第一届熊赛分析与方程部分试题

Eufisky posted @ 2018年7月17日 18:14 in 数学分析 with tags 熊赛逆逆 , 766 阅读

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%\title{中国科学技术大学\\

%2015年硕士学位研究生入学考试试题}

%\author{(线性代数与解析几何)}

%\date{}

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\lfoot{科目名称：分析与方程}

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{\erhao \CJKfamily{kd}{第一届Xionger网络数学竞赛} }\\

\vspace{0.3cm}

{\sanhao \textbf{分析与方程部分试题解答} }\\

\vspace{0.2cm}

{\sihao 整理编辑: 马明辉 \qquad

2018年6月8日}

%{\sanhao \textbf{解答：Eufisky (Xiongge)}}\\

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\item Suppose $S_n=a_1+\cdots+a_n$, and $T_n=S_1+\cdots+S_n$. If $\lim\limits_{n\to\infty} \frac{1}{n}T_n=s$ and $na_n$ is bounded, then $\lim\limits_{n\to\infty} S_n=s.$

\begin{proof}

We may assume that $\lim\limits_{n\to\infty} \frac{1}{n}T_n=0$, or we can replace $a_1$ by $a_1-s$. For $\forall \varepsilon>0$, there is a large number $N$, $\forall n\ge N$, we have $|T_n|\le \varepsilon n.$ Since

$$T_{n+k}-T_n=S_{n+1}+\cdots+S_{n+k}$$

$$=kS_n+(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}),$$

we have

kS_n=T_{n+k}-T_n-(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}).

Suppose $|na_n|\le C$, then

$$k|S_n|\le (2n+k)\varepsilon+C\left(\frac{k}{n+1}+\frac{k-1}{n+2}+\cdots+\frac{1}{n+k}\right)\le (2n+k)\varepsilon+C\frac{k^2}{n},$$

or equivalent,

$$|S_n|\le \left(\frac{2n}{k}+1\right)\varepsilon+C\frac{k}{n}.$$

Take $k=[\sqrt{\varepsilon}n]>\frac{1}{2}\sqrt{\varepsilon}n$, then we have

$$|S_n|\le \left(\frac{4}{\sqrt{\varepsilon}}+1\right)\varepsilon+C\sqrt{\varepsilon}=\left(4+\sqrt{\varepsilon}+C\right)\sqrt{\varepsilon}.$$

\end{proof}

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\item Suppose $f$ is a real value function on $\mathbb{R}$, and $f(x+y)=f(x)+f(y)$ for $\forall x, y\in \mathbb{R}$. If the set $\{(x,f(x)): x\in\mathbb{R}\}$ is not dense in $\mathbb{R}^2$, then $f$ is continuous.

\begin{proof}

In fact, we have $f(x)\equiv f(1)x$. It is not hard to see $f(rx)=rf(x)$ for $x\in\mathbb{R}, r\in \mathbb{Q}$. We denote $c=f(1)$, then $f(r)=cr$ for $r\in \mathbb{Q}$. If $f(x)\equiv cx$ is false, then there exists a number $x_0\in \mathbb{R}-\mathbb{Q}$, s.t. $y_0=f(x_0)\ne cx_0$. Then we have

$$f(x_0s+r)=y_0s+cr=c(x_0s+r)+(y_0-cx_0)s, \forall s, t\in \mathbb{Q}.$$

Fix $y\in\mathbb{R}$. For $\forall \varepsilon>0$, there is a number $s\in\mathbb{Q}$, s.t. $|s(y_0-cx_0)-y|<\varepsilon$, then $|s|< \frac{|y|+\varepsilon}{|y_0-cx_0|}\le \frac{|y|+1}{|y_0-cx_0|}=C$. Take $r\in\mathbb{Q}$ s.t. $|x_0+r|<\varepsilon$, we have

$$f(x_0+r)=c(x_0+r)+(y_0-cx_0),$$

and

$$f(s(x_0+r))=sc(x_0+r)+s(y_0-cx_0),$$

then $|s(x_0+r)|\le C\varepsilon$ and

$$|f(s(x_0+r))-y|\le |sc(x_0+r)|+|s(y_0-cx_0)-y|< (Cc+1)\varepsilon.$$

Fix $(x,y)\in\mathbb{R}^2$. For $\forall \varepsilon>0$, there exists a number $a\in \mathbb{R}$ s.t. $|a|<\varepsilon$, and $|f(a)-(y-cx)|< \varepsilon$. Take $r\in \mathbb{Q}$ s.t. $|x-r|<\varepsilon$, we have

$$f(a+r)=f(a)+cr=f(a)+cx+c(r-x)$$

then $|(a+r)-x|<2\varepsilon$ and

$$|f(a+r)-y|=|f(a)-(y-cx)|+|c(r-x)|< (1+c)\varepsilon.$$

It is a contradiction.

