실해석학 HW4
A. [과제 수령 및 확인 리포트]
확정 문항 리스트
- Chapter 7: 1, 2, 3, 5, 7, 9, 11, 15, 16, 18
- Chapter 8: 1, 2, 3, 5, 9, 11, 13, 14, 15
총 문항 수: \(10+9=\boxed{19}\)
B. [문항별 상세 솔루션]
Chapter 7 — Problem 1
[문제 원문 및 번역]
Original: Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.
번역: 유계 함수들의 모든 균등수렴 함수열은 균등유계임을 증명하여라.
[요청 사항]
\[f\_n \subset \text{bounded functions},\quad f\_n\to f\ \text{uniformly on }E \implies \exists M<\infty:\ \vert f\_n(x) \vert \le M\quad(\forall n,\forall x\in E)\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n:E\to \mathbb C,\quad f\_n\ \text{bounded}\)
Def: \(\quad f\_n\ \text{uniformly bounded} \iff \exists M<\infty\ \text{s.t.}\ \vert f\_n(x) \vert <M\quad(\forall x\in E,\forall n)\) (\(\because\) 교재 Ch.7 Def. 7.19)
\(f\_n\to f\ \text{uniformly} \implies \forall \varepsilon>0,\exists N:\ n,m\ge N\Rightarrow \vert f\_n(x)-f\_m(x) \vert \le\varepsilon\quad(\forall x\in E)\) (\(\because\) 교재 Ch.7 Thm. 7.8 Cauchy criterion)
\[\varepsilon=1 \implies \exists N:\ n\ge N\Rightarrow \vert f\_n(x)-f\_N(x) \vert \le 1\quad(\forall x\in E)\] \[\because f\_1,\dots,f\_N\ \text{bounded}\] \[\forall k=1,\dots,N,\quad \exists M\_k<\infty:\ \vert f\_k(x) \vert \le M\_k\quad(\forall x\in E)\] \[M\_0:=\max{M\_1,\dots,M\_N}<\infty\] \[n\le N\implies \vert f\_n(x) \vert \le M\_0\]\(n>N\implies \vert f\_n(x) \vert \le \vert f\_n(x)-f\_N(x) \vert +\vert f\_N(x) \vert \le 1+M\_N\) (\(\because \vert a+b \vert \le \vert a \vert +\vert b \vert\))
\[M:=\max{M\_0,1+M\_N}\] \[\therefore \vert f\_n(x) \vert \le M\quad(\forall n,\forall x\in E)\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\_n\to f\ \text{uniformly and each }f\_n\text{ bounded} \implies f\_n\text{ is uniformly bounded.}}\]Chapter 7 — Problem 2
[문제 원문 및 번역]
Original: If ({f_n}) and ({g_n}) converge uniformly on a set (E), prove that ({f_n+g_n}) converges uniformly on (E). If, in addition, ({f_n}) and ({g_n}) are sequences of bounded functions, prove that ({f_ng_n}) converges uniformly on (E).
번역: \(f\_n\)과 \(g\_n\)이 집합 \(E\)에서 균등수렴하면, \(f\_n+g\_n\)도 \(E\)에서 균등수렴함을 증명하여라. 추가로 \(f\_n\), \(g\_n\)이 유계 함수열이면, \(f\_n g\_n\)도 \(E\)에서 균등수렴함을 증명하여라.
[요청 사항]
\[f\_n\to f,\quad g\_n\to g\ \text{uniformly} \implies f\_n+g\_n\to f+g\ \text{uniformly}\] \[+\quad f\_n,g\_n\ \text{bounded} \implies f\_n g\_n\to fg\ \text{uniformly}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n,g\_n:E\to\mathbb C,\quad f\_n\to f,\quad g\_n\to g\)
Def: \(\quad h\_n\to h\ \text{uniformly} \iff \sup_{x\in E}\vert h\_n(x)-h(x) \vert \to0\) (\(\because\) 교재 Ch.7 Thm. 7.9)
(i) Sum
\[\vert (f\_n+g\_n)(x)-(f+g)(x) \vert \le \vert f\_n(x)-f(x) \vert +\vert g\_n(x)-g(x) \vert\] \[\sup\_E \vert (f\_n+g\_n)-(f+g) \vert \le \sup\_E\vert f\_n-f \vert +\sup\_E\vert g\_n-g \vert\] \[f\_n\to f\ \text{uniformly}\implies \sup\_E\vert f\_n-f \vert \to0\] \[g\_n\to g\ \text{uniformly}\implies \sup\_E\vert g\_n-g \vert \to0\] \[\therefore \sup\_E \vert (f\_n+g\_n)-(f+g) \vert \to0\]\(\implies f\_n+g\_n\to f+g\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
(ii) Product
\[f\_n g\_n-fg=f\_n g\_n-fg\_n+fg\_n-fg\] \[=f\_n(g\_n-g)+g(f\_n-f)\] \[\vert f\_n g\_n-fg \vert \le \vert f\_n \vert \vert g\_n-g \vert +\vert g \vert \vert f\_n-f \vert\] \[\because f\_n,g\_n\ \text{bounded functions and uniformly convergent}\]\(\implies f\_n\ \text{uniformly bounded}\) (\(\because\) Problem 1 result)
\[\exists A,B<\infty:\quad \vert f\_n(x) \vert \le A,\quad \vert g(x) \vert \le B\quad(\forall n,\forall x)\] \[\therefore \vert f\_n g\_n-fg \vert \le A\vert g\_n-g \vert +B\vert f\_n-f \vert\] \[\sup\_E\vert f\_n g\_n-fg \vert \le A\sup\_E\vert g\_n-g \vert +B\sup\_E\vert f\_n-f \vert \to0\]\(\therefore f\_n g\_n\to fg\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\_n+g\_n\to f+g\ \text{uniformly},\qquad f\_n g\_n\to fg\ \text{uniformly if the sequences are bounded.}}\]Chapter 7 — Problem 3
[문제 원문 및 번역]
Original: Construct sequences ({f_n},{g_n}) which converge uniformly on some set (E), but such that ({f_ng_n}) does not converge uniformly on (E) — of course, ({f_ng_n}) must converge on (E).
번역: 어떤 집합 \(E\) 위에서 \(f\_n,g\_n\)은 균등수렴하지만, \(f\_n g\_n\)은 \(E\)에서 균등수렴하지 않는 예를 구성하여라. 물론 \(f\_n g\_n\)은 \(E\)에서 점별수렴해야 한다.
[요청 사항]
\[\exists E,\ f\_n,g\_n: \quad f\_n\to f,\ g\_n\to g\ \text{uniformly}\] \[\text{but}\quad f\_n g\_n\to fg\ \text{pointwise, not uniformly}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad E=\mathbb R\)
Def: \(\quad f\_n(x)=x+\frac1n,\qquad g\_n(x)=x+\frac1n\)
\[f(x)=x,\qquad g(x)=x\](i) ({f_n}), ({g_n}) uniformly converge
\[\vert f\_n(x)-f(x) \vert =\left\vert x+\frac1n-x \right\vert =\frac1n\] \[\sup_{x\in\mathbb R}\vert f\_n(x)-f(x) \vert =\frac1n\to0\]\(\therefore f\_n\to f\ \text{uniformly on }\mathbb R\) (\(\because\) 교재 Ch.7 Thm. 7.9)
\[\vert g\_n(x)-g(x) \vert =\frac1n\] \[\sup_{x\in\mathbb R}\vert g\_n(x)-g(x) \vert =\frac1n\to0\]\(\therefore g\_n\to g\ \text{uniformly on }\mathbb R\) (\(\because\) 교재 Ch.7 Thm. 7.9)
(ii) Product converges pointwise
\[f\_n(x)g\_n(x)=\left(x+\frac1n\right)^{2} =x^{2}+\frac{2x}{n}+\frac1{n^{2}}\] \[\forall x\in\mathbb R,\quad \frac{2x}{n}+\frac1{n^{2}}\to0\] \[\therefore f\_n(x)g\_n(x)\to x^{2}=f(x)g(x)\]\(\implies f\_n g\_n\to fg\ \text{pointwise}\) (\(\because\) 교재 Ch.7 Def. 7.1 pointwise convergence)
(iii) Product does not converge uniformly
\[\vert f\_n(x)g\_n(x)-x^{2} \vert =\left\vert \frac{2x}{n}+\frac1{n^{2}} \right\vert\] \[x=n^{2}\implies \left\vert \frac{2x}{n}+\frac1{n^{2}} \right\vert = \left\vert \frac{2n^{2}}{n}+\frac1{n^{2}} \right\vert = 2n+\frac1{n^{2}}\] \[\therefore \sup_{x\in\mathbb R}\vert f\_n(x)g\_n(x)-x^{2} \vert =\infty\] \[\implies \sup\_E\vert f\_n g\_n-fg \vert \not\to0\]\(\therefore f\_n g\_n\not\to fg\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad E=\mathbb R,\quad f\_n(x)=g\_n(x)=x+\frac1n.}\] \[\boxed{f\_n\to x,\quad g\_n\to x\ \text{uniformly},\qquad f\_n g\_n\to x^{2}\ \text{pointwise but not uniformly}.}\]Chapter 7 — Problem 5
[문제 원문 및 번역]
Original: Let \(f\_n(x)= \begin{cases} 0,& x<\dfrac1{n+1},[2mm] \sin^2\dfrac{\pi}{x},& \dfrac1{n+1}\le x\le \dfrac1n,[2mm] 0,& \dfrac1n<x. \end{cases}\) Show that ({f_n}) converges to a continuous function, but not uniformly. Use the series (\sum f_n) to show that absolute convergence, even for all (x), does not imply uniform convergence.
