c)<1 \implies M_0=0\] \[\therefore f=0\text{ on }[c,x_0]\] \[\implies f=0\text{ on }[a,x_0] \implies x_0\in S\] \[x_0>c=\sup S \quad \Rightarrow \Leftarrow\] \[\therefore c=b\] \[\therefore f=0\text{ on }[a,b]\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f(x)=0\quad \forall x\in[a,b]. }\]Ch.5 - Problem 27
문제 원문 및 번역
Original. Let \(\phi\) be a real function defined on a rectangle \(R\) in the plane, given by \(a\le x\le b,\qquad \alpha\le y\le \beta.\) A solution of the initial-value problem \(y'=\phi(x,y),\qquad y(a)=c\qquad(\alpha\le c\le\beta)\) is, by definition, a differentiable function \(f\) on \([a,b]\) such that \(f(a)=c\), \(\alpha\le f(x)\le\beta\), and \(f'(x)=\phi(x,f(x))\qquad(a<x\le b).\) Prove that such a problem has at most one solution if there is a constant \(A\) such that \(\vert \phi(x,y_2)-\phi(x,y_1) \vert \le A\vert y_2-y_1 \vert\) whenever \((x,y_1)\in R\) and \((x,y_2)\in R\).
번역. \(\phi\)가 평면의 직사각형 \(R\) 위에 정의된 실함수이고, \(a\le x\le b,\qquad \alpha\le y\le\beta\) 라고 하자. 초기값 문제 \(y'=\phi(x,y),\qquad y(a)=c\qquad(\alpha\le c\le\beta)\) 의 해란, 정의상 \([a,b]\)에서 미분가능한 함수 \(f\)로서 \(f(a)=c\), \(\alpha\le f(x)\le\beta\), 그리고 \(f'(x)=\phi(x,f(x))\qquad(a<x\le b)\) 를 만족하는 함수이다. 만약 어떤 상수 \(A\)가 존재하여 \(R\)의 임의의 두 점 \((x,y_1),(x,y_2)\)에 대해 \(\vert \phi(x,y_2)-\phi(x,y_1) \vert \le A\vert y_2-y_1 \vert\) 가 성립하면, 이 초기값 문제의 해가 많아야 하나임을 증명하여라.
요청 사항
\[\text{Lipschitz condition in }y \implies \text{IVP uniqueness}\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad f,g\text{ are two solutions}\] \[\implies f(a)=g(a)=c\] \[\text{Def:}\quad h(x)=f(x)-g(x)\] \[\text{Step 1:}\quad h(a)=f(a)-g(a)=0\]\(\text{Step 2:}\quad h'(x)=f'(x)-g'(x)\) \(\because \text{교재 Thm 5.3(a) derivative linearity}\)
\[= \phi(x,f(x))-\phi(x,g(x))\]\(\text{Step 3:}\quad \vert h'(x) \vert = \vert \phi(x,f(x))-\phi(x,g(x)) \vert \le A\vert f(x)-g(x) \vert = A\vert h(x) \vert\) \(\because \text{문제의 Lipschitz condition}\)
\(\text{Step 4:}\quad h(a)=0,\quad \vert h' \vert \le A\vert h \vert \implies h(x)=0\quad \forall x\in[a,b]\) \(\because \text{Ch.5 Problem 26 result}\)
\[\therefore f(x)-g(x)=0 \quad \forall x\in[a,b]\] \[\therefore f=g\] \[\therefore \text{at most one solution}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad \text{The initial-value problem has at most one solution.} }\] \[\boxed{\text{Chapter 5 complete.}}\]Ch.6 - Problem 1
문제 원문 및 번역
Original. Suppose \(\alpha\) increases on \([a,b]\), \(a\le x_0\le b\), \(\alpha\) is continuous at \(x_0\), \(f(x_0)=1\), and \(f(x)=0\) if \(x\ne x_0\). Prove that \(f\in\mathscr R(\alpha)\) and that \(\int_a^b f,d\alpha=0.\)
번역. \(\alpha\)가 \([a,b]\)에서 증가함수이고, \(a\le x_0\le b\), \(\alpha\)가 \(x_0\)에서 연속이며, \(f(x_0)=1\), \(x\ne x_0\)이면 \(f(x)=0\)이라고 하자. \(f\in\mathscr R(\alpha)\)이고 \(\int_a^b f,d\alpha=0\) 임을 증명하여라.
요청 사항
\(\alpha\uparrow,\quad \alpha\text{ continuous at }x_0,\quad f=\mathbf 1_{x_0}\) \(\implies f\in\mathscr R(\alpha),\qquad \int_a^b f,d\alpha=0\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad f(x)= \begin{cases} 1,&x=x_0,\\ 0,&x\ne x_0. \end{cases}\] \[\text{Step 1:}\quad 0\le f\le1 \implies f\text{ bounded on }[a,b]\]\(\text{Step 2:}\quad \alpha\text{ continuous at }x_0 \implies \forall \varepsilon>0,\ \exists u<v,\ x_0\in[u,v]\subset[a,b]\) \(\text{such that}\quad \alpha(v)-\alpha(u)<\varepsilon\) \(\because \text{continuity of }\alpha\text{ at }x_0\)
\[\text{Step 3:}\quad P=\{a,u,v,b\}\text{ refined if necessary}\] \[\text{On intervals not containing }x_0:\quad M_i=m_i=0\] \[\text{On interval containing }x_0:\quad M_i=1,\quad m_i=0\] \[\text{Step 4:}\quad U(P,f,\alpha)-L(P,f,\alpha) \le 1\cdot[\alpha(v)-\alpha(u)] < \varepsilon\]\(\implies f\in\mathscr R(\alpha)\) \(\because \text{교재 Thm 6.6: }f\in\mathscr R(\alpha)\iff \forall\varepsilon>0,\exists P,\ U(P,f,\alpha)-L(P,f,\alpha)<\varepsilon\)
\[\text{Step 5:}\quad m_i=0\quad \forall i \implies L(P,f,\alpha)=0\quad \forall P\]\(\implies \underline{\int_a^b} f,d\alpha = \sup_P L(P,f,\alpha) = 0\) \(\because \text{교재 Def 6.2 lower Stieltjes integral}\)
\(\text{Step 6:}\quad f\in\mathscr R(\alpha) \implies \int_a^b f,d\alpha = \underline{\int_a^b} f,d\alpha = \overline{\int_a^b} f,d\alpha\) \(\because \text{교재 Def 6.2}\)
\[\therefore \int_a^b f,d\alpha=0\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f\in\mathscr R(\alpha),\qquad \int_a^b f,d\alpha=0. }\]Ch.6 - Problem 2
문제 원문 및 번역
Original. Suppose \(f\ge0\), \(f\) is continuous on \([a,b]\), and \(\int_a^b f(x),dx=0.\) Prove that \(f(x)=0\) for all \(x\in[a,b]\). Compare this with Exercise 1.