\end{proof}

\item Suppose $\Omega\subset \mathbb{C}$ is a domain, and $u_n=Re f_n$, where $f_n\in H(\Omega)$. If $\{u_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$, and $\{f_n(z_0)\}$ is convergence for some $z_0\in \Omega$. Prove that $\{f_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$.

\begin{proof}

Since the compact set has a finite open covering property, we only need to consider the case $\Omega=\mathbb{D}$. We can obtain the conclusion by the Borel-Carath$\acute{\text{e}}$odory lemma: Suppose $f\in H(\mathbb{D})$. Let $M(r)=\max\limits_{|z|=r} |f(z)|$, $A(r)=\max\limits_{|z|=r} Re f(z)$, then for $0<r<R<1$, we have

$$M(r)\le \frac{2r}{R-r}A(R)+\frac{R+r}{R-r}|f(0)|.$$

\end{proof}

\item Suppose $\Omega\subset \mathbb{C}$ is a convex domain, and $f\in H(\Omega)$. If $Re f'(z)\ge 0$ for $\forall z\in\Omega$ and $f$ is not constant function, then $f$ is injective.

\begin{proof}

Consider $e^{-f'(z)}\in H(\Omega)$, then $|e^{-f'(z)}|=e^{-Re f'(z)}\le 1$. If $Re f'(z_0)=0$ for some $z_0\in\Omega$, then $e^{-f'(z)}=const$, or $f'(z)=const$ by the maximum principle. That is to say $f(z)=cz$ with $c\ne 0$, then $f$ is injective. If $Re f'(z)> 0$ for $\forall z\in\Omega$, then for $\forall z_1, z_2\in\Omega$ with $z_1\ne z_2$, we have

$$f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(z)dz=(z_2-z_1)\int_{0}^{1} f'(z_1+t(z_2-z_1))dt,$$

then

$$Re\frac{f(z_2)-f(z_1)}{z_2-z_1}=\int_{0}^{1} Re f'(z_1+t(z_2-z_1))dt>0.$$

\end{proof}

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\item Prove the linear span of $t^ne^{-t}, n=0,1,2,\cdots$ is dense in $L^2(0,\infty)$.

\begin{proof}

Let $M$ be the closed linear span of $t^ne^{-t}, n=0,1,2,\cdots$. Take any $\varphi\in M^{\bot}$, we have

$$\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots.$$

Let $z$ be a complex with $\Im\ z>-1$, and

$$f(z)=\int_{0}^{\infty} e^{izt}e^{-t}\varphi(t)dt.$$

Since $|\frac{e^z-1}{z}|\le C\max\{1,e^{|z|}\}$, we have

$$f'(z)=\lim\limits_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim\limits_{h\to 0} \int_{0}^{\infty} \frac{e^{iht}-1}{h}e^{izt}e^{-t}\varphi(t)dt$$

$$=\int_{0}^{\infty} ite^{izt}e^{-t}\varphi(t)dt$$

by the dominated convergence theorem. That means $f$ is analytic. Similarly, we have

$$f^{(n)}(z)=\int_{0}^{\infty} i^nt^ne^{izt}e^{-t}\varphi(t)dt.$$

Since $f^{(n)}(0)=i^n\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots$, we have $f(z)\equiv 0$, that means $e^{izt}e^{-t}\in M$. According to the Weierstrass approximation theorem, every continuous periodic function $h(t)$ is the uniform limit of trigonometric polynomials, we can get $h(t)e^{-t}\in M$. Let $g(t)$ be a continuous function with compact support, and $g_1(t)=g(t)e^t$. Denote by $h(t)$ a $T$ periodic function such that

$$h(t)\equiv g_1(t), t\in [0,T],$$

where $T$ is large enough so that the support of $g_1(t)$ is contained in the interval $[0,T]$. Then

$$|g_1(t)-h(t)|\le ||g_1||_{L^{\infty}}\chi_{(T,\infty)}(t),$$

so that

$$|g(t)-h(t)e^{-t}|\le ||g_1||_{L^{\infty}}e^{-t}\chi_{(T,\infty)}(t).$$

Let $T\to\infty$, we can get $g(t)\in M$. Since the set of all continuous functions with compact support is dense in $L^2(0,\infty)$, we have $M=L^2(0,\infty)$.