번역: 위와 같이 \(f\_n\)을 정의하자. \(f\_n\)이 어떤 연속함수로 수렴하지만 균등수렴하지 않음을 보여라. 또한 급수 \(\sum f\_n\)을 이용하여, 모든 \(x\)에서 절대수렴하더라도 균등수렴을 함의하지 않음을 보여라.
[요청 사항]
\[f\_n\to f\ \text{pointwise},\quad f\in C,\quad f\_n\not\to f\ \text{uniformly}\] \[\sum f\_n(x)\ \text{absolutely converges }\forall x \quad\not\implies\quad \sum f\_n\ \text{uniformly converges}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad I\_n=\left[\frac1{n+1},\frac1n\right]\)
\[f\_n(x)=0\quad(x\notin I\_n),\qquad f\_n(x)=\sin^{2}\frac{\pi}{x}\quad(x\in I\_n)\]Def: \(\quad f(x):=\lim_{n\to\infty}f\_n(x)\) (\(\because\) 교재 Ch.7 Def. 7.1 pointwise convergence)
(i) Pointwise limit
\[x\ne0 \implies \exists N:\ n\ge N\Rightarrow x\notin I\_n\] \[\therefore n\ge N\Rightarrow f\_n(x)=0\] \[x=0\implies 0\notin I\_n\Rightarrow f\_n(0)=0\] \[\therefore \forall x,\quad f\_n(x)\to0\] \[\implies f(x)=0\] \[f\equiv0\implies f\ \text{continuous}\](ii) Not uniform
\[x\_n:=\frac{2}{2n+1}\] \[\frac1{n+1}<\frac{2}{2n+1}<\frac1n \implies x\_n\in I\_n\] \[\frac{\pi}{x\_n} =\frac{2n+1}{2}\pi =\left(n+\frac12\right)\pi\] \[f\_n(x\_n) = \sin^{2}\left(n+\frac12\right)\pi =1\] \[\therefore \sup\_x \vert f\_n(x)-0 \vert \ge1\] \[\because 0\le f\_n(x)\le1 \implies \sup\_x \vert f\_n(x) \vert =1\] \[\sup\_x\vert f\_n-f \vert =1\not\to0\]\(\therefore f\_n\not\to f\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
(iii) Absolute convergence ($\not\Rightarrow$) uniform convergence
\[I\_n^\circ\cap I\_m^\circ=\varnothing\quad(n\ne m)\] \[\therefore \forall x,\quad \#{n:f\_n(x)\ne0}\le1\] \[\sum_{n=1}^{\infty}\vert f\_n(x) \vert = \sum_{n=1}^{\infty}f\_n(x) \le1\] \[\therefore \sum f\_n(x)\ \text{absolutely converges for every }x\]하지만 tail에 대해,
\[\sup\_x\left\vert \sum_{k=n}^{m}f\_k(x) \right\vert \ge \sup\_x f\_n(x)=1\] \[\therefore \sum f\_n\ \text{is not uniformly Cauchy}\]\(\implies \sum f\_n\ \text{does not converge uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.8 Cauchy criterion)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\_n\to0\ \text{pointwise},\quad 0\ \text{continuous},\quad f\_n\not\to0\ \text{uniformly}.}\] \[\boxed{\boldsymbol{Ans.}\quad \sum f\_n(x)\ \text{converges absolutely for every }x,\ \text{but not uniformly}.}\]Chapter 7 — Problem 7
[문제 원문 및 번역]
Original: For (n=1,2,3,\ldots), (x) real, put \(f\_n(x)=\frac{x}{1+nx^{2}}.\) Show that ({f_n}) converges uniformly to a function (f), and that the equation \(f'(x)=\lim_{n\to\infty}f\_n'(x)\) is correct if (x\ne0), but false if (x=0).
번역: \(n=1,2,3,\ldots\), 실수 \(x\)에 대해 \(f\_n(x)=\frac{x}{1+nx^{2}}\) 라 하자. \(f\_n\)이 어떤 함수 \(f\)로 균등수렴함을 보이고, \(f'(x)=\lim_{n\to\infty}f\_n'(x)\) 가 \(x\ne0\)에서는 맞지만 \(x=0\)에서는 틀림을 보여라.
[요청 사항]
\[f\_n\to f\ \text{uniformly}\] \[x\ne0:\quad f'(x)=\lim f\_n'(x)\] \[x=0:\quad f'(0)\ne\lim f\_n'(0)\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n(x)=\frac{x}{1+nx^{2}}\)
(i) Pointwise limit
\[x=0\implies f\_n(0)=0\] \[x\ne0\implies \frac{x}{1+nx^{2}}\to0\] \[\therefore f\_n(x)\to0\quad(\forall x)\]Def: \(\quad f(x)=0\)
(ii) Uniform convergence
\[\vert f\_n(x) \vert =\frac{\vert x \vert}{1+nx^{2}}\] \[y:=\vert x \vert \ge0 \implies \vert f\_n(x) \vert =\frac{y}{1+ny^{2}}\] \[\phi\_n(y):=\frac{y}{1+ny^{2}}\] \[\phi\_n'(y)=\frac{1-ny^{2}}{(1+ny^{2})^{2}}\] \[\phi\_n'(y)=0 \iff y=\frac1{\sqrt n}\] \[\sup_{x\in\mathbb R}\vert f\_n(x) \vert = \phi\_n\left(\frac1{\sqrt n}\right) = \frac{1/\sqrt n}{1+n(1/n)} = \frac1{2\sqrt n} \to0\]\(\therefore f\_n\to0\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
(iii) Derivatives
\[f\_n'(x) = \frac{(1+nx^{2})-x(2nx)}{(1+nx^{2})^{2}} = \frac{1-nx^{2}}{(1+nx^{2})^{2}}\] \[f(x)=0\implies f'(x)=0\quad(\forall x)\]만약 \(x\ne0\)이면,
\[\lim_{n\to\infty}f\_n'(x) = \lim_{n\to\infty} \frac{1-nx^{2}}{(1+nx^{2})^{2}}\] \[\sim \frac{-nx^{2}}{n^{2}x^{4}} = -\frac1{nx^{2}} \to0\] \[\therefore x\ne0:\quad \lim f\_n'(x)=0=f'(x)\]만약 \(x=0\)이면,
\[f\_n'(0)=\frac{1-0}{(1+0)^{2}}=1\] \[\lim_{n\to\infty}f\_n'(0)=1\] \[f'(0)=0\] \[\therefore \lim f\_n'(0)=1\ne0=f'(0)\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\_n\to f\equiv0\ \text{uniformly}.}\] \[\boxed{\boldsymbol{Ans.}\quad x\ne0:\ f'(x)=\lim f\_n'(x),\qquad x=0:\ f'(0)\ne\lim f\_n'(0).}\]Chapter 7 — Problem 9
[문제 원문 및 번역]
Original: Let ({f_n}) be a sequence of continuous functions which converges uniformly to a function (f) on a set (E). Prove that \(\lim_{n\to\infty} f\_n(x\_n)=f(x)\) for every sequence of points (x_n\in E) such that (x_n\to x), and (x\in E). Is the converse of this true?
번역: \(f\_n\)이 \(E\) 위의 연속함수열이고 \(f\)로 균등수렴한다고 하자. \(x\_n\in E\), \(x\_n\to x\), \(x\in E\)인 모든 점열에 대해 \(\lim_{n\to\infty} f\_n(x\_n)=f(x)\) 임을 증명하여라. 이 명제의 역은 참인가?
[요청 사항]
\[f\_n\in C(E),\quad f\_n\to f\ \text{uniformly} \implies x\_n\to x\Rightarrow f\_n(x\_n)\to f(x)\] \[\text{Converse?}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad x\_n\in E,\quad x\_n\to x,\quad x\in E\)
\[f\_n\to f\ \text{uniformly},\quad f\_n\in C(E)\]\(\implies f\in C(E)\) (\(\because\) 교재 Ch.7 Thm. 7.12)
\[\vert f\_n(x\_n)-f(x) \vert \le \vert f\_n(x\_n)-f(x\_n) \vert +\vert f(x\_n)-f(x) \vert\]\(f\_n\to f\ \text{uniformly} \implies \sup_{y\in E}\vert f\_n(y)-f(y) \vert \to0\) (\(\because\) 교재 Ch.7 Thm. 7.9)
\[\therefore \vert f\_n(x\_n)-f(x\_n) \vert \le \sup\_E\vert f\_n-f \vert \to0\] \[f\in C(E),\quad x\_n\to x \implies f(x\_n)\to f(x)\] \[\therefore \vert f(x\_n)-f(x) \vert \to0\] \[\therefore f\_n(x\_n)\to f(x)\]Converse
Claim: 역은 거짓이다.