번역. \(f\ge0\), \(f\)가 \([a,b]\)에서 연속이고 \(\int_a^b f(x),dx=0\) 라고 하자. 모든 \(x\in[a,b]\)에 대해 \(f(x)=0\)임을 증명하여라. 이것을 연습문제 1과 비교하여라.
요청 사항
\[f\ge0,\quad f\in C[a,b],\quad \int_a^b f,dx=0 \implies f\equiv0\]풀이 과정 (Symbolic & Referenced)
\(\text{Def:}\quad \alpha(x)=x\) \(\implies \int_a^b f,dx=\int_a^b f,d\alpha\) \(\because \text{교재 Def 6.2: Riemann integral is special case }\alpha(x)=x\)
\[\text{Step 1:}\quad \text{Assume }\exists x_0\in[a,b]\text{ such that }f(x_0)>0\] \[\text{Def:}\quad \eta=\frac{f(x_0)}2>0\] \[\text{Step 2:}\quad f\text{ continuous at }x_0 \implies \exists \delta>0: \vert x-x_0 \vert <\delta\implies \vert f(x)-f(x_0) \vert <\eta\] \[\implies f(x)>f(x_0)-\eta=\eta\] \[\text{Step 3:}\quad I=[u,v]\subset[a,b]\cap(x_0-\delta,x_0+\delta),\quad v-u>0\] \[\implies f(x)\ge\eta\quad \forall x\in I\]\(\text{Step 4:}\quad \int_a^b f,dx = \int_a^u f,dx+\int_u^v f,dx+\int_v^b f,dx\) \(\because \text{교재 Thm 6.12(c): additivity over intervals}\)
\(\text{Step 5:}\quad f\ge0 \implies \int_a^u f,dx\ge0,\quad \int_v^b f,dx\ge0\) \(\because \text{교재 Thm 6.12(b): order preservation}\)
\(\text{Step 6:}\quad f\ge \eta \text{ on }[u,v] \implies \int_u^v f,dx\ge \int_u^v \eta,dx=\eta(v-u)>0\) \(\because \text{교재 Thm 6.12(b), Def 6.1}\)
\[\implies \int_a^b f,dx>0\] \[\Rightarrow \Leftarrow \quad \int_a^b f,dx=0\] \[\therefore \nexists x_0:f(x_0)>0\] \[f\ge0 \implies f(x)=0\quad \forall x\in[a,b]\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f(x)=0\quad \forall x\in[a,b]. }\] \[\boxed{ \boldsymbol{Compare.}\quad \text{Ch.6-1에서는 한 점에서만 }1\text{인 함수도 적분값 }0;\text{ 여기서는 연속성 때문에 한 점의 양수값 }\implies\text{ 양의 길이 구간에서 양수.} }\]Ch.6 - Problem 3
문제 원문 및 번역
Original. Define three functions \(\beta_1,\beta_2,\beta_3\) as follows: \(\beta_j(x)=0\) if \(x<0\), \(\beta_j(x)=1\) if \(x>0\), for \(j=1,2,3\); and \(\beta_1(0)=0,\qquad \beta_2(0)=1,\qquad \beta_3(0)=\frac12.\) Let \(f\) be a bounded function on \([-1,1]\).
(a) Prove that \(f\in\mathscr R(\beta_1)\) if and only if \(f(0+)=f(0)\), and that then \(\int f,d\beta_1=f(0).\)
(b) State and prove a similar result for \(\beta_2\).
(c) Prove that \(f\in\mathscr R(\beta_3)\) if and only if \(f\) is continuous at \(0\).
(d) If \(f\) is continuous at \(0\), prove that \(\int f,d\beta_1=\int f,d\beta_2=\int f,d\beta_3=f(0).\)
번역. 세 함수 \(\beta_1,\beta_2,\beta_3\)를 다음과 같이 정의한다. \(j=1,2,3\)에 대해 \(x<0\)이면 \(\beta_j(x)=0\), \(x>0\)이면 \(\beta_j(x)=1\), 그리고 \(\beta_1(0)=0,\qquad \beta_2(0)=1,\qquad \beta_3(0)=\frac12.\) \(f\)는 \([-1,1]\) 위의 유계함수라고 하자.
(a) \(f\in\mathscr R(\beta_1)\)일 필요충분조건은 \(f(0+)=f(0)\)임을 증명하고, 그때 \(\int f,d\beta_1=f(0)\) 임을 보여라.
(b) \(\beta_2\)에 대해 유사한 결과를 진술하고 증명하여라.
(c) \(f\in\mathscr R(\beta_3)\)일 필요충분조건은 \(f\)가 \(0\)에서 연속임을 증명하여라.
(d) \(f\)가 \(0\)에서 연속이면 \(\int f,d\beta_1=\int f,d\beta_2=\int f,d\beta_3=f(0)\) 임을 증명하여라.