\end{proof}

\item Let $A$ is a unital commutative Banach algebra that is generated by $\{1,x\}$ for some $x\in A$. Then the complement set of $\sigma(x)$ is connected.

\begin{proof}

Let us decompose $\sigma(x)^c$ into its connected components, obtaining an unbounded component $\Omega_{\infty}$ together with a sequence of holes $\Omega_1, \Omega_2, \cdots,$

$$\sigma(x)^c=\Omega_{\infty}\cup\Omega_1\cup\Omega_2\cup\cdots.$$

Let $\Omega=\Omega_1\cup\Omega_2\cup\cdots$. If $\sigma(x)^c$ is not connected, the $\Omega\ne \emptyset$. Suppose $\lambda\in\Omega$, then for arbitrary polynomial $p(z)$, since $p(z)$ is analytic, we have

$$|p(\lambda)|\le \max_{z\in \sigma(x)} |p(z)|=\max\limits_{\omega\in Sp(A)} |\omega(p(x))|\le ||p(x)||$$

by the maximum principle and Gelfand theorem. If we defind

$$\omega: p(x)\mapsto p(\lambda),$$

then $\omega$ is bounded on $\{p(x)\}$. Since $\{p(x)\}$ is dense in $A$, $\omega$ have unique extension on $A$, and $\omega(xy)=\omega(x)\omega(y)$, that means $\omega\in Sp(A)$. Then $\lambda=\omega(x)\in\sigma(x)$, it is a contradiction.

\end{proof}

\item Suppose $X$ is a compact Hausdorff space. $\Omega$ is a family of colsed connected subset of $X$, and $\Omega$ is totally order with respect to inclusion relation. Then $Y=\cap\{A: A\in\Omega\}$ is connected.

\begin{proof}

If $Y$ is not connected, then are open set $B$ and $C$, with $B\cap C=\emptyset$, $B\cap Y\ne \emptyset$ and $C\cap Y\ne \emptyset$. Consider the set $Y_1=\cap\{A-(B\cup C): A\in\Omega\}$, then $Y_1=Y-(B\cup C)=\emptyset$. Since $A$ is connected, if $A-(B\cup C)=\emptyset$, or $A\subset B\cup C$, then $A\subset B$, or $A\subset C$, it is impossible. Thus $A-(B\cup C)\ne\emptyset$. Since $A-(B\cup C)$ is compact, and finite intersection is not empty, then $Y_1\ne\emptyset$. It is a contradiction.

\end{proof}

\item Suppose the measurable set $A\subset \mathbb{R}$ with $0<m(A)<\infty$. Let $f(x,r)=m(A\cap[x-r,x+r])/2r$, then there exists $x\in\mathbb{R}$ s.t.

$$0<\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)<1.$$

\begin{proof}

Since $0<m(A)<\infty$, there are interval $I_1, I_2$ with $|I_1|=|l_2|=2r_0$ s.t. $m(A\cap I_1)>\frac{1}{2}|I_1|$, $m(A\cap I_2)<\frac{1}{2}|I_2|$.

Since $f(x,r)$ is continuous about $x$, there exists $x_0$ with $f(x_0,r_0)=\frac{1}{2}$. Since we have

$$m(A\cap [x_0-r_0,x_0+r_0])=m(A\cap [x_0-r_0, x_0])+m(A\cap [x_0, x_0+r_0])=r_0,$$

there exists $x_1\in [x_0-\frac{r_0}{2},x_0+\frac{r_0}{2}]$ with $f(x_1,\frac{r_0}{2})=\frac{1}{2}$. Or equivalently, there exists $x_1\in\mathbb{R}$ with

$$|x_1-x_0|\le\frac{r_0}{2}, f(x_1,\frac{r_0}{2})=\frac{1}{2}.$$

Similarly, there exists $x_n\in\mathbb{R}$ with

$$|x_n-x_{n-1}|\le\frac{r_0}{2^n},\quad f\left(x_n,\frac{r_0}{2^n}\right)=\frac{1}{2}.$$