Def: \(\quad E=(0,1),\qquad f\_n(x)=x^n\)
\[\forall x\in(0,1),\quad x^n\to0\] \[f(x)=0\]모든 \(x\_n\to x\in(0,1)\)에 대해,
\[\exists r<1,\exists N:\ n\ge N\Rightarrow 0<x\_n\le r\] \[\therefore 0\le f\_n(x\_n)=x\_n^n\le r^n\to0=f(x)\] \[\implies f\_n(x\_n)\to f(x)\]하지만
\[\sup_{x\in(0,1)}\vert x^n-0 \vert =1\] \[\therefore \sup\_E\vert f\_n-f \vert \not\to0\]\(\implies f\_n\not\to f\ \text{uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.9)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\_n\to f\ \text{uniformly and }f\_n\in C \implies x\_n\to x\Rightarrow f\_n(x\_n)\to f(x).}\] \[\boxed{\boldsymbol{Ans.}\quad \text{Converse is false; }E=(0,1),\ f\_n(x)=x^n.}\]Chapter 7 — Problem 11
[문제 원문 및 번역]
Original: Suppose ({f_n},{g_n}) are defined on (E), and (a) (\sum f_n) has uniformly bounded partial sums; (b) (g_n\to0) uniformly on (E); (c) (g_1(x)\ge g_2(x)\ge g_3(x)\ge\cdots) for every (x\in E). Prove that (\sum f_ng_n) converges uniformly on (E). Hint: Compare with Theorem 3.42.
번역: \(f\_n,g\_n\)이 \(E\) 위에서 정의되어 있고, (a) \(\sum f\_n\)의 부분합들이 균등유계이며, (b) \(g\_n\to0\)이 \(E\)에서 균등수렴하고, (c) 모든 \(x\in E\)에 대해 \(g\_1(x)\ge g\_2(x)\ge g\_3(x)\ge\cdots\)라 하자. 그러면 \(\sum f\_n g\_n\)이 \(E\)에서 균등수렴함을 증명하여라.
[요청 사항]
\[\sum f\_n\ \text{partial sums uniformly bounded},\quad g\_n\downarrow0,\quad g\_n\to0\ \text{uniformly}\] \[\implies \sum f\_n g\_n\ \text{uniformly convergent}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad S\_N(x)=\sum_{k=1}^{N}f\_k(x)\)
(a) \(\implies \exists M<\infty:\quad \vert S\_N(x) \vert \le M \quad(\forall N,\forall x\in E)\)
Def: \(\quad B\_k(x)=\sum_{j=n}^{k}f\_j(x)=S\_k(x)-S\_{n-1}(x)\)
\[\vert B\_k(x) \vert \le \vert S\_k(x) \vert +\vert S\_{n-1}(x) \vert \le2M\](b),(c) \(\implies g\_n(x)\ge g\_{n+1}(x)\ge\cdots\ge0\)
since \(g\_n(x)\downarrow \lim g\_n(x)=0\)
\(m\ge n\)에 대해,
\(\sum_{k=n}^{m}f\_k(x)g\_k(x) = \sum_{k=n}^{m-1}B\_k(x)\bigl(g\_k(x)-g\_{k+1}(x)\bigr) +B\_m(x)g\_m(x)\) (\(\because\) Abel summation / 교재 Thm. 3.42 비교)
\[\left\vert \sum_{k=n}^{m}f\_k(x)g\_k(x) \right\vert \le \sum_{k=n}^{m-1}\vert B\_k(x) \vert \bigl(g\_k(x)-g\_{k+1}(x)\bigr) +\vert B\_m(x) \vert g\_m(x)\] \[\le 2M\sum_{k=n}^{m-1}(g\_k(x)-g\_{k+1}(x)) +2Mg\_m(x)\] \[= 2M(g\_n(x)-g\_m(x))+2Mg\_m(x) = 2Mg\_n(x)\] \[\therefore \sup_{x\in E} \left\vert \sum_{k=n}^{m}f\_k(x)g\_k(x) \right\vert \le 2M\sup_{x\in E}g\_n(x)\]\(g\_n\to0\ \text{uniformly} \implies \sup\_E\vert g\_n \vert \to0\) (\(\because\) 교재 Ch.7 Thm. 7.9)
\[\therefore \sup\_E \left\vert \sum_{k=n}^{m}f\_k g\_k \right\vert \to0\] \[\implies \sum f\_n g\_n\ \text{uniformly Cauchy}\]\(\therefore \sum f\_n g\_n\ \text{converges uniformly}\) (\(\because\) 교재 Ch.7 Thm. 7.8)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \sum f\_n g\_n\ \text{converges uniformly on }E.}\]Chapter 7 — Problem 15
[문제 원문 및 번역]
Original: Suppose (f) is a real continuous function on (\mathbb R^1), (f_n(t)=f(nt)) for (n=1,2,3,\ldots), and ({f_n}) is equicontinuous on ([0,1]). What conclusion can you draw about (f)?
번역: \(f\)가 \(\mathbb R^1\) 위의 실연속함수이고, \(f\_n(t)=f(nt)\), \(n=1,2,3,\ldots\)라 하자. \(f\_n\)이 \([0,1]\)에서 동등연속이면 \(f\)에 대해 어떤 결론을 내릴 수 있는가?
[요청 사항]
\[f\_n(t)=f(nt),\quad f\_n\ \text{equicontinuous on }[0,1]\] \[\implies ?\quad f\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n(t)=f(nt)\)
Def: \(\quad f\_n\ \text{equicontinuous on }[0,1]\)
\(\iff \forall\varepsilon>0,\exists\delta>0: \vert s-t \vert <\delta \Rightarrow \vert f\_n(s)-f\_n(t) \vert <\varepsilon \quad(\forall n)\) (\(\because\) 교재 Ch.7 Def. 7.22)
\(x,y\ge0\)라 하자.
\[\exists n\ \text{large}:\quad \frac{x}{n},\frac{y}{n}\in[0,1],\qquad \left\vert \frac{x}{n}-\frac{y}{n} \right\vert = \frac{\vert x-y \vert}{n} <\delta\] \[\therefore \vert f\_n(x/n)-f\_n(y/n) \vert <\varepsilon\] \[f\_n(x/n)=f\left(n\cdot \frac{x}{n}\right)=f(x)\] \[f\_n(y/n)=f\left(n\cdot \frac{y}{n}\right)=f(y)\] \[\therefore \vert f(x)-f(y) \vert <\varepsilon\] \[\forall\varepsilon>0,\quad \vert f(x)-f(y) \vert <\varepsilon\] \[\implies f(x)=f(y)\] \[\therefore f\ \text{is constant on }[0,\infty)\]\((-\infty,0)\)에 대한 강제적인 결론은 없다:
\[f(x)= \begin{cases} x,&x<0,\ 0,&x\ge0 \end{cases}\] \[\implies f\_n(t)=f(nt)=0\quad(0\le t\le1)\] \[\therefore f\_n\ \text{equicontinuous on }[0,1]\] \[\text{but }f\ \text{not constant on }\mathbb R\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f\ \text{must be constant on }[0,\infty),\ \text{but need not be constant on all of }\mathbb R.}\]Chapter 7 — Problem 16
[문제 원문 및 번역]
Original: Suppose ({f_n}) is an equicontinuous sequence of functions on a compact set (K), and ({f_n}) converges pointwise on (K). Prove that ({f_n}) converges uniformly on (K).
번역: \(f\_n\)이 compact set \(K\) 위의 동등연속 함수열이고, \(K\)에서 점별수렴한다고 하자. 그러면 \(f\_n\)이 \(K\)에서 균등수렴함을 증명하여라.
[요청 사항]
\[K\ \text{compact},\quad f\_n\ \text{equicontinuous},\quad f\_n\to f\ \text{pointwise}\] \[\implies f\_n\to f\ \text{uniformly}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n\to f\ \text{pointwise on }K\)
\[\varepsilon>0\]equicontinuity에 의해,
\(\exists\delta>0: d(x,y)<\delta \Rightarrow \vert f\_n(x)-f\_n(y) \vert <\frac{\varepsilon}{3} \quad(\forall n)\) (\(\because\) 교재 Ch.7 Def. 7.22)
\(K\)가 compact이므로,
\(\exists x\_1,\dots,x\_m\in K: K\subset \bigcup_{j=1}^{m}V(x\_j,\delta)\) (\(\because\) compactness finite subcover)
각 \(x\_j\)에서의 점별수렴:
\[\forall j,\quad f\_n(x\_j)\to f(x\_j)\] \[\therefore \exists N:\ n,m\ge N \Rightarrow \vert f\_n(x\_j)-f\_m(x\_j) \vert <\frac{\varepsilon}{3} \quad(j=1,\dots,m)\]임의의 \(x\in K\)에 대해,
\[\exists j:\ d(x,x\_j)<\delta\] \[\vert f\_n(x)-f\_m(x) \vert \le \vert f\_n(x)-f\_n(x\_j) \vert + \vert f\_n(x\_j)-f\_m(x\_j) \vert + \vert f\_m(x\_j)-f\_m(x) \vert\] \[<\frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon\] \[\therefore n,m\ge N\Rightarrow \sup_{x\in K}\vert f\_n(x)-f\_m(x) \vert <\varepsilon\] \[\implies f\_n\ \text{uniformly Cauchy}\]\(\therefore f\_n\to f\ \text{uniformly on }K\) (\(\because\) 교재 Ch.7 Thm. 7.8)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \text{Equicontinuity}+\text{compactness}+\text{pointwise convergence} \implies \text{uniform convergence}.}\]Chapter 7 — Problem 18
[문제 원문 및 번역]
Original: Let ({f_n}) be a uniformly bounded sequence of functions which are Riemann-integrable on ([a,b]), and put \(F\_n(x)=\int\_a^x f\_n(t),dt \qquad (a\le x\le b).\) Prove that there exists a subsequence ({F_{n_k}}) which converges uniformly on ([a,b]).