요청 사항
\(\beta_j\text{ jump at }0,\quad f\text{ bounded}\) \(\mathscr R(\beta_j)\text{ 조건과 적분값 계산}\)
풀이 과정 (Symbolic & Referenced)
\(\text{Def:}\quad P=\{-1=x_0<\cdots<x_n=1\}\) \(\Delta\beta_i=\beta(x_i)-\beta(x_{i-1})\) \(U(P,f,\beta)=\sum M_i\Delta\beta_i,\quad L(P,f,\beta)=\sum m_i\Delta\beta_i\) \(\because \text{교재 Def 6.2}\)
(a) \(\beta_1(0)=0\)
\[\text{Step 1:}\quad \beta_1(x)=0\ (x\le0),\quad \beta_1(x)=1\ (x>0)\] \[\implies \Delta\beta_i=0 \text{ except for the unique interval }[x_{i-1},x_i]\text{ with }x_{i-1}\le0<x_i\] \[\text{Step 2:}\quad \text{If }P\ni0,\quad \Delta\beta\text{ only on }[0,x_i]\]\(\implies U(P,f,\beta_1)=\sup_{0\le x\le x_i}f(x)\) \(L(P,f,\beta_1)=\inf_{0\le x\le x_i}f(x)\)
\(\text{Step 3:}\quad f\in\mathscr R(\beta_1) \iff \forall\varepsilon>0,\exists t>0: \sup_{0\le x\le t}f(x)-\inf_{0\le x\le t}f(x)<\varepsilon\) \(\because \text{교재 Thm 6.6}\)
\[\iff f(x)\to f(0)\text{ as }x\to0+\] \[\iff f(0+)=f(0)\] \[\text{Step 4:}\quad f(0+)=f(0) \implies \sup_{0\le x\le t}f(x),\inf_{0\le x\le t}f(x)\to f(0)\]\(\implies \int_{-1}^{1}f,d\beta_1=f(0)\) \(\because \text{교재 Def 6.2 upper/lower equal common value}\)
(b) \(\beta_2(0)=1\)
\[\text{Step 5:}\quad \beta_2(x)=0\ (x<0),\quad \beta_2(x)=1\ (x\ge0)\] \[\implies \Delta\beta_i=0 \text{ except for interval }[x_{i-1},x_i]\text{ with }x_{i-1}<0\le x_i\] \[\text{If }P\ni0,\quad \Delta\beta\text{ only on }[x_{i-1},0]\]\(\implies U(P,f,\beta_2)=\sup_{x_{i-1}\le x\le0}f(x)\) \(L(P,f,\beta_2)=\inf_{x_{i-1}\le x\le0}f(x)\)
\[\text{Step 6:}\quad f\in\mathscr R(\beta_2) \iff \forall\varepsilon>0,\exists t>0: \sup_{-t\le x\le0}f(x)-\inf_{-t\le x\le0}f(x)<\varepsilon\] \[\iff f(0-)=f(0)\] \[\text{Step 7:}\quad f(0-)=f(0) \implies \int_{-1}^{1}f,d\beta_2=f(0)\] \[\therefore \boxed{ f\in\mathscr R(\beta_2)\iff f(0-)=f(0),\quad \int f,d\beta_2=f(0) }\](c) \(\beta_3(0)=\frac12\)
\[\text{Step 8:}\quad \beta_3(x)=0\ (x<0),\quad \beta_3(0)=\frac12,\quad \beta_3(x)=1\ (x>0)\] \[\implies \text{jump }1/2\text{ from left to }0,\quad \text{jump }1/2\text{ from }0\text{ to right}\] \[\text{Step 9:}\quad P\ni0,\quad [-t,0],[0,t]\text{ are active intervals}\]\(U(P,f,\beta_3)-L(P,f,\beta_3)\) \(= \frac12 \left[ \sup_{-t\le x\le0}f-\inf_{-t\le x\le0}f \right] + \frac12 \left[ \sup_{0\le x\le t}f-\inf_{0\le x\le t}f \right]\)
\(\text{Step 10:}\quad f\in\mathscr R(\beta_3) \iff \text{both oscillations}\to0\) \(\because \text{교재 Thm 6.6}\)
\[\iff f(0-)=f(0)\text{ and }f(0+)=f(0)\] \[\iff f\text{ continuous at }0\](d)
\[\text{Step 11:}\quad f\text{ continuous at }0 \implies f(0-)=f(0)=f(0+)\] \[\implies f\in\mathscr R(\beta_1)\cap\mathscr R(\beta_2)\cap\mathscr R(\beta_3)\] \[\text{Step 12:}\quad \int f,d\beta_1=f(0),\quad \int f,d\beta_2=f(0)\]\(\text{Step 13:}\quad \int f,d\beta_3 = \frac12 f(0)+\frac12 f(0) = f(0)\) \(\because \text{교재 Def 6.2 + jump contributions; cf. Thm 6.15 unit step}\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f\in\mathscr R(\beta_1) \iff f(0+)=f(0), \qquad \int f,d\beta_1=f(0). }\] \[\boxed{ \boldsymbol{Ans.}\quad f\in\mathscr R(\beta_2) \iff f(0-)=f(0), \qquad \int f,d\beta_2=f(0). }\] \[\boxed{ \boldsymbol{Ans.}\quad f\in\mathscr R(\beta_3) \iff f\text{ is continuous at }0. }\] \[\boxed{ \boldsymbol{Ans.}\quad f\text{ continuous at }0 \implies \int f,d\beta_1 = \int f,d\beta_2 = \int f,d\beta_3 = f(0). }\]Ch.6 - Problem 4
문제 원문 및 번역
Original. If \(f(x)=0\) for all irrational \(x\), \(f(x)=1\) for all rational \(x\), prove that \(f\notin\mathscr R\) on \([a,b]\) for any \(a<b\).