Let $x=\lim\limits_{n\to\infty} x_n$, then $|x-x_n|=\left|\sum\limits_{k=n+1}^{\infty} (x_k-x_{k-1})\right|\le \sum\limits_{k=n+1}^{\infty} |x_k-x_{k-1}|\le\frac{r_0}{2^n}$. For any $r<r_0$, there exists unique $N\ge 1$ s.t. $\frac{r_0}{2^N}\le r<\frac{r_0}{2^{N-1}}$. Then we have

$$\left[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}\right]\subset\left[x-\frac{r_0}{2^N},x+\frac{r_0}{2^N}\right]\subset[x-r,x+r],$$

and

\begin{align*}

f(x,r)&=\frac{m\left( A\cap \left[ x-r,x+r \right] \right)}{2r}\ge \frac{m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)}{2\times\frac{r_0}{2^{N-1}}}\\

&=\frac{1}{4}f\left( x_{N+1},\frac{r_0}{2^{N+1}} \right) =\frac{1}{8}.

\end{align*}

On the other hand, we have

\begin{align*}

m\left( A\cap \left[ x-r,x+r \right] \right) &\le m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)

&+m\left( \left[ x-r,x+r \right] -\left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)

&=\frac{r_0}{2^{N+1}}+2r-\frac{2r_0}{2^{N+1}}=2r-\frac{r_0}{2^{N+1}}\le 2r-\frac{1}{4}r=\frac{7}{4}r,

\end{align*}

then

$$f(x,r)=\frac{m(A\cap[x-r,x+r])}{2r}\le\frac{7}{8}.$$

That is to say

$$\frac{1}{8}\le\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)\le\frac{7}{8}.$$

\end{proof}

\item Suppose $\{f_n\}_{n=1}^{\infty}$ is a bounded sequence in $L^p$ with $1\le p<\infty$. If $f_n\to f$ a.e., then $f\in L^p$ and

$$\lim\limits_{n\to\infty}\int |f_n|^p-|f_n-f|^p=\int |f|^p.$$

\begin{proof}

We denote $M=\mathop{\sup}_{n\ge 1}\int |f_n|^p<\infty$.Since $f_n\to f$ a.e., we have $|f_n|^p\to|f|^p$ a.e. and by Fatou Lemma

$$\int |f|^p\le\mathop{\underline{\lim}}\limits_{n\to\infty}\int |f_n|^p\le M<\infty,$$

that is to say $f\in L^p$. For $\forall a,b\ge 0$, we have

$$|a^p-b^p|=p\xi^{p-1}|a-b|\le p\max\{a,b\}^{p-1}|a-b|$$

by Lagrange Mean Value Theorem, where $\xi$ is a real number between $a$ and $b$. Then we obtain

$$||f_n|^p-|f_n-f|^p|\le p\max\{|f_n|,|f_n-f|\}^{p-1}|f|.$$

Fixed $\varepsilon>0$. Suppose $A$ is a measurable set with $m(A)<\infty$, and the follow inequality holds

$$\int_{A^c} |f|^p\le \varepsilon.$$

There is a $\delta>0$ such that

$$\int_B |f|^p<\varepsilon \text{ whenever } m(B)<\delta$$

by absolute continuity. Since $f_n\to f$ a.e. on $A$ and $m(A)<\infty$, we can find a measurable subset $a\subset A$ such $m(A\setminus a)<\delta$ and $f_n\to f$ uniformly on $a$ by Egorov Theorem. Then we have

$$\int_a |f_n|^p-|f_n-f|^p\to \int_a |f|^p,$$

as $n\to\infty$, and

$$\int_{a^c} |f|^p=\int_{A\setminus a} |f|^p+\int_{A^c} |f|^p< 2\varepsilon.$$

Since the function $\max\{|f_n|,|f_n-f|\}^{p-1}\in L^{p'}$, and $$||\max\{|f_n|,|f_n-f|\}^{p-1}||_{p'}=||\max\{|f_n|,|f_n-f|\}||_p^{p-1}\le (3M^p)^{\frac{1}{p'}}.$$

We have

\begin{align*}

\int_{a^c} ||f_n|^p-|f_n-f|^p|&\le p\int_{a^c}\max\{|f_n|,|f_n-f|\}^{p-1}|f|\\

&\le p(3M)^{\frac{1}{p'}}(\int_{a^c} |f|^p)^{\frac{1}{p}}

\le p(3M)^{\frac{1}{p'}}(2\varepsilon)^{\frac{1}{p}},

\end{align*}

by H\"{o}lder inequality. Thus we obtain

$$\mathop{\overline{\lim}}\limits_{n\to\infty}\left|\int |f_n|^p-|f_n-f|^p-\int |f|^p\right|\le (1+p(3M)^{\frac{1}{p'}})(2\varepsilon)^{\frac{1}{p}}.$$