번역: \(f\_n\)이 \([a,b]\)에서 Riemann 적분가능하고 균등유계인 함수열이라 하자. \(F\_n(x)=\int\_a^x f\_n(t),dt \qquad (a\le x\le b).\) 로 정의한다. 그러면 \(F\_n\)의 어떤 부분수열 \(F\_{n\_k}\)이 \([a,b]\)에서 균등수렴함을 증명하여라.
[요청 사항]
\[\vert f\_n(x) \vert \le M,\quad f\_n\in\mathcal R[a,b]\] \[F\_n(x)=\int\_a^x f\_n(t),dt\] \[\implies \exists F\_{n\_k}:\quad F\_{n\_k}\ \text{uniformly convergent on }[a,b]\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f\_n\ \text{uniformly bounded}\)
\(\exists M<\infty:\quad \vert f\_n(t) \vert \le M \quad(\forall n,\forall t\in[a,b])\) (\(\because\) 교재 Ch.7 Def. 7.19)
\[F\_n(x)=\int\_a^x f\_n(t),dt\](i) Pointwise boundedness of ({F_n})
\[\vert F\_n(x) \vert = \left\vert \int\_a^x f\_n(t),dt \right\vert \le \int\_a^x \vert f\_n(t) \vert ,dt \le M(x-a) \le M(b-a)\] \[\therefore F\_n\ \text{uniformly bounded on }[a,b]\] \[\implies F\_n\ \text{pointwise bounded}\](ii) Equicontinuity of ({F_n})
\(x,y\in[a,b]\)에 대해, \(x<y\)라 가정하자.
\[\vert F\_n(y)-F\_n(x) \vert = \left\vert \int\_x^y f\_n(t),dt \right\vert \le \int\_x^y \vert f\_n(t) \vert ,dt \le M(y-x)\] \[\therefore \vert F\_n(y)-F\_n(x) \vert \le M\vert y-x \vert\]주어진 \(\varepsilon>0\)에 대해,
\[\delta:=\frac{\varepsilon}{M+1}\] \[\vert x-y \vert <\delta \implies \vert F\_n(x)-F\_n(y) \vert \le M\vert x-y \vert < \varepsilon\]\(\therefore F\_n\ \text{equicontinuous on }[a,b]\) (\(\because\) 교재 Ch.7 Def. 7.22)
(iii) Apply Arzelà–Ascoli form
\[[a,b]\ \text{compact}\] \[F\_n\ \text{pointwise bounded and equicontinuous}\]\(\implies F\_n\ \text{contains a uniformly convergent subsequence}\) (\(\because\) 교재 Ch.7 Thm. 7.25)
\[\therefore \exists F\_{n\_k}:\quad F\_{n\_k}\to F\ \text{uniformly on }[a,b]\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \exists n\_k:\ F\_{n\_k}\ \text{converges uniformly on }[a,b].}\]Chapter 8 — Problem 1
[문제 원문 및 번역]
Original: Define \(f(x)= \begin{cases} e^{-1/x^2},& x\ne0,\ 0,& x=0. \end{cases}\) Prove that (f) has derivatives of all orders at (x=0), and that \(f^{(n)}(0)=0\qquad(n=1,2,3,\ldots).\)
번역: 다음과 같이 정의하자. \(f(x)= \begin{cases} e^{-1/x^{2}},& x\ne0,\ 0,& x=0. \end{cases}\) \(f\)가 \(x=0\)에서 모든 차수의 도함수를 가지며, \(f^{(n)}(0)=0\qquad(n=1,2,3,\ldots)\) 임을 증명하여라.
[요청 사항]
\[f^{(n)}(0)\ \text{exists}\quad(\forall n\ge1)\] \[f^{(n)}(0)=0\quad(\forall n\ge1)\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f(0)=0,\qquad f(x)=e^{-1/x^{2}}\quad(x\ne0)\)
Key limit:\quad \(\lim_{x\to0}x^{-m}e^{-1/x^{2}}=0\qquad(m=0,1,2,\ldots)\)
Let
\[u=\frac1{x^{2}}\] \[x\to0\implies u\to+\infty\]\(\vert x \vert ^{-m}e^{-1/x^{2}} = u^{m/2}e^{-u} \to0\) (\(\because\) 교재 Ch.8 Thm. 8.6(f): \(u^\alpha e^{-u}\to0\))
Step 1: (\(f'(0)\))
\[f'(0) = \lim_{x\to0}\frac{f(x)-f(0)}{x} = \lim_{x\to0}\frac{e^{-1/x^{2}}}{x}\] \[= \lim_{x\to0}x^{-1}e^{-1/x^{2}} =0\] \[\therefore f'(0)=0\]Step 2: 형태 귀납
Claim:\quad \(x\ne0\Rightarrow f^{(n)}(x)=P\_n\left(\frac1x\right)e^{-1/x^{2}}\)
여기서 \(P\_n\)은 polynomial이다.
\[n=0:\quad f(x)=1\cdot e^{-1/x^{2}}\] \[P\_0(t)=1\]가정하자.
\[f^{(n)}(x)=P\_n(1/x)e^{-1/x^{2}}\]그러면
\[f^{(n+1)}(x) = \frac{d}{dx}\left[P\_n(1/x)e^{-1/x^{2}}\right]\] \[= -\frac1{x^{2}}P\_n'(1/x)e^{-1/x^{2}} + P\_n(1/x)\cdot \frac{2}{x^{3}}e^{-1/x^{2}}\] \[= \left[-x^{-2}P\_n'(1/x)+2x^{-3}P\_n(1/x)\right]e^{-1/x^{2}}\] \[= P\_{n+1}(1/x)e^{-1/x^{2}}\] \[\therefore f^{(n)}(x)=P\_n(1/x)e^{-1/x^{2}} \quad(x\ne0)\]Step 3: (\(f^{(n)}(0)=0\))
Induction hypothesis:
\[f^{(n-1)}(0)=0\]그러면
\[f^{(n)}(0) = \lim_{x\to0} \frac{f^{(n-1)}(x)-f^{(n-1)}(0)}{x}\] \[= \lim_{x\to0} \frac{P\_{n-1}(1/x)e^{-1/x^{2}}}{x}\] \[= \lim_{x\to0} x^{-1}P\_{n-1}(1/x)e^{-1/x^{2}}\]\(P\_{n-1}\)이 polynomial이므로,
\[\exists C,m:\quad \left\vert x^{-1}P\_{n-1}(1/x) \right\vert \le C\vert x \vert ^{-m} \quad(0<\vert x \vert <1)\] \[\therefore \left\vert x^{-1}P\_{n-1}(1/x)e^{-1/x^{2}} \right\vert \le C\vert x \vert ^{-m}e^{-1/x^{2}} \to0\](\(\because\) 교재 Ch.8 Thm. 8.6(f))
\[\therefore f^{(n)}(0)=0\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f^{(n)}(0)\ \text{exists for every }n\ge1,\qquad f^{(n)}(0)=0.}\]Chapter 8 — Problem 2
[문제 원문 및 번역]
Original: Let (a_{ij}) be the number in the (i)-th row and (j)-th column of the array \(\begin{matrix} -1&0&0&0&\cdots\ \frac12&-1&0&0&\cdots\ \frac14&\frac12&-1&0&\cdots\ \frac18&\frac14&\frac12&-1&\cdots\ \cdots&\cdots&\cdots&\cdots \end{matrix}\) so that \(a\_{ij}= \begin{cases} 0,& i<j,\ -1,& i=j,\ 2^{j-i-1},& i>j. \end{cases}\) Prove that \(\sum\_i\sum\_j a\_{ij}=-2,\qquad \sum\_j\sum\_i a\_{ij}=0.\)
번역: \(a\_{ij}\)를 위 배열의 \(i\)번째 행, \(j\)번째 열의 수라 하자. 즉, \(a\_{ij}= \begin{cases} 0,& i<j,\ -1,& i=j,\ 2^{j-i-1},& i>j. \end{cases}\) 다음을 증명하여라. \(\sum\_i\sum\_j a\_{ij}=-2,\qquad \sum\_j\sum\_i a\_{ij}=0.\)
[요청 사항]
\[\text{iterated sums differ}\] \[\sum\_i\left(\sum\_j a\_{ij}\right)=-2,\qquad \sum\_j\left(\sum\_i a\_{ij}\right)=0\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad a\_{ij}= \begin{cases} 0,& i<j,\ -1,& i=j,\ 2^{j-i-1},& i>j. \end{cases}\)
주의: double series order는 절대수렴이 없으면 중요할 수 있다. (\(\because\) 교재 Ch.8 Thm. 8.3은 interchange를 위해 \(\sum\_i b\_i<\infty\)를 요구한다.)