번역. 모든 무리수 \(x\)에 대해 \(f(x)=0\), 모든 유리수 \(x\)에 대해 \(f(x)=1\)이라 하자. 임의의 \(a<b\)에 대해 \(f\notin\mathscr R\) on \([a,b]\)임을 증명하여라.
요청 사항
\[f=\mathbf 1_{\mathbb Q},\quad [a,b],\ a<b \implies f\notin\mathscr R\]풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad f(x)= \begin{cases} 1,&x\in\mathbb Q,\\ 0,&x\notin\mathbb Q. \end{cases}\] \[\text{Def:}\quad P=\{a=x_0<x_1<\cdots<x_n=b\}\]\(\text{Step 1:}\quad [x_{i-1},x_i]\text{ contains rational and irrational points}\) \(\because \mathbb Q,\ \mathbb R\setminus\mathbb Q\text{ are dense in }\mathbb R\)
\[\text{Step 2:}\quad M_i=\sup_{x_{i-1}\le x\le x_i}f(x)=1,\qquad m_i=\inf_{x_{i-1}\le x\le x_i}f(x)=0\]\(\text{Step 3:}\quad U(P,f)=\sum_{i=1}^n M_i\Delta x_i = \sum_{i=1}^n 1\cdot(x_i-x_{i-1}) = b-a\) \(\because \text{교재 Def 6.1 upper Riemann sum}\)
\(\text{Step 4:}\quad L(P,f)=\sum_{i=1}^n m_i\Delta x_i = \sum_{i=1}^n0\cdot(x_i-x_{i-1}) = 0\) \(\because \text{교재 Def 6.1 lower Riemann sum}\)
\[\text{Step 5:}\quad U(P,f)-L(P,f)=b-a>0 \quad \forall P\] \[\implies \nexists P:\ U(P,f)-L(P,f)<\varepsilon \quad \text{for }0<\varepsilon<b-a\]\(\therefore f\notin\mathscr R\) \(\because \text{교재 Thm 6.6: integrability criterion}\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f\notin\mathscr R[a,b]\quad \forall a<b. }\]Ch.6 - Problem 6
문제 원문 및 번역
Original. Let \(P\) be the Cantor set constructed in Sec. 2.44. Let \(f\) be a bounded real function on \([0,1]\) which is continuous at every point outside \(P\). Prove that \(f\in\mathscr R\) on \([0,1]\).
Hint: \(P\) can be covered by finitely many segments whose total length can be made as small as desired. Proceed as in Theorem 6.10.
번역. \(P\)를 Sec. 2.44에서 구성된 칸토어 집합이라 하자. \(f\)가 \([0,1]\) 위의 유계 실함수이고, \(P\) 밖의 모든 점에서 연속이라고 하자. \(f\in\mathscr R\) on \([0,1]\)임을 증명하여라.
힌트: \(P\)는 전체 길이가 원하는 만큼 작아지는 유한 개의 구간들로 덮을 수 있다. 정리 6.10처럼 진행하여라.
요청 사항
\(f\text{ bounded},\quad f\text{ continuous on }[0,1]\setminus P\) \(P=\text{Cantor set},\quad m(P)=0 \implies f\in\mathscr R[0,1]\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad \vert f(x) \vert \le M\quad (0\le x\le1)\]\(\text{Step 1:}\quad \forall\varepsilon>0,\quad P\subset \bigcup_{j=1}^N [u_j,v_j], \qquad \sum_{j=1}^N(v_j-u_j)<\frac{\varepsilon}{4M}\) \(\because \text{Cantor set }P\text{ can be covered by finitely many intervals of arbitrarily small total length}\)
\[\text{Def:}\quad E=\bigcup_{j=1}^N [u_j,v_j],\qquad K=[0,1]\setminus \bigcup_{j=1}^N(u_j,v_j)\] \[\text{Step 2:}\quad K\text{ compact},\quad f\text{ continuous at every point of }K\]\(\implies f\text{ uniformly continuous on }K\) \(\because \text{교재 Thm 4.19 uniform continuity on compact sets}\)
\[\implies \exists\delta>0:\quad s,t\in K,\ \vert s-t \vert <\delta \implies \vert f(s)-f(t) \vert <\frac{\varepsilon}{2}\]\(\text{Step 3:}\quad \text{Choose partition }Q\) \(\text{such that every }u_j,v_j\in Q,\quad \Delta x_i<\delta\text{ if }[x_{i-1},x_i]\not\subset E\)
\[\text{Step 4:}\quad \text{If }[x_{i-1},x_i]\subset E: \quad M_i-m_i\le2M\] \[\text{If }[x_{i-1},x_i]\not\subset E: \quad M_i-m_i<\frac{\varepsilon}{2}\] \[\because \text{uniform continuity on }K,\ \text{same method as 교재 Thm 6.10}\]\(\text{Step 5:}\quad U(Q,f)-L(Q,f) = \sum_i(M_i-m_i)\Delta x_i\) \(\le 2M\sum_{j=1}^N(v_j-u_j) + \frac{\varepsilon}{2}\cdot 1\)
\[< 2M\cdot\frac{\varepsilon}{4M} + \frac{\varepsilon}{2} = \varepsilon\] \[\text{Step 6:}\quad \forall\varepsilon>0,\exists Q:\ U(Q,f)-L(Q,f)<\varepsilon\]\(\therefore f\in\mathscr R[0,1]\) \(\because \text{교재 Thm 6.6}\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad f\in\mathscr R[0,1]. }\]Ch.6 - Problem 7
문제 원문 및 번역
Original. Suppose \(f\) is a real function on \((0,1]\) and \(f\in\mathscr R\) on \([c,1]\) for every \(c>0\). Define \(\int_0^1 f(x),dx=\lim_{c\to0}\int_c^1 f(x),dx\) if this limit exists and is finite.