\end{proof}

\item Suppose $D_n(t)$ are the Dirichlet kernels, and $F_N(t)$ is the $N$-th Fej$\acute{\text{e}}$r kernel given by

$$F_N(t)=\frac{D_0(t)+\cdots+D_{N-1}(t)}{N}.$$

Let $L_N(t)=\min\left(N,\frac{\pi^2}{Nt^2}\right)$. Prove

$$F_N(t)=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}\le L_N(t)$$

and $\int_{\mathbb{T}} L_N(t)dt\le 4\pi$. If $f\in L^1(\mathbb{T})$ and the $N$-th Ces$\grave{\text{a}}$ro mean of Fourier series is

$$\sigma_N(f)(x)=\frac{S_0(f)(x)+\cdots+S_{N-1}(f)(x)}{N},$$

then $\sigma_N(f)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.

\begin{proof}

Since $D_N(t)=\sum\limits_{n=-N}^{N}e^{int}=\frac{\sin(N+\frac{1}{2})t}{\sin\frac{t}{2}}$, we have

$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}.$$

Since $|D_N(t)|\le 2N+1$, we can get $F_N(t)\le N$. For $0<x<\frac{\pi}{2}$, we have $\sin x\ge\frac{2}{\pi}x$, then

$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}\le\frac{1}{N}\frac{1}{\sin^2\frac{t}{2}}\le\frac{\pi^2}{Nt^2}.$$

That mens $F_N(t)\le L_N(t)$. And

$$\int_{\mathbb{T}} L_N(t)dt=2\int_0^{\pi}L_N(t)dt=2\int_0^{\frac{\pi}{N}}Ndt+2\int_{\frac{\pi}{N}}^{\pi}\frac{\pi^2}{Nt^2}dt=4\pi-\frac{2\pi}{N} \le 4\pi.$$

Since $\int_{\mathbb{T}}F_N(t)=1$ and $F_N(t)\le L_N(t)$, we can get $\{F_N(t)\}$ is an approximation to the identity, then $\sigma_N(f)(x)=(f*F_N)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.

\end{proof}

\item Suppose the sequence $\{a_n\}$ satisfying $a_{n+1}=(4n-2)a_n+a_{n-1}$. Prove that $\{a_n\}$ is convergence if and only if

$$(e-1)a_0+(e+1)a_1=0.$$

\begin{proof}

If $\{a_n\}$ is convengence and not vanishing, then it is obvious that $a_na_{n+1}<0$. We can assume $a_0>0$, $a_1<0$, then $a_{2n}>0$, $a_{2n+1}<0$. Let $b_n=4n-2$ and

$$a_n=p_{n-2}a_1+q_{n-2}a_0, n\ge 2.$$

Since $a_2=b_1a_1+a_0$, and $a_3=(1+b_1b_2)a_1+b_2a_0$, we can get

$$p_0=b_1, p_1=1+b_1b_2, q_0=1, q_1=b_2.$$

Since $a_{n+2}=b_{n+1}a_{n+1}+a_n=b_{n+1}(p_{n-1}a_1+q_{n-1}a_0)+(p_{n-2}a_1+q_{n-2}a_0)=(b_{n+1}p_{n-1}+p_{n-2})a_1+(b_{n+1}q_{n-1}+q_{n-2})a_0$, we can get

$$p_n=b_{n+1}p_{n-1}+p_{n-2}, q_n=b_{n+1}q_{n-1}+q_{n-2}.$$

That is to say

$$\frac{p_n}{q_n}=\left[b_1,b_2,\cdots,b_{n+1}\right]=b_1+\cfrac{1}{b_2+\cfrac{1}{\cdots+\cfrac{1}{b_{n+1}}}},$$

and we have $\frac{p_n}{q_n}\to [b_1,b_2,\cdots]$ as $n\to\infty$. Since

$$a_{2n+1}=p_{2n-1}a_1+q_{2n-1}a_0<0, a_{2n+2}=p_{2n}a_1+q_{2n}a_0>0,$$

we have

$$\frac{p_{2n}}{q_{2n}}(-a_1)<a_0<\frac{p_{2n-1}}{q_{2n-1}}(-a_1),$$

Let $n\to\infty$, we can get

$$a_0+[b_1,b_2,\cdots]a_1=0.$$

On the contrary, if $a_0+[b_1,b_2,\cdots]a_1=0$, since $\left|[b_1,b_2,\cdots]-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}$, we can get

$$|a_n|=|a_1|\cdot\left|p_{n-2}-q_{n-2}[b_1,b_2,\cdots]\right|\le\frac{|a_1|}{q_{n-1}}\to 0.$$