Step 1: Row sums
\(i\)를 고정하자.
\[\sum_{j=1}^{\infty}a\_{ij} = \sum_{j=1}^{i-1}2^{j-i-1}+(-1)+\sum_{j=i+1}^{\infty}0\] \[= \sum_{j=1}^{i-1}2^{j-i-1}-1\]\(k=i-j\)라 하자. 그러면 \(j=1,\dots,i-1\Rightarrow k=i-1,\dots,1\)이다.
\[\sum_{j=1}^{i-1}2^{j-i-1} = \sum_{k=1}^{i-1}2^{-k-1}\] \[= \frac12\sum_{k=1}^{i-1}2^{-k} = \frac12(1-2^{-(i-1)}) = \frac12-2^{-i}\] \[\therefore \sum_{j=1}^{\infty}a\_{ij} = \left(\frac12-2^{-i}\right)-1 = -\frac12-2^{-i}\]따라서
\[\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a\_{ij} = \sum_{i=1}^{\infty} \left(-\frac12-2^{-i}\right)\]이것은 \(-\infty\)로 발산하며, \(-2\)가 아니다.
\[\therefore \text{given exponent }2^{j-i-1}\text{ cannot yield the printed target }-2.\]표시된 배열은 대신 다음을 제공한다.
\[a\_{ij}=2^{j-i}\quad(i>j)\]왜냐하면
\[a\_{21}=\frac12=2^{-1},\quad a\_{31}=\frac14=2^{-2},\quad a\_{32}=\frac12=2^{-1}\]따라서 배열과 일치하는 정의를 사용하자:
\[a\_{ij}= \begin{cases} 0,& i<j,\ -1,& i=j,\ 2^{j-i},& i>j. \end{cases}\]그러면
\[\sum_{j=1}^{\infty}a\_{ij} = \sum_{j=1}^{i-1}2^{j-i}-1\] \[= \sum_{k=1}^{i-1}2^{-k}-1 = (1-2^{-(i-1)})-1 = -2^{-(i-1)}\]\(\therefore \sum\_i\sum\_j a\_{ij} = \sum_{i=1}^{\infty}-2^{-(i-1)} = -\sum_{r=0}^{\infty}2^{-r} = -2\) (\(\because\) geometric series; 교재 Ch.3 Thm. 3.26이 Ch.7 Ex.7.3에서 인용됨)
Step 2: Column sums
\(j\)를 고정하자.
\[\sum_{i=1}^{\infty}a\_{ij} = \sum_{i=1}^{j-1}0+(-1)+\sum_{i=j+1}^{\infty}2^{j-i}\] \[= -1+\sum_{r=1}^{\infty}2^{-r}\] \[= -1+1 =0\] \[\therefore \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a\_{ij} = \sum_{j=1}^{\infty}0 =0\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \sum\_i\sum\_j a\_{ij}=-2,\qquad \sum\_j\sum\_i a\_{ij}=0.}\] \[\boxed{\text{Note: the array corresponds to }a\_{ij}=2^{j-i}\ (i>j). \text{ The printed }2^{j-i-1}\text{ is inconsistent with the array and target.}}\]Chapter 8 — Problem 3
[문제 원문 및 번역]
Original: Prove that \(\sum\_i\sum\_j a\_{ij}=\sum\_j\sum\_i a\_{ij}\) if (a_{ij}\ge0) for all (i) and (j) — the case (+\infty=+\infty) may occur.
번역: 모든 \(i,j\)에 대해 \(a\_{ij}\ge0\)이면, \(\sum\_i\sum\_j a\_{ij}=\sum\_j\sum\_i a\_{ij}\) 임을 증명하여라. 단, \(+\infty=+\infty\)인 경우도 허용한다.
[요청 사항]
\[a\_{ij}\ge0 \implies \sum\_i\sum\_j a\_{ij} = \sum\_j\sum\_i a\_{ij}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad a\_{ij}\ge0\)
유한한 rectangle에 대해,
\(S\_{mn}:=\sum_{i=1}^{m}\sum_{j=1}^{n}a\_{ij} = \sum_{j=1}^{n}\sum_{i=1}^{m}a\_{ij}\) (\(\because\) finite sums commute)
\(a\_{ij}\ge0\)이므로,
\[m\_1\le m\_2,\ n\_1\le n\_2 \implies S\_{m\_1 n\_1}\le S\_{m\_2 n\_2}\]따라서
\[A:=\sup_{m,n}S\_{mn}\in[0,+\infty]\]Step 1: Compare row-first sum
고정된 \(m\)에 대해,
\[\sum_{i=1}^{m}\sum_{j=1}^{\infty}a\_{ij} = \sum_{i=1}^{m}\sup\_n\sum_{j=1}^{n}a\_{ij}\] \[= \sup\_n\sum_{i=1}^{m}\sum_{j=1}^{n}a\_{ij} = \sup\_n S\_{mn} \le A\] \[\therefore \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a\_{ij} = \sup\_m\sum_{i=1}^{m}\sum_{j=1}^{\infty}a\_{ij} \le A\]역으로,
\[S\_{mn}\le \sum_{i=1}^{m}\sum_{j=1}^{\infty}a\_{ij} \le \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a\_{ij}\] \[\therefore A\le \sum\_i\sum\_j a\_{ij}\] \[\therefore \sum\_i\sum\_j a\_{ij}=A\]Step 2: Compare column-first sum
유사하게,
\[\sum_{j=1}^{n}\sum_{i=1}^{\infty}a\_{ij} = \sup\_m S\_{mn} \le A\] \[\therefore \sum\_j\sum\_i a\_{ij}\le A\]또한,
\[S\_{mn}\le \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a\_{ij}\] \[\therefore A\le \sum\_j\sum\_i a\_{ij}\] \[\therefore \sum\_j\sum\_i a\_{ij}=A\]따라서
\[\sum\_i\sum\_j a\_{ij} = A = \sum\_j\sum\_i a\_{ij}\]\(\therefore \sum\_i\sum\_j a\_{ij} = \sum\_j\sum\_i a\_{ij}\) (\(\because\) nonnegative monotone partial sums; Ch.8 Thm. 8.3 interchange condition과 비교)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad a\_{ij}\ge0\ \forall i,j \implies \sum\_i\sum\_j a\_{ij} = \sum\_j\sum\_i a\_{ij},}\] \[\boxed{\text{including the case }+\infty=+\infty.}\]Chapter 8 — Problem 5
[문제 원문 및 번역]
Original: Find the following limits: \((a)\quad \lim_{x\to0}\frac{e-(1+x)^{1/x}}{x},\) \((b)\quad \lim_{n\to\infty}\frac{n}{\log n}\left(n^{1/n}-1\right),\) \((c)\quad \lim_{x\to0}\frac{\tan x-x}{x(1-\cos x)},\) \((d)\quad \lim_{x\to0}\frac{x-\sin x}{\tan x-x}.\)
번역: 다음 극한들을 구하여라. \((a)\quad \lim_{x\to0}\frac{e-(1+x)^{1/x}}{x},\) \((b)\quad \lim_{n\to\infty}\frac{n}{\log n}\left(n^{1/n}-1\right),\) \((c)\quad \lim_{x\to0}\frac{\tan x-x}{x(1-\cos x)},\) \((d)\quad \lim_{x\to0}\frac{x-\sin x}{\tan x-x}.\)
[요청 사항]
\[\text{(a),(b),(c),(d)}\ \text{각 극한값}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad e^x=\exp x,\quad \log x=L(x)\) (\(\because\) 교재 Ch.8 exponential/log definitions and Thm. 8.6)
(a)
\[(1+x)^{1/x} = \exp\left(\frac{\log(1+x)}{x}\right)\] \[\log(1+x) = x-\frac{x^{2}}{2}+O(x^{3})\] \[\therefore \frac{\log(1+x)}{x} = 1-\frac{x}{2}+O(x^{2})\] \[(1+x)^{1/x} = e\cdot \exp\left(-\frac{x}{2}+O(x^{2})\right)\] \[= e\left(1-\frac{x}{2}+O(x^{2})\right)\] \[e-(1+x)^{1/x} = e-e\left(1-\frac{x}{2}+O(x^{2})\right) = \frac e2x+O(x^{2})\]\(\therefore \lim_{x\to0} \frac{e-(1+x)^{1/x}}{x} = \frac e2\) (\(\because\) 교재 Ch.8 Thm. 8.1 power series differentiation; Thm. 8.6 exponential)
(b)
\[n^{1/n}=e^{(\log n)/n}\]Let
\[u\_n=\frac{\log n}{n}\]\(n\to\infty\implies u\_n\to0\) (\(\because \log n/n\to0\); 교재 Ch.8 Eq. (45): \(\log x\)는 powers보다 느리게 증가한다.)
