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date{}
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{This is the Title}
author{Name}
address{School of Mathematical Sciences\
University of Science and Technology of China\
Hefei, 230026\ P.R. China\}
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egin{abstract}
Briefly describe the topic.
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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 $limlimits_{n oinfty} frac{1}{n}T_n=s$ and $na_n$ is bounded, then $limlimits_{n oinfty} S_n=s.$
egin{proof}
We may assume that $limlimits_{n oinfty} frac{1}{n}T_n=0$, or we can replace $a_1$ by $a_1-s$. For $forall varepsilon>0$, there is a large number $N$, $forall nge N$, 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
$$k|S_n|le (2n+k)varepsilon+C(frac{k}{n+1}+frac{k-1}{n+2}+cdots+frac{1}{n+k})le (2n+k)varepsilon+Cfrac{k^2}{n},$$
or equivalent,
$$|S_n|le (frac{2n}{k}+1)varepsilon+Cfrac{k}{n}.$$
Take $k=[sqrt{varepsilon}n]>frac{1}{2}sqrt{varepsilon}n$, then we have
$$|S_n|le (frac{4}{sqrt{varepsilon}}+1)varepsilon+Csqrt{varepsilon}=(4+sqrt{varepsilon}+C)sqrt{varepsilon}.$$
end{proof}
2. Suppose $f$ is a real value function on $mathbb{R}$, and $f(x+y)=f(x)+f(y)$ for $forall x, yin mathbb{R}$. If the set ${(x,f(x)): xinmathbb{R}}$ is not dense in $mathbb{R}^2$, then $f$ is continuous.
egin{proof}
In fact, we have $f(x)equiv f(1)x$. It is not hard to see $f(rx)=rf(x)$ for $xinmathbb{R}, rin mathbb{Q}$. We denote $c=f(1)$, then $f(r)=cr$ for $rin mathbb{Q}$. If $f(x)equiv cx$ is false, then there exists a number $x_0in mathbb{R}-mathbb{Q}$, s.t. $y_0=f(x_0)
e cx_0$. Then we have
$$f(x_0s+r)=y_0s+cr=c(x_0s+r)+(y_0-cx_0)s, forall s, tin mathbb{Q}.$$
Fix $yinmathbb{R}$. For $forall varepsilon>0$, there is a number $sinmathbb{Q}$, 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 $rinmathbb{Q}$ 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 $|s(x_0+r)|le Cvarepsilon$ and
$$|f(s(x_0+r))-y|le |sc(x_0+r)|+|s(y_0-cx_0)-y|< (Cc+1)varepsilon.$$
Fix $(x,y)inmathbb{R}^2$. For $forall varepsilon>0$, there exists a number $ain mathbb{R}$ s.t. $|a|<varepsilon$, and $|f(a)-(y-cx)|< varepsilon$. Take $rin mathbb{Q}$ s.t. $|x-r|<varepsilon$, we have
$$f(a+r)=f(a)+cr=f(a)+cx+c(r-x)$$
then $|(a+r)-x|<2varepsilon$ and
$$|f(a+r)-y|=|f(a)-(y-cx)|+|c(r-x)|< (1+c)varepsilon.$$
It is a contradiction.
end{proof}
3. Suppose $Omegasubset mathbb{C}$ is a domain, and $u_n=Re f_n$, where $f_nin H(Omega)$. If ${u_n}$ is uniform convergence on arbitrary compact subset of $Omega$, and ${f_n(z_0)}$ is convergence for some $z_0in Omega$. Prove that ${f_n}$ is uniform convergence on arbitrary compact subset of $Omega$.
egin{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$acute{ ext{e}}$odory lemma: Suppose $fin H(mathbb{D})$. Let $M(r)=maxlimits_{|z|=r} |f(z)|$, $A(r)=maxlimits_{|z|=r} Re f(z)$, 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 $Omegasubset mathbb{C}$ is a convex domain, and $fin H(Omega)$. If $Re f'(z)ge 0$ for $forall zinOmega$ and $f$ is not constant function, then $f$ is injective.
egin{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 $z_0inOmega$, then $e^{-f'(z)}=const$, or $f'(z)=const$ by the maximum principle. That is to say $f(z)=cz$ with $c
e 0$, then $f$ is injective. If $Re f'(z)> 0$ for $forall zinOmega$, then for $forall z_1, z_2inOmega$ with $z_1
e z_2$, 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
$$Refrac{f(z_2)-f(z_1)}{z_2-z_1}=int_{0}^{1} Re f'(z_1+t(z_2-z_1))dt>0.$$
end{proof}
5. Prove the linear span of $t^ne^{-t}, n=0,1,2,cdots$ is dense in $L^2(0,infty)$.
egin{proof}
Let $M$ be the closed linear span of $t^ne^{-t}, n=0,1,2,cdots$. Take any $varphiin M^{ot}$, 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 $|frac{e^z-1}{z}|le Cmax{1,e^{|z|}}$, we have
$$f'(z)=limlimits_{h o 0}frac{f(z+h)-f(z)}{h}=limlimits_{h o 0} int_{0}^{infty} frac{e^{iht}-1}{h}e^{izt}e^{-t}varphi(t)dt$$
$$=int_{0}^{infty} ite^{izt}e^{-t}varphi(t)dt$$