(a) If \(f\in\mathscr R\) on \([0,1]\), show that this definition of the integral agrees with the old one.
(b) Construct a function \(f\) such that the above limit exists, although it fails to exist with \(\vert f \vert\) in place of \(f\).
번역. \(f\)가 \((0,1]\) 위의 실함수이고, 모든 \(c>0\)에 대해 \(f\in\mathscr R\) on \([c,1]\)라고 하자. 다음 극한이 존재하고 유한하면 \(\int_0^1 f(x),dx=\lim_{c\to0}\int_c^1 f(x),dx\) 로 정의한다.
(a) 만약 \(f\in\mathscr R\) on \([0,1]\)이면, 이 정의가 기존의 적분 정의와 일치함을 보여라.
(b) 위 극한은 존재하지만, \(f\) 대신 \(\vert f \vert\)를 넣으면 존재하지 않는 함수 \(f\)를 구성하여라.
요청 사항
\[(a)\quad f\in\mathscr R[0,1]\implies \lim_{c\to0+}\int_c^1 f=\int_0^1 f\] \[(b)\quad \int_0^1 f\text{ improper converges},\quad \int_0^1 \vert f \vert \text{ improper diverges}\]풀이 과정 (Symbolic & Referenced)
(a)
\(\text{Def:}\quad f\in\mathscr R[0,1]\implies f\text{ bounded}\) \(\exists M>0:\ \vert f(x) \vert \le M\)
\(\text{Step 1:}\quad \int_0^1 f,dx = \int_0^c f,dx+\int_c^1 f,dx\) \(\because \text{교재 Thm 6.12(c): interval additivity}\)
\(\text{Step 2:}\quad \left\vert\int_0^c f,dx\right\vert \le M(c-0)=Mc\) \(\because \text{교재 Thm 6.12(d): }\vert\int f,d\alpha\vert\le M[\alpha(b)-\alpha(a)]\)
\[\text{Step 3:}\quad c\to0+ \implies \left\vert\int_0^c f,dx\right\vert\le Mc\to0\] \[\implies \int_c^1 f,dx = \int_0^1 f,dx-\int_0^c f,dx \to \int_0^1 f,dx\] \[\therefore \lim_{c\to0+}\int_c^1 f,dx = \int_0^1 f,dx\](b)
\[\text{Def:}\quad f(x)=\frac{\sin(1/x)}{x} \qquad (0<x\le1)\] \[\text{Step 4:}\quad f\text{ continuous on }[c,1]\quad \forall c>0\]\(\implies f\in\mathscr R[c,1]\) \(\because \text{교재 Thm 6.8: continuous functions are Riemann integrable}\)
\[\text{Step 5:}\quad u=\frac1x \implies dx=-\frac1{u^2}du,\quad \frac1x=u\] \[\int_c^1\frac{\sin(1/x)}x,dx = \int_{1/c}^{1}u\sin u\left(-\frac1{u^2}\right)du = \int_1^{1/c}\frac{\sin u}{u},du\] \[\text{Step 6:}\quad \lim_{c\to0+}\int_c^1 f(x),dx = \int_1^\infty\frac{\sin u}{u},du\]\(\text{exists}\) \(\because \text{Dirichlet test: }\frac1u\downarrow0,\ \int \sin u\text{ bounded}\)
\[\text{Step 7:}\quad \int_c^1 \vert f(x) \vert,dx = \int_1^{1/c}\frac{\vert\sin u\vert}{u},du\] \[\text{Step 8:}\quad \int_1^\infty\frac{\vert\sin u\vert}{u},du=\infty\]\(\because \int_{k\pi+\pi/6}^{k\pi+5\pi/6}\frac{\vert\sin u\vert}{u},du \ge \frac12\int_{k\pi+\pi/6}^{k\pi+5\pi/6}\frac{du}{u}\) \(\ge C\cdot\frac1{k+1} \quad (C>0)\)
\[\sum_{k=1}^{\infty}\frac1{k+1}=\infty \implies \int_1^\infty\frac{\vert\sin u\vert}{u},du=\infty\] \[\therefore \lim_{c\to0+}\int_c^1 \vert f(x) \vert,dx \text{ does not exist as a finite limit}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad (a)\ f\in\mathscr R[0,1]\implies \lim_{c\to0+}\int_c^1 f,dx = \int_0^1 f,dx. }\] \[\boxed{ \boldsymbol{Ans.}\quad (b)\ f(x)=\frac{\sin(1/x)}x: \quad \int_0^1 f,dx\text{ converges improperly, but } \int_0^1 \vert f \vert,dx\text{ diverges.} }\]Ch.6 - Problem 8
문제 원문 및 번역
Original. Suppose \(f\in\mathscr R\) on \([a,b]\) for every \(b>a\), where \(a\) is fixed. Define \(\int_a^\infty f(x),dx=\lim_{b\to\infty}\int_a^b f(x),dx\) if this limit exists and is finite. In that case, we say that the integral on the left converges. If it also converges after \(f\) has been replaced by \(\vert f \vert\), it is said to converge absolutely.
Assume that \(f(x)\ge0\) and that \(f\) decreases monotonically on \([1,\infty)\). Prove that \(\int_1^\infty f(x),dx\) converges if and only if \(\sum_{n=1}^{\infty} f(n)\) converges. This is the integral test for convergence of series.
번역. 고정된 \(a\)에 대해 모든 \(b>a\)에서 \(f\in\mathscr R\) on \([a,b]\)라고 하자. 다음 극한이 존재하고 유한하면 \(\int_a^\infty f(x),dx=\lim_{b\to\infty}\int_a^b f(x),dx\) 로 정의한다. 이때 좌변의 적분이 수렴한다고 한다. \(f\)를 \(\vert f \vert\)로 바꾸어도 수렴하면 절대수렴한다고 한다.