Since

$$\frac{e-1}{e+1}=[0,2,6,10,\cdots]=\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{\cdots}}}},$$

we can get $[b_1,b_2,\cdots]=\frac{e+1}{e-1}$, then $a_0+[b_1,b_2,\cdots]a_1=0$ is equivalent to

$$(e-1)a_0+(e+1)a_1=0.$$

\end{proof}

\item Suppose $\{x_n\}$ satisfying $x_1=1$, $x_{n+1}=x_n+\frac{1}{S_n}$, where $S_n=x_1+\cdots+x_n$. Prove that\\

(a) $x_n^2-2\ln S_n$ is increasing and $x_n^2-2\ln S_{n-1}$ is decreasing for $n\ge 2$.\\

(b) $x_n^2-2\ln n-\ln\ln n$ is convengence.\\

\begin{proof}

For $n\ge 2$, we have

\begin{align*}

&\left( x_{n+1}^{2}-\text{2}\ln S_n \right) -\left( x_{n}^{2}-\text{2}\ln S_{n-1} \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}+\text{2}\ln \frac{S_{n-1}}{S_n}

&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}+\text{2}\ln \left( 1-\frac{x_n}{S_n} \right) <\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_n}{S_n}+\frac{x_{n}^{2}}{2S_{n}^{2}} \right) =\frac{1-x_{n}^{2}}{2S_{n}^{2}}\le 0.

\end{align*}

Similarly,

\begin{align*}

&\left( x_{n+1}^{2}-\text{2}\ln S_{n+1} \right) -\left( x_{n}^{2}-\text{2}\ln S_n \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}-\text{2}\ln \frac{S_{n+1}}{S_n}

&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-\text{2}\ln \left( 1+\frac{x_{n+1}}{S_n} \right) >\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_{n+1}}{S_n}-\frac{x_{n+1}^{2}}{2S_{n}^{2}}+\frac{x_{n+1}^{3}}{3S_{n}^{3}} \right)

&=\frac{x_{n+1}^{2}-1}{S_{n}^{2}}-\frac{2}{3}\frac{x_{n+1}^{3}}{S_{n}^{3}}=\frac{x_{n+1}^{2}}{S_{n}^{2}}\left( 1-\frac{1}{x_{n+1}^{2}}-\frac{2x_{n+1}}{3S_n} \right).

\end{align*}

Since $x_1=1, x_2=2$, and $S_1=1, S_2=3$, we have

$$x_{n+1}-S_n=x_n+\frac{1}{S_n}-S_n=\frac{1}{S_n}-S_{n-1}\le \frac{1}{S_2}-S_1=-\frac{1}{2}<0,$$

that means $\frac{2x_{n+1}}{3S_n}\le\frac{2}{3}$, then

$$1-\frac{1}{x_{n+1}^2}-\frac{2x_{n+1}}{3S_n}>\frac{1}{3}-\frac{1}{x_{n+1}^2}>0.$$

That implys (a).\\

Since $x_n^2-2\ln S_n$ is increasing, we can get $x_n^2-2\ln S_n\ge x_2^2-2\ln S_2=4-2\ln 3>1$. Since $x_n\ge 1$, we have $S_n\ge n$, then

$$x_n^2\ge 1+2\ln S_n\ge 1+2\ln n.$$

Since $x_n^2-2\ln S_{n-1}$ is decreasing, we can get $x_n^2-2\ln S_n\le x_{n+1}^2-2\ln S_n\le x_2^2-2\ln S_2=4$, then

$$x_n^2\le 4+2\ln S_n.$$

Since $S_n\ge n$, we can get $x_n\le 1+1+\frac{1}{2}+\cdots+\frac{1}{n-1}\le 2+\ln n$, then $S_n\le nx_n\le n(2+\ln n)$, and