\[e^{u\_n}-1=u\_n+O(u\_n^{2})\]\(\frac{n}{\log n}\left(n^{1/n}-1\right) = \frac1{u\_n}(e^{u\_n}-1) \to1\) (\(\because\) 교재 Ch.8 exponential derivative \((e^x)'=e^x,\ e^0=1\))
(c)
\[\tan x=\frac{\sin x}{\cos x}\] \[\sin x=x-\frac{x^{3}}{6}+O(x^{5})\] \[\cos x=1-\frac{x^{2}}{2}+O(x^{4})\] \[\tan x = x+\frac{x^{3}}{3}+O(x^{5})\] \[\tan x-x=\frac{x^{3}}{3}+O(x^{5})\] \[1-\cos x=\frac{x^{2}}{2}+O(x^{4})\] \[x(1-\cos x)=\frac{x^{3}}{2}+O(x^{5})\]\(\therefore \frac{\tan x-x}{x(1-\cos x)} \to \frac{1/3}{1/2} = \frac23\) (\(\because\) 교재 Ch.8 trigonometric functions from \(E(ix)\), power series Thm. 8.1)
(d)
\[x-\sin x = x-\left(x-\frac{x^{3}}{6}+O(x^{5})\right) = \frac{x^{3}}{6}+O(x^{5})\] \[\tan x-x = \frac{x^{3}}{3}+O(x^{5})\]\(\therefore \frac{x-\sin x}{\tan x-x} \to \frac{1/6}{1/3} = \frac12\) (\(\because\) 교재 Ch.8 trigonometric power-series framework)
[최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad (a)=\frac e2,\qquad (b)=1,\qquad (c)=\frac23,\qquad (d)=\frac12.}\]Chapter 8 — Problem 9
[문제 원문 및 번역]
Original: (a) Put \(s\_N=1+\frac12+\cdots+\frac1N.\) Prove that \(\lim_{N\to\infty}(s\_N-\log N)\) exists. The limit, often denoted by (\gamma), is called Euler’s constant. (b) Roughly how large must (m) be so that (N=10^m) satisfies (s_N>100)?
번역: (a) \(s\_N=1+\frac12+\cdots+\frac1N.\) 라 하자. \(\lim_{N\to\infty}(s\_N-\log N)\) 이 존재함을 증명하여라. 이 극한은 보통 \(\gamma\)로 표시하며 Euler 상수라 부른다. (b) \(N=10^m\)이 \(s\_N>100\)을 만족하려면 \(m\)은 대략 얼마나 커야 하는가?
[요청 사항]
\[\text{(a)}\quad s\_N-\log N\ \text{converges}\] \[\text{(b)}\quad N=10^m,\quad s\_N>100\Rightarrow m\approx ?\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad s\_N=\sum_{k=1}^{N}\frac1k\)
Def: \(\quad \log x=\int\_1^x\frac{dt}{t}\) (\(\because\) 교재 Ch.8 Eq. (39): \(L(y)=\int\_1^y dx/x\))
(a) Existence
\[a\_N:=s\_N-\log N\] \[a\_{N+1}-a\_N = \frac1{N+1}-\log(N+1)+\log N\] \[= \frac1{N+1}-\log\left(1+\frac1N\right)\] \[\log\left(1+\frac1N\right) = \int\_1^{1+1/N}\frac{dt}{t}\]\(1\le t\le1+1/N\)에 대해,
\[\frac1t\ge \frac1{1+1/N}=\frac{N}{N+1}\] \[\therefore \log\left(1+\frac1N\right) \ge \frac1N\cdot\frac{N}{N+1} = \frac1{N+1}\] \[\implies a\_{N+1}-a\_N\le0\] \[\therefore a\_N\ \text{decreasing}\]또한,
\[\log N = \sum_{k=1}^{N-1}\int\_k^{k+1}\frac{dt}{t}\]\(t\in[k,k+1]\)에 대해,
\[\frac1t\le\frac1k\] \[\therefore \int\_k^{k+1}\frac{dt}{t}\le\frac1k\] \[\log N \le \sum_{k=1}^{N-1}\frac1k =s\_{N-1} <s\_N\] \[\therefore a\_N=s\_N-\log N>0\] \[a\_N\ \text{decreasing and bounded below by }0\]\(\therefore a\_N\to\gamma\quad\text{exists}\) (\(\because\) monotone convergence theorem for real sequences)
(b) Estimate (m)
\[s\_N=\log N+\gamma+o(1)\] \[N=10^m \implies \log N=m\log10\] \[s\_N>100 \approx m\log10+\gamma>100\] \[m> \frac{100-\gamma}{\log10}\]문제 진술에서 알려진 바와 같이:
\[\gamma\approx0.5772\] \[\log10\approx2.302585\] \[m\gtrsim \frac{99.4228}{2.302585} \approx43.18\] \[\therefore m\approx44\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \lim_{N\to\infty}(s\_N-\log N)=\gamma\ \text{exists}.}\] \[\boxed{\boldsymbol{Ans.}\quad N=10^m,\ s\_N>100\ \text{requires roughly }m=44.}\]Chapter 8 — Problem 11
[문제 원문 및 번역]
Original: Suppose (f\in\mathcal R) on ((0,A]) for all (A<\infty), and (f(x)\to1) as (x\to+\infty). Prove that \(\lim_{t\to0}t\int\_0^\infty e^{-tx}f(x),dx=1.\)
번역: 모든 \(A<\infty\)에 대해 \(f\in\mathcal R\) on \((0,A]\)이고, \(x\to+\infty\)일 때 \(f(x)\to1\)이라 하자. 다음을 증명하여라. \(\lim_{t\to0}t\int\_0^\infty e^{-tx}f(x),dx=1.\)
[요청 사항]
\[f(x)\to1\quad(x\to\infty)\] \[\implies t\int\_0^\infty e^{-tx}f(x),dx\to1 \quad(t\to0^+)\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad I(t):=t\int\_0^\infty e^{-tx}f(x),dx\)
Def: \(\quad u=tx\Rightarrow x=\frac ut,\quad dx=\frac{du}{t}\)
\[I(t) = t\int\_0^\infty e^{-tx}f(x),dx = t\int\_0^\infty e^{-u}f(u/t)\frac{du}{t}\] \[= \int\_0^\infty e^{-u}f(u/t),du\]\(1=\int\_0^\infty e^{-u},du\) (\(\because \Gamma(1)=\int\_0^\infty e^{-u}du=1\); 교재 Ch.8 Def. 8.17, Thm. 8.18)
따라서
\[I(t)-1 = \int\_0^\infty e^{-u}{f(u/t)-1},du\]\(\varepsilon>0\)라 하자.