\(f(x)\ge0\)이고 \(f\)가 \([1,\infty)\)에서 단조감소한다고 가정하자. 다음 적분 \(\int_1^\infty f(x),dx\) 이 수렴할 필요충분조건은 다음 급수 \(\sum_{n=1}^{\infty} f(n)\) 이 수렴하는 것임을 증명하여라. 이것은 급수 수렴의 적분판정법이다.
요청 사항
\(f\ge0,\quad f\downarrow\text{ on }[1,\infty)\) \(\int_1^\infty f(x),dx<\infty \iff \sum_{n=1}^{\infty}f(n)<\infty\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad S_N=\sum_{n=1}^{N}f(n),\qquad I_N=\int_1^N f(x),dx\]\(\text{Step 1:}\quad n\le x\le n+1 \implies f(n)\ge f(x)\ge f(n+1)\) \(\because f\text{ monotonically decreasing}\)
\(\text{Step 2:}\quad \int_n^{n+1}f(n+1),dx \le \int_n^{n+1}f(x),dx \le \int_n^{n+1}f(n),dx\) \(\because \text{교재 Thm 6.12(b): order preservation}\)
\[\implies f(n+1) \le \int_n^{n+1}f(x),dx \le f(n)\] \[\text{Step 3:}\quad \sum_{n=1}^{N-1}f(n+1) \le \sum_{n=1}^{N-1}\int_n^{n+1}f(x),dx \le \sum_{n=1}^{N-1}f(n)\]\(\implies \sum_{n=2}^{N}f(n) \le \int_1^N f(x),dx \le \sum_{n=1}^{N-1}f(n)\) \(\because \text{교재 Thm 6.12(c): additivity over intervals}\)
\[\text{Step 4:}\quad I_N\le S_{N-1}\le S_N\] \[\text{Step 5:}\quad S_N = f(1)+\sum_{n=2}^{N}f(n) \le f(1)+I_N\] \[\therefore I_N\le S_N\le f(1)+I_N\]\(\text{Step 6:}\quad f\ge0 \implies I_N\text{ increasing in }N,\quad S_N\text{ increasing in }N\) \(\because \text{교재 Thm 6.12(b),(c)}\)
(\(\Rightarrow\))
\[\int_1^\infty f(x),dx\text{ converges} \implies \sup_N I_N<\infty\] \[S_N\le f(1)+I_N \implies \sup_N S_N<\infty\] \[S_N\uparrow,\ \sup_N S_N<\infty \implies S_N\to S<\infty\] \[\therefore \sum_{n=1}^{\infty}f(n)\text{ converges}\](\(\Leftarrow\))
\[\sum_{n=1}^{\infty}f(n)\text{ converges} \implies \sup_N S_N<\infty\] \[I_N\le S_N \implies \sup_N I_N<\infty\] \[I_N\uparrow,\ \sup_N I_N<\infty \implies I_N\to I<\infty\] \[\therefore \int_1^\infty f(x),dx\text{ converges}\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad \int_1^\infty f(x),dx\text{ converges} \iff \sum_{n=1}^{\infty}f(n)\text{ converges}. }\]Ch.6 - Problem 11
문제 원문 및 번역
Original. Let \(\alpha\) be a fixed increasing function on \([a,b]\). For \(u\in\mathscr R(\alpha)\), define \(\vert u \vert_2= \left\{ \int_a^b \vert u \vert^2,d\alpha \right\}^{1/2}.\) Suppose \(f,g,h\in\mathscr R(\alpha)\), and prove the triangle inequality \(\vert f-h \vert_2\le \vert f-g \vert_2+\vert g-h \vert_2\) as a consequence of the Schwarz inequality, as in the proof of Theorem 1.37.
번역. \(\alpha\)를 \([a,b]\) 위의 고정된 증가함수라 하자. \(u\in\mathscr R(\alpha)\)에 대해 \(\vert u \vert_2= \left\{ \int_a^b \vert u \vert^2,d\alpha \right\}^{1/2}\) 로 정의한다. \(f,g,h\in\mathscr R(\alpha)\)라고 하자. Schwarz 부등식의 결과로서, 정리 1.37의 증명처럼 다음 삼각부등식을 증명하여라. \(\vert f-h \vert_2\le \vert f-g \vert_2+\vert g-h \vert_2.\)
요청 사항
\(\vert u \vert_2=\left(\int_a^b \vert u \vert^2,d\alpha\right)^{1/2}\) \(\implies \vert f-h \vert_2\le \vert f-g \vert_2+\vert g-h \vert_2\)
풀이 과정 (Symbolic & Referenced)
\[\text{Def:}\quad u=f-g,\qquad v=g-h\] \[\implies f-h=(f-g)+(g-h)=u+v\]\(\text{Step 1:}\quad u,v\in\mathscr R(\alpha)\) \(\because f,g,h\in\mathscr R(\alpha),\quad \text{교재 Thm 6.12(a): linearity}\)
\(\text{Step 2:}\quad \vert u \vert,\vert v \vert,u^2,v^2,uv\in\mathscr R(\alpha)\) \(\because \text{교재 Thm 6.13(a),(b): products and absolute values}\)
\[\text{Step 3:}\quad \vert u+v \vert_2^2 = \int_a^b \vert u+v \vert^2,d\alpha\]실수값 함수이므로 \(\vert u+v \vert^2=(u+v)^2=u^2+2uv+v^2\)
\(\therefore \vert u+v \vert_2^2 = \int_a^b u^2,d\alpha + 2\int_a^b uv,d\alpha + \int_a^b v^2,d\alpha\) \(\because \text{교재 Thm 6.12(a): linearity}\)
\(\text{Step 4:}\quad \int_a^b uv,d\alpha \le \left(\int_a^b u^2,d\alpha\right)^{1/2} \left(\int_a^b v^2,d\alpha\right)^{1/2}\) \(\because \text{교재 Exercise 6.10(c): Schwarz inequality }(p=q=2)\)
\[\text{Step 5:}\quad \vert u+v \vert_2^2 \le \vert u \vert_2^2+2\vert u \vert_2\vert v \vert_2+\vert v \vert_2^2 = (\vert u \vert_2+\vert v \vert_2)^2\] \[\text{Step 6:}\quad \vert u+v \vert_2\ge0,\quad \vert u \vert_2+\vert v \vert_2\ge0\] \[\implies \vert u+v \vert_2\le \vert u \vert_2+\vert v \vert_2\] \[u=f-g,\quad v=g-h\] \[\therefore \vert f-h \vert_2\le\vert f-g \vert_2+\vert g-h \vert_2\]최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad \vert f-h \vert_2\le \vert f-g \vert_2+\vert g-h \vert_2. }\]Ch.6 - Problem 12
문제 원문 및 번역
Original. With the notations of Exercise 11, suppose \(f\in\mathscr R(\alpha)\) and \(\varepsilon>0\). Prove that there exists a continuous function \(g\) on \([a,b]\) such that \(\vert f-g \vert_2<\varepsilon.\) Hint: Let $P={x_0,\dots,x_n}$ be a suitable partition of \([a,b]\), define \(g(t)=f(x_i)\quad\text{if }x_{i-1}<t<x_i,\) and then modify \(g\) in a suitable way.