$$x_n^2\le 4+2\ln n+2\ln(2+\ln n).$$

Thus, it is obvious that

$$\frac{x_n}{\sqrt{\ln n}}\to\sqrt{2},$$

and we have

$$\frac{S_n}{n\sqrt{\ln n}}\to \sqrt{2}$$

by Stolz formula. Since $x_n^2-2\ln S_n\le 4$, we can get $x_n^2-2\ln S_n$ is convengence, then

$$x_n^2-2\ln n-\ln\ln n=x_n^2-2\ln S_n+2\ln\frac{S_n}{n\sqrt{\ln n}}$$

is convengence. That implys (b).\\

Since $x_n^2=2\ln n+\ln\ln n+a+o(1)$, we can get

$$\frac{x_n^2}{2\ln n}-1=\frac{\ln \ln n}{2\ln n}+\frac{a+o(1)}{2\ln n},$$

then

\begin{align*}

\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)&=\left(\cfrac{x_n}{\sqrt{2\ln n}}+1\right)^{-1}\cfrac{\ln n}{\ln \ln n}\left(\frac{x_n^2}{2\ln n}-1\right)\\

&\to \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.

\end{align*}

Moreover, we have

$$\lim\limits_{n\to\infty} \ln \ln n\left(\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)-\frac{1}{4}\right)=\frac{a}{4}.$$

\end{proof}

\item Suppose $f\in C[0,\infty)$ and for $\forall a\ge0$, we have

$$\lim\limits_{x\to\infty} f(x+a)-f(x)=0.$$

Then there exist $g\in C[0,\infty)$ and $h\in C^1[0,\infty)$ with $f=g+h$, such that

$$\lim\limits_{x\to\infty} g(x)=0, \text{ } \lim\limits_{x\to\infty} h'(x)=0.$$

\begin{proof}

In fact, $f$ is uniformly continuous. Otherwise, there are two sequences $\{x_n\}_{n=1}^{\infty}$, $\{y_n\}_{n=1}^{\infty}$ and a positive $\varepsilon$, such that

$$x_n, y_n\to\infty, |x_n-y_n|\to 0, |f(x_n)-f(y_n)|\ge\varepsilon_0$$

as $n\to\infty$. Consider the functions

$$\varphi_n(x)=f(x_n+x)-f(x_n)$$

and

$$\phi_n(x)=f(y_n+x)-f(y_n)$$

defined on the interval [0,1]. Then we have

$$\varphi(x), \phi(x)\to 0$$

as $n\to\infty$ due to $\lim\limits_{x\to\infty} f(x+a)-f(x)=0$. For $\forall 0<\varepsilon<\cfrac{1}{2}$, there is a set $A_{\varepsilon}\in[0,1]$ such that $m([0,1]\setminus A_{\varepsilon})<\varepsilon$ and $\varphi_n, \phi_n\to 0$ uniformly on $A_{\varepsilon}$ by Egorov Theorem. Take a integer $N$ such that $\forall n\ge N$ and $\forall x\in A_{\varepsilon}$, we have $|x_n-y_n|<1-2\varepsilon$ and

$$|\varphi_n(x)|\le\cfrac{\varepsilon_0}{3},\qquad |\phi_n(x)|\le\cfrac{\varepsilon_0}{3}.$$

Since $m((x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon}))=m(x_n+A_{\varepsilon})+m(y_n+A_{\varepsilon})-m((x_n+A_{\varepsilon})\cup(y_n+A_{\varepsilon}))\ge 2(1-A_{\varepsilon})-(1+|x_n-y_n|)=1-2\varepsilon-|x_n-y_n|>0$, there is a point $x\in(x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon})$. We have $x-x_n,x-y_n\in A_{\varepsilon}$, and then

$$|\varphi(x-x_n)|=|f(x)-f(x_n)|\le\cfrac{\varepsilon_0}{3}, \quad |\phi(x-y_n)|=|f(x)-f(y_n)|\le\cfrac{\varepsilon_0}{3},$$

thus $|f(x_n)-f(y_n)|\le \cfrac{2}{3}\varepsilon_0$, it is a contradiction.

Let $h(x)=\int_{x}^{x+1} f(t)dt$, and $g(x)=f(x)-h(x)$, then we have

$$h'(x)=f(x+1)-f(x)\to 0$$

$$g(x)=\int_{0}^{1} f(x)-f(x+t)\to 0 \text{ as } n\to\infty.$$

\end{proof}

\end{enumerate}

\end{document}