\[f(x)\to1 \implies \exists B>0:\ x>B\Rightarrow \vert f(x)-1 \vert <\varepsilon\]분할:
\[\vert I(t)-1 \vert \le \int\_0^{Bt}e^{-u}\vert f(u/t)-1 \vert ,du + \int\_{Bt}^{\infty}e^{-u}\vert f(u/t)-1 \vert ,du\]\(u>Bt\)에 대해,
\[\frac ut>B \implies \vert f(u/t)-1 \vert <\varepsilon\] \[\therefore \int\_{Bt}^{\infty}e^{-u}\vert f(u/t)-1 \vert ,du \le \varepsilon\int\_{Bt}^{\infty}e^{-u},du \le\varepsilon\]\(f\in\mathcal R\) on \((0,B]\)이므로,
\[f\ \text{bounded on }(0,B]\quad\Rightarrow\quad \exists M:\ \vert f(x)-1 \vert \le M\quad(0<x\le B)\] \[\therefore \int\_0^{Bt}e^{-u}\vert f(u/t)-1 \vert ,du \le M\int\_0^{Bt}e^{-u},du \le MBt\to0\] \[\therefore \limsup_{t\to0^+}\vert I(t)-1 \vert \le\varepsilon\] \[\because \varepsilon>0\ \text{arbitrary}\] \[\therefore I(t)\to1\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \lim_{t\to0^+}t\int\_0^\infty e^{-tx}f(x),dx=1.}\]Chapter 8 — Problem 13
[문제 원문 및 번역]
Original: Put (f(x)=x) if (0\le x<2\pi), and apply Parseval’s theorem to conclude that \(\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6}.\)
번역: \(0\le x<2\pi\)에서 \(f(x)=x\)라 하고, Parseval 정리를 적용하여 다음을 결론지어라. \(\sum_{n=1}^{\infty}\frac1{n^{2}}=\frac{\pi^{2}}{6}.\)
[요청 사항]
\[f(x)=x\quad(0\le x<2\pi)\] \[\text{Use Parseval} \implies \sum_{n=1}^{\infty}\frac1{n^{2}}=\frac{\pi^{2}}{6}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f(x)=x\quad(0\le x<2\pi),\quad f(x+2\pi)=f(x)\)
\([0,2\pi]\)에서 real Fourier series를 사용하자:
\[f(x)\sim \frac{a\_0}{2}+\sum_{n=1}^{\infty}(a\_n\cos nx+b\_n\sin nx)\] \[a\_0=\frac1\pi\int\_0^{2\pi}x,dx = \frac1\pi\cdot\frac{(2\pi)^{2}}{2} = 2\pi\] \[\therefore \frac{a\_0}{2}=\pi\]\(n\ge1\)에 대해,
\[a\_n=\frac1\pi\int\_0^{2\pi}x\cos nx,dx\] \[\int x\cos nx,dx = \frac{x\sin nx}{n}+\frac{\cos nx}{n^{2}}\] \[a\_n = \frac1\pi \left[ \frac{x\sin nx}{n}+\frac{\cos nx}{n^{2}} \right]_{0}^{2\pi} = 0\] \[b\_n=\frac1\pi\int\_0^{2\pi}x\sin nx,dx\] \[\int x\sin nx,dx = -\frac{x\cos nx}{n}+\frac{\sin nx}{n^{2}}\] \[b\_n = \frac1\pi \left[ -\frac{x\cos nx}{n}+\frac{\sin nx}{n^{2}} \right]_{0}^{2\pi} = \frac1\pi\left(-\frac{2\pi}{n}\right) = -\frac2n\]따라서
\[x\sim \pi-\sum_{n=1}^{\infty}\frac2n\sin nx\]Parseval real form:
\(\frac1\pi\int\_0^{2\pi}\vert f(x) \vert ^{2}dx = \frac{a\_0^{2}}{2} + \sum_{n=1}^{\infty}(a\_n^{2}+b\_n^{2})\) (\(\because\) 교재 Ch.8 Parseval Thm. 8.16)
좌변:
\[\frac1\pi\int\_0^{2\pi}x^{2}dx = \frac1\pi\cdot\frac{(2\pi)^{3}}{3} = \frac{8\pi^{2}}{3}\]우변:
\[\frac{a\_0^{2}}{2} + \sum_{n=1}^{\infty}b\_n^{2} = \frac{(2\pi)^{2}}{2} + \sum_{n=1}^{\infty}\frac4{n^{2}}\] \[= 2\pi^{2}+4\sum_{n=1}^{\infty}\frac1{n^{2}}\]그러므로
\[\frac{8\pi^{2}}{3} = 2\pi^{2} + 4\sum_{n=1}^{\infty}\frac1{n^{2}}\] \[4\sum_{n=1}^{\infty}\frac1{n^{2}} = \frac{8\pi^{2}}{3}-2\pi^{2} = \frac{2\pi^{2}}{3}\] \[\therefore \sum_{n=1}^{\infty}\frac1{n^{2}} = \frac{\pi^{2}}{6}\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad \displaystyle \sum_{n=1}^{\infty}\frac1{n^{2}} = \frac{\pi^{2}}{6}.}\]Chapter 8 — Problem 14
[문제 원문 및 번역]
Original: If \(f(x)=(\pi-|x|)^2\qquad(-\pi\le x\le\pi),\) prove that \(f(x)=\frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac4{n^2}\cos nx\) and deduce that \(\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6},\qquad \sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{90}.\)
번역: \(f(x)=(\pi-\vert x \vert)^{2}\qquad(-\pi\le x\le\pi),\) 라 하자. 다음을 증명하여라. \(f(x)=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac4{n^{2}}\cos nx\) 그리고 다음을 도출하여라. \(\sum_{n=1}^{\infty}\frac1{n^{2}}=\frac{\pi^{2}}{6},\qquad \sum_{n=1}^{\infty}\frac1{n^{4}}=\frac{\pi^{4}}{90}.\)
[요청 사항]
\[f(x)=(\pi-\vert x \vert)^{2}\] \[\implies f(x)=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac4{n^{2}}\cos nx\] \[\implies \sum\frac1{n^{2}}=\frac{\pi^{2}}{6},\quad \sum\frac1{n^{4}}=\frac{\pi^{4}}{90}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad f(x)=(\pi-\vert x \vert)^{2},\quad -\pi\le x\le\pi\)
\[f(-x)=f(x) \implies b\_n=0\] \[f(x)\sim \frac{a\_0}{2}+\sum_{n=1}^{\infty}a\_n\cos nx\]Step 1: (\(a\_0\))
\[a\_0=\frac1\pi\int\_{-\pi}^{\pi}f(x),dx = \frac2\pi\int\_0^\pi(\pi-x)^{2}dx\] \[= \frac2\pi\cdot\frac{\pi^{3}}{3} = \frac{2\pi^{2}}{3}\] \[\therefore \frac{a\_0}{2}=\frac{\pi^{2}}{3}\]Step 2: (\(a\_n\))
\[a\_n=\frac1\pi\int\_{-\pi}^{\pi}f(x)\cos nx,dx = \frac2\pi\int\_0^\pi(\pi-x)^{2}\cos nx,dx\]\(u=\pi-x\)라 하자.
\[x=\pi-u,\quad dx=-du\] \[\int\_0^\pi(\pi-x)^{2}\cos nx,dx = \int\_0^\pi u^{2}\cos(n(\pi-u)),du\] \[\cos(n(\pi-u)) = \cos(n\pi)\cos(nu)+\sin(n\pi)\sin(nu) = (-1)^n\cos(nu)\] \[\therefore a\_n=\frac2\pi(-1)^n\int\_0^\pi u^{2}\cos(nu),du\]계산:
\[\int u^{2}\cos(nu),du = \frac{u^{2}\sin(nu)}{n} + \frac{2u\cos(nu)}{n^{2}} - \frac{2\sin(nu)}{n^{3}}\] \[\int\_0^\pi u^{2}\cos(nu),du = \frac{2\pi\cos(n\pi)}{n^{2}} = \frac{2\pi(-1)^n}{n^{2}}\] \[a\_n= \frac2\pi(-1)^n \cdot \frac{2\pi(-1)^n}{n^{2}} = \frac4{n^{2}}\] \[\therefore f(x)\sim \frac{\pi^{2}}{3} + \sum_{n=1}^{\infty}\frac4{n^{2}}\cos nx\]\(f\)가 연속이고 piecewise smooth이므로,
\(\text{Fourier series converges to }f(x)\) (\(\because\) 교재 Ch.8 Thm. 8.14 pointwise convergence condition)
\[\therefore f(x)= \frac{\pi^{2}}{3} + \sum_{n=1}^{\infty}\frac4{n^{2}}\cos nx\]Step 3: Deduce ($\sum 1/n^2$)
\(x=0\)을 대입하자.
\[f(0)=\pi^{2}\] \[\pi^{2} = \frac{\pi^{2}}{3} + 4\sum_{n=1}^{\infty}\frac1{n^{2}}\] \[4\sum_{n=1}^{\infty}\frac1{n^{2}} = \frac{2\pi^{2}}{3}\] \[\therefore \sum_{n=1}^{\infty}\frac1{n^{2}} = \frac{\pi^{2}}{6}\]Step 4: Deduce ($\sum 1/n^4$)
Parseval:
\(\frac1\pi\int\_{-\pi}^{\pi}\vert f(x) \vert ^{2}dx = \frac{a\_0^{2}}{2} + \sum_{n=1}^{\infty}a\_n^{2}\) (\(\because\) 교재 Ch.8 Parseval Thm. 8.16)
좌변:
\[\frac1\pi\int\_{-\pi}^{\pi}(\pi-\vert x \vert)^{4}dx = \frac2\pi\int\_0^\pi(\pi-x)^{4}dx\] \[= \frac2\pi\cdot\frac{\pi^{5}}{5} = \frac{2\pi^{4}}{5}\]우변:
\[\frac{a\_0^{2}}{2} + \sum_{n=1}^{\infty}a\_n^{2} = \frac12\left(\frac{2\pi^{2}}{3}\right)^{2} + \sum_{n=1}^{\infty}\left(\frac4{n^{2}}\right)^{2}\] \[= \frac{2\pi^{4}}{9} + 16\sum_{n=1}^{\infty}\frac1{n^{4}}\]따라서
\[\frac{2\pi^{4}}{5} = \frac{2\pi^{4}}{9} + 16\sum_{n=1}^{\infty}\frac1{n^{4}}\] \[16\sum_{n=1}^{\infty}\frac1{n^{4}} = 2\pi^{4}\left(\frac15-\frac19\right) = 2\pi^{4}\cdot\frac4{45} = \frac{8\pi^{4}}{45}\] \[\therefore \sum_{n=1}^{\infty}\frac1{n^{4}} = \frac{\pi^{4}}{90}\][최종 답안]
\[\boxed{\boldsymbol{Ans.}\quad f(x)=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac4{n^{2}}\cos nx.}\] \[\boxed{\boldsymbol{Ans.}\quad \sum_{n=1}^{\infty}\frac1{n^{2}}=\frac{\pi^{2}}{6},\qquad \sum_{n=1}^{\infty}\frac1{n^{4}}=\frac{\pi^{4}}{90}.}\]Chapter 8 — Problem 15
[문제 원문 및 번역]
Original: With (D_n) as defined in (77), put \(K\_N(x)=\frac1{N+1}\sum_{n=0}^{N}D\_n(x).\) Prove that \(K\_N(x)=\frac1{N+1}\cdot\frac{1-\cos (N+1)x}{1-\cos x},\) and that \((a)\quad K\_N\ge0,\qquad (b)\quad \frac1{2\pi}\int\_{-\pi}^{\pi}K\_N(x),dx=1,\) \((c)\quad K\_N(x)\le\frac1{N+1}\cdot\frac{2}{1-\cos\delta} \quad\text{if }0<\delta\le |x|\le\pi.\) If (s_N=s_N(f;x)) is the (N)-th partial sum of the Fourier series of (f), consider the arithmetic means \(\sigma\_N=\frac{s\_0+s\_1+\cdots+s\_N}{N+1}.\) Prove that \(\sigma\_N(f;x)=\frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t)K\_N(t),dt,\) and hence prove Fejér’s theorem: If (f) is continuous, with period (2\pi), then \(\sigma\_N(f;x)\to f(x)\) uniformly on ([-\pi,\pi]).
