第一届熊赛试题解答

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\textup{2010} Mathematics Subject Classification}
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\begin{document}
\date{}

\title
{This is the Title}

\author{Name}
\address{School of Mathematical Sciences\\
University of Science and Technology of China\\
Hefei, 230026\\ P.R. China\\}
\email{email}

 

 

\begin{abstract}
Briefly describe the topic.
\end{abstract}

\subjclass[2010]{}

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\section{Introduction}
This section is the introduction to the paper.\\\\

1. 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
$$k|S_n|\le (2n+k)\varepsilon+C(\frac{k}{n+1}+\frac{k-1}{n+2}+\cdots+\frac{1}{n+k})\le (2n+k)\varepsilon+C\frac{k^2}{n},$$
or equivalent,
$$|S_n|\le (\frac{2n}{k}+1)\varepsilon+C\frac{k}{n}.$$
Take $k=[\sqrt{\varepsilon}n]>\frac{1}{2}\sqrt{\varepsilon}n$, then we have
$$|S_n|\le (\frac{4}{\sqrt{\varepsilon}}+1)\varepsilon+C\sqrt{\varepsilon}=(4+\sqrt{\varepsilon}+C)\sqrt{\varepsilon}.$$

\end{proof}

2. 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}

3. 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}

4. 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}

5. 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}

6. 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}

7. 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}

8. 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
$$|x_n-x_{n-1}|\le\frac{r_0}{2^n}, f(x_n,\frac{r_0}{2^n})=\frac{1}{2}.$$
Let $x=\lim\limits_{n\to\infty} x_n$, then $|x-x_n|=|\sum\limits_{k=n+1}^{\infty} (x_k-x_{k-1})|\le \sum\limits_{k=n+1}^{\infty} |x_k-x_{k-1}|\le\frac{r_0}{2^n}$. 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
$$[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}]\subset[x-\frac{r_0}{2^N},x+\frac{r_0}{2^N}]\subset[x-r,x+r],$$
and
$$f(x,r)=\frac{m(A\cap[x-r,x+r])}{2r}\ge\frac{m(A\cap[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}])}{2\frac{r_0}{2^{N-1}}}$$
$$=\frac{1}{4}f(x_{N+1},\frac{r_0}{2^{N+1}})=\frac{1}{8}.$$
On the other hand, we have
$$m(A\cap[x-r,x+r])\le m(A\cap[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}])$$
$$+m([x-r,x+r]-[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}])$$
$$=\frac{r_0}{2^{N+1}}+2r-\frac{2r_0}{2^{N+1}}=2r-\frac{r_0}{2^{N+1}}\le 2r-\frac{1}{4}r=\frac{7}{4}r,$$
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}

9. 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
$\int_{a^c} ||f_n|^p-|f_n-f|^p|\le p\int_{a^c}\max\{|f_n|,|f_n-f|\}^{p-1}|f|
\le p(3M)^{\frac{1}{p'}}(\int_{a^c} |f|^p)^{\frac{1}{p}}\\
\le p(3M)^{\frac{1}{p'}}(2\varepsilon)^{\frac{1}{p}},$ by H\"{o}lder inequality. Thus we obtain
$$\mathop{\overline{\lim}}\limits_{n\to\infty}|\int |f_n|^p-|f_n-f|^p-\int |f|^p|\le (1+p(3M)^{\frac{1}{p'}})(2\varepsilon)^{\frac{1}{p}}.$$

\end{proof}

10. 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 $L_N(t)=\min(N,\frac{\pi^2}{Nt^2})$. 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}

11. 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
$$\frac{p_n}{q_n}=[b_1,b_2,\cdots,b_{n+1}]=b_1+\frac{1}{b_2+\frac{1}{\cdots+\frac{1}{b_{n+1}}}},$$
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 $|[b_1,b_2,\cdots]-\frac{p_n}{q_n}|<\frac{1}{q_nq_{n+1}}$, we can get
$$|a_n|=|a_1|\cdot|p_{n-2}-q_{n-2}[b_1,b_2,\cdots]|\le\frac{|a_1|}{q_{n-1}}\to 0.$$
Since
$$\frac{e-1}{e+1}=[0,2,6,10,\cdots]=\frac{1}{2+\frac{1}{6+\frac{1}{10+\frac{1}{\cdots}}}},$$
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}

12. 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.\\
(c) $\lim\limits_{n\to\infty} \cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)=\frac{1}{4}$.
\begin{proof}
For $n\ge 2$, we have
$$(x_{n+1}^2-2\ln S_n)-(x_n^2-2\ln S_{n-1})=(x_n+\frac{1}{S_n})^2-x_n^2+2\ln\frac{S_{n-1}}{S_n}$$
$$=\frac{2x_n}{S_n}+\frac{1}{S_n^2}+2\ln(1-\frac{x_n}{S_n})<\frac{2x_n}{S_n}+\frac{1}{S_n^2}-2(\frac{x_n}{S_n}+\frac{x_n^2}{2S_n^2})=\frac{1-x_n^2}{2S_n^2}\le 0.$$
Similarly,
$$(x_{n+1}^2-2\ln S_{n+1})-(x_n^2-2\ln S_n)=(x_n+\frac{1}{S_n})^2-x_n^2-2\ln\frac{S_{n+1}}{S_n}$$
$$=\frac{2x_n}{S_n}+\frac{1}{S_n^2}-2\ln(1+\frac{x_{n+1}}{S_n})>\frac{2x_n}{S_n}+\frac{1}{S_n^2}-2(\frac{x_{n+1}}{S_n}-\frac{x_{n+1}^2}{2S_n^2}+\frac{x_{n+1}^3}{3S_n^3})$$
$$=\frac{x_{n+1}^2-1}{S_n^2}-\frac{2}{3}\frac{x_{n+1}^3}{S_n^3}=\frac{x_{n+1}^2}{S_n^2}(1-\frac{1}{x_{n+1}^2}-\frac{2x_{n+1}}{3S_n}).$$
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
$$\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)=(\cfrac{x_n}{\sqrt{2\ln n}}+1)^{-1}\cfrac{\ln n}{\ln \ln n}\left(\frac{x_n^2}{2\ln n}-1\right)$$
$$\to \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.$$
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}

13. 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
$$|\varphi_n(x)|\le\cfrac{\varepsilon_0}{3},|\phi_n(x)|\le\cfrac{\varepsilon_0}{3}.$$
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
$$|\varphi(x-x_n)|=|f(x)-f(x_n)|\le\cfrac{\varepsilon_0}{3}, |\phi(x-y_n)|=|f(x)-f(y_n)|\le\cfrac{\varepsilon_0}{3},$$
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
$$h'(x)=f(x+1)-f(x)\to 0$$
as $x\to\infty$. Since $f$ is uniformly continous, there is positive $M$ such $\forall x,y\ge 0$ with $|x-y|\le 1$, we have $|f(x)-f(y)|\le M$. Since $f(x)-f(x+t)\to 0$ as $x\to\infty$ and $|f(x)-f(x+t)|\le M$ for $\forall t\in [0,1]$, by DCT, we have
$$g(x)=\int_{0}^{1} f(x)-f(x+t)\to 0 \text{ as } n\to\infty.$$

\end{proof}

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%\bibitem{AF12}%
%Antunes, P., Freitas, P.: Optimal spectral rectangles and lattice ellipses. \emph{Proc. Royal Soc. London Ser. A.} \textbf{469} (2012), 20120492.

 

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\end{document}