번역. 연습문제 11의 표기법을 사용하자. \(f\in\mathscr R(\alpha)\)이고 \(\varepsilon>0\)라고 하자. \([a,b]\) 위의 연속함수 \(g\)가 존재하여 \(\vert f-g \vert_2<\varepsilon\) 임을 증명하여라.
힌트: \(P=\{x_0,\dots,x_n\}\)을 \([a,b]\)의 적절한 분할로 잡고, \(g(t)=f(x_i)\quad\text{if }x_{i-1}<t<x_i\) 로 정의한 뒤, \(g\)를 적절히 수정하여라.
요청 사항
\[f\in\mathscr R(\alpha),\quad \varepsilon>0 \implies \exists g\in C[a,b]: \vert f-g \vert_2<\varepsilon\]풀이 과정 (Symbolic & Referenced)
\(\text{Def:}\quad \vert f(x) \vert \le M\quad (a\le x\le b)\) \(\because f\in\mathscr R(\alpha)\implies f\text{ bounded} \quad \because \text{교재 Def 6.2}\)
\(\text{Step 1:}\quad f^2\in\mathscr R(\alpha)\) \(\because \text{교재 Thm 6.13(a): product }ff=f^2\in\mathscr R(\alpha)\)
\(\text{Step 2:}\quad \exists P=\{x_0,\dots,x_n\}: U(P,f,\alpha)-L(P,f,\alpha)<\eta\) \(\because \text{교재 Thm 6.6}\)
보다 직접적으로 \(L^2\)-오차를 제어하기 위해, 각 구간에서 대표값 \(c_i=f(x_i)\)를 택한다.
\[\text{Def:}\quad s(t)=f(x_i)\quad (x_{i-1}<t<x_i)\] \[\text{Step 3:}\quad \vert s(t) \vert \le M,\quad \vert f(t)-s(t) \vert \le M_i-m_i \quad (x_{i-1}<t<x_i)\]\(\implies \vert f(t)-s(t) \vert^2\le 2M(M_i-m_i)\) \(\because \vert f(t)-s(t) \vert\le2M,\quad r^2\le 2Mr\ (0\le r\le2M)\)
\[\text{Step 4:}\quad \int_a^b \vert f-s \vert^2,d\alpha \le 2M\sum_{i=1}^n(M_i-m_i)\Delta\alpha_i = 2M[U(P,f,\alpha)-L(P,f,\alpha)]\] \[\text{Step 5:}\quad \text{Choose }P\text{ so that } U(P,f,\alpha)-L(P,f,\alpha)<\frac{\varepsilon^2}{8M}\] \[\implies \vert f-s \vert_2^2<\frac{\varepsilon^2}{4} \implies \vert f-s \vert_2<\frac\varepsilon2\]\[\text{Now modify }s\text{ to a continuous }g.\] \[\text{Step 6:}\quad s\text{ has finitely many jumps at }x_1,\dots,x_{n-1}\]
각 \(x_i\) 주변에 서로소 작은 구간 \(I_i=(x_i-\delta_i,x_i+\delta_i)\) 를 택하여 \(\sum_i[\alpha(x_i+\delta_i)-\alpha(x_i-\delta_i)] < \frac{\varepsilon^2}{16M^2}\) 가 되게 한다.
\[\text{단, }\alpha\text{의 jump가 있는 }x_i\text{에서는 분할점을 조정하여 피함}\] \[\because \alpha\text{ increasing has at most countably many discontinuities; finite partition points can be chosen at continuity points}\] \[\text{Step 7:}\quad g(t)=s(t)\text{ outside }\bigcup I_i\] \[g\text{ is linearly interpolated on each }I_i\] \[\implies g\in C[a,b],\quad \vert g(t) \vert \le M,\quad \vert s(t)-g(t) \vert \le2M\] \[\text{Step 8:}\quad \int_a^b \vert s-g \vert^2,d\alpha = \int_{\cup I_i}\vert s-g \vert^2,d\alpha \le 4M^2\sum_i[\alpha(x_i+\delta_i)-\alpha(x_i-\delta_i)] < \frac{\varepsilon^2}{4}\] \[\implies \vert s-g \vert_2<\frac\varepsilon2\]\(\text{Step 9:}\quad \vert f-g \vert_2 \le \vert f-s \vert_2+\vert s-g \vert_2 < \frac\varepsilon2+\frac\varepsilon2 = \varepsilon\) \(\because \text{Ch.6 Problem 11 triangle inequality}\)
최종 답안
\[\boxed{ \boldsymbol{Ans.}\quad \forall\varepsilon>0,\ \exists g\in C[a,b]\text{ such that } \vert f-g \vert_2<\varepsilon. }\]Ch.6 - Problem 19
문제 원문 및 번역
Original. Let \(\gamma_1\) be a curve in \(\mathbb R^k\), defined on \([a,b]\); let \(\phi\) be a continuous one-to-one mapping of \([c,d]\) onto \([a,b]\), such that \(\phi(c)=a\); and define \(\gamma_2(s)=\gamma_1(\phi(s)).\) Prove that \(\gamma_2\) is an arc, a closed curve, or a rectifiable curve if and only if the same is true of \(\gamma_1\). Prove that \(\gamma_2\) and \(\gamma_1\) have the same length.