번역: (77)에서 정의된 \(D\_n\)에 대해 \(K\_N(x)=\frac1{N+1}\sum_{n=0}^{N}D\_n(x).\) 라 하자. 다음을 증명하여라. \(K\_N(x)=\frac1{N+1}\cdot\frac{1-\cos (N+1)x}{1-\cos x},\) 또한 \((a)\quad K\_N\ge0,\qquad (b)\quad \frac1{2\pi}\int\_{-\pi}^{\pi}K\_N(x),dx=1,\) \((c)\quad K\_N(x)\le\frac1{N+1}\cdot\frac{2}{1-\cos\delta} \quad\text{if }0<\delta\le \vert x \vert \le\pi.\) \(s\_N=s\_N(f;x)\)가 \(f\)의 Fourier 급수의 \(N\)-번째 부분합일 때, 산술평균 \(\sigma\_N=\frac{s\_0+s\_1+\cdots+s\_N}{N+1}.\) 을 고려하라. 다음을 증명하고, \(\sigma\_N(f;x)=\frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t)K\_N(t),dt,\) 따라서 Fejér 정리, 즉 \(f\)가 주기 \(2\pi\)인 연속함수이면 \(\sigma\_N(f;x)\to f(x)\) 가 \([-\pi,\pi]\)에서 균등수렴함을 증명하여라.
[요청 사항]
\[K\_N\ \text{formula},\quad K\_N\ge0,\quad \int K\_N=2\pi,\quad K\_N\ \text{tail bound}\] \[\sigma\_N(f;x)=\frac1{2\pi}\int f(x-t)K\_N(t)dt\] \[f\in C,\ 2\pi\text{-periodic} \implies \sigma\_N(f;\cdot)\to f\ \text{uniformly}\][풀이 과정 (Symbolic & Referenced)]
Def: \(\quad D\_n(x)=\sum_{k=-n}^{n}e^{ikx} = \frac{\sin(n+\frac12)x}{\sin(x/2)}\) (\(\because\) 교재 Ch.8 Eq. (77) Dirichlet kernel)
Def: \(\quad K\_N(x)=\frac1{N+1}\sum_{n=0}^{N}D\_n(x)\)
Step 1: Formula for (\(K\_N\))
\[D\_n(x)=\sum_{k=-n}^{n}e^{ikx}\] \[\sum_{n=0}^{N}D\_n(x) = \sum_{n=0}^{N}\sum_{k=-n}^{n}e^{ikx}\] \[= \sum_{k=-N}^{N}(N+1-\vert k \vert)e^{ikx}\]또한,
\[\left\vert \sum_{j=0}^{N}e^{ijx} \right\vert ^{2} = \sum_{j=0}^{N}\sum_{l=0}^{N}e^{i(j-l)x} = \sum_{k=-N}^{N}(N+1-\vert k \vert)e^{ikx}\] \[\therefore K\_N(x) = \frac1{N+1} \left\vert \sum_{j=0}^{N}e^{ijx} \right\vert ^{2}\]Geometric sum:
\[\sum_{j=0}^{N}e^{ijx} = \frac{1-e^{i(N+1)x}}{1-e^{ix}}\] \[\left\vert \sum_{j=0}^{N}e^{ijx} \right\vert ^{2} = \frac{\vert 1-e^{i(N+1)x} \vert ^{2}}{\vert 1-e^{ix} \vert ^{2}}\] \[\vert 1-e^{i\theta} \vert ^{2} = (1-e^{i\theta})(1-e^{-i\theta}) = 2-2\cos\theta = 2(1-\cos\theta)\] \[\therefore K\_N(x) = \frac1{N+1} \cdot \frac{2(1-\cos(N+1)x)}{2(1-\cos x)}\] \[= \frac1{N+1} \cdot \frac{1-\cos(N+1)x}{1-\cos x}\]Step 2: (\(K\_N\ge0\))
\[K\_N(x) = \frac1{N+1} \left\vert \sum_{j=0}^{N}e^{ijx} \right\vert ^{2} \ge0\] \[\therefore K\_N\ge0\]Step 3: Integral normalization
\[K\_N(x)=\frac1{N+1}\sum_{n=0}^{N}D\_n(x)\]\(\frac1{2\pi}\int\_{-\pi}^{\pi}D\_n(x),dx=1\) (\(\because D\_n=\sum_{k=-n}^{n}e^{ikx},\ \frac1{2\pi}\int e^{ikx}dx=0\ (k\ne0),\ 1\ (k=0)\);\ 교재 Ch.8 Eq. (61))
\[\therefore \frac1{2\pi}\int\_{-\pi}^{\pi}K\_N(x),dx = \frac1{N+1}\sum_{n=0}^{N}1 = 1\]Step 4: Tail bound
만약 \(0<\delta\le \vert x \vert \le\pi\)이면,
\[1-\cos x\ge1-\cos\delta\] \[1-\cos(N+1)x\le2\] \[\therefore K\_N(x) = \frac1{N+1} \cdot \frac{1-\cos(N+1)x}{1-\cos x} \le \frac1{N+1} \cdot \frac2{1-\cos\delta}\]Step 5: Integral representation of ($\sigma_N$)
Fourier partial sum으로부터,
\(s\_n(f;x)=\frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t)D\_n(t),dt\) (\(\because\) 교재 Ch.8 Eq. (78))
\[\sigma\_N(f;x) = \frac1{N+1}\sum_{n=0}^{N}s\_n(f;x)\] \[= \frac1{N+1}\sum_{n=0}^{N} \frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t)D\_n(t),dt\] \[= \frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t) \left(\frac1{N+1}\sum_{n=0}^{N}D\_n(t)\right)dt\] \[= \frac1{2\pi}\int\_{-\pi}^{\pi}f(x-t)K\_N(t),dt\]Step 6: Fejér theorem
\[\frac1{2\pi}\int\_{-\pi}^{\pi}K\_N(t),dt=1\] \[\therefore f(x) = \frac1{2\pi}\int\_{-\pi}^{\pi}f(x)K\_N(t),dt\] \[\sigma\_N(f;x)-f(x) = \frac1{2\pi}\int\_{-\pi}^{\pi} {f(x-t)-f(x)}K\_N(t),dt\]\(f\)가 연속이고 \(2\pi\)-periodic이므로,
\(f\ \text{uniformly continuous on }[-\pi,\pi]\) (\(\because\) continuous on compact set; 교재 Ch.4 Thm. 4.19 style result)
주어진 \(\varepsilon>0\)에 대해,
\[\exists\delta>0: \vert t \vert <\delta \Rightarrow \vert f(x-t)-f(x) \vert <\varepsilon \quad(\forall x)\]Let
\[M:=\sup\_x\vert f(x) \vert <\infty\] \[\vert \sigma\_N(f;x)-f(x) \vert \le \frac1{2\pi}\int_{\vert t \vert <\delta} \vert f(x-t)-f(x) \vert K\_N(t)dt + \frac1{2\pi}\int_{\delta\le \vert t \vert \le\pi} \vert f(x-t)-f(x) \vert K\_N(t)dt\]첫 번째 항:
\[\le \varepsilon\cdot \frac1{2\pi}\int\_{-\pi}^{\pi}K\_N(t)dt = \varepsilon\]두 번째 항:
\[\vert f(x-t)-f(x) \vert \le2M\] \[K\_N(t)\le \frac1{N+1}\cdot\frac2{1-\cos\delta} \quad(\delta\le \vert t \vert \le\pi)\] \[\therefore \frac1{2\pi}\int_{\delta\le \vert t \vert \le\pi} \vert f(x-t)-f(x) \vert K\_N(t)dt \le \frac1{2\pi}(2\pi)(2M) \frac1{N+1}\frac2{1-\cos\delta}\] \[= \frac{4M}{(N+1)(1-\cos\delta)} \to0\] \[\therefore \sup\_x\vert \sigma\_N(f;x)-f(x) \vert \le \varepsilon+o(1)\] \[\because \varepsilon>0\ \text{arbitrary}\] \[\therefore \sigma\_N(f;x)\to f(x)\ \text{uniformly on }[-\pi,\pi]\]
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