번역. \(\gamma_1\)을 \([a,b]\)에서 정의된 \(\mathbb R^k\) 안의 곡선이라 하자. \(\phi\)는 \([c,d]\)에서 \([a,b]\) 위로 가는 연속 일대일 대응이고 \(\phi(c)=a\)라고 하자. \(\gamma_2(s)=\gamma_1(\phi(s))\) 로 정의한다. \(\gamma_2\)가 호, 폐곡선, 또는 길이유한곡선일 필요충분조건은 \(\gamma_1\)도 각각 그러한 것임을 증명하여라. 또한 \(\gamma_2\)와 \(\gamma_1\)의 길이가 같음을 증명하여라.
요청 사항
\(\gamma_2=\gamma_1\circ\phi,\quad \phi:[c,d]\to[a,b]\text{ continuous bijection},\quad \phi(c)=a\) \(\gamma_2\text{ arc/closed/rectifiable} \iff \gamma_1\text{ arc/closed/rectifiable}\) \(\Lambda(\gamma_2)=\Lambda(\gamma_1)\)
풀이 과정 (Symbolic & Referenced)
\(\text{Def:}\quad \Lambda(\gamma)=\sup_P\Lambda(P,\gamma)\) \(\Lambda(P,\gamma)=\sum_{i=1}^{n}\vert\gamma(x_i)-\gamma(x_{i-1})\vert\) \(\because \text{교재 Def 6.26: curve length and rectifiability}\)
\[\text{Step 1:}\quad \phi:[c,d]\to[a,b]\text{ continuous one-to-one onto},\quad \phi(c)=a\]\(\implies \phi(d)=b\) \(\because \phi([c,d])=[a,b],\ \phi(c)=a,\ \phi\text{ one-to-one continuous}\)
\[\text{Step 2:}\quad \phi\text{ continuous one-to-one on interval} \implies \phi\text{ strictly monotone}\] \[\phi(c)=a,\ \phi(d)=b \implies \phi\text{ strictly increasing}\]Arc
\(\text{Step 3:}\quad \gamma_1\text{ arc} \iff \gamma_1\text{ one-to-one}\) \(\because \text{교재 Def 6.26}\)
\[\gamma_2(s_1)=\gamma_2(s_2) \iff \gamma_1(\phi(s_1))=\gamma_1(\phi(s_2))\] \[\gamma_1\text{ one-to-one} \implies \phi(s_1)=\phi(s_2) \implies s_1=s_2\] \[\therefore \gamma_1\text{ arc}\implies\gamma_2\text{ arc}\] \[\gamma_2\text{ arc},\quad \gamma_1(t_1)=\gamma_1(t_2)\] \[\exists s_1,s_2:\ \phi(s_j)=t_j \quad \because \phi\text{ onto}\] \[\gamma_2(s_1)=\gamma_2(s_2) \implies s_1=s_2 \implies t_1=t_2\] \[\therefore \gamma_2\text{ arc}\implies\gamma_1\text{ arc}\] \[\therefore \gamma_2\text{ arc}\iff\gamma_1\text{ arc}\]Closed curve
\[\text{Step 4:}\quad \gamma_2(c)=\gamma_1(\phi(c))=\gamma_1(a)\] \[\gamma_2(d)=\gamma_1(\phi(d))=\gamma_1(b)\] \[\therefore \gamma_2(c)=\gamma_2(d) \iff \gamma_1(a)=\gamma_1(b)\]\(\therefore \gamma_2\text{ closed} \iff \gamma_1\text{ closed}\) \(\because \text{교재 Def 6.26: closed curve}\)
Rectifiable and same length
\[\text{Step 5:}\quad Q=\{c=s_0<s_1<\cdots<s_n=d\}\] \[\text{Def:}\quad P=\{\phi(s_0),\phi(s_1),\dots,\phi(s_n)\}\] \[\phi\text{ strictly increasing} \implies P=\{a=t_0<t_1<\cdots<t_n=b\}\text{ partition of }[a,b]\] \[\text{Step 6:}\quad \Lambda(Q,\gamma_2) = \sum_{i=1}^n\vert\gamma_2(s_i)-\gamma_2(s_{i-1})\vert\] \[= \sum_{i=1}^n\vert\gamma_1(\phi(s_i))-\gamma_1(\phi(s_{i-1}))\vert\] \[= \sum_{i=1}^n\vert\gamma_1(t_i)-\gamma_1(t_{i-1})\vert = \Lambda(P,\gamma_1)\]\(\text{Step 7:}\quad Q\leftrightarrow P\text{ bijection between partitions}\) \(\because \phi\text{ strictly increasing bijection}\)
\[\implies \sup_Q\Lambda(Q,\gamma_2) = \sup_P\Lambda(P,\gamma_1)\] \[\therefore \Lambda(\gamma_2)=\Lambda(\gamma_1)\]\(\text{Step 8:}\quad \gamma_2\text{ rectifiable} \iff \Lambda(\gamma_2)<\infty \iff \Lambda(\gamma_1)<\infty \iff \gamma_1\text{ rectifiable}\) \(\because \text{교재 Def 6.26}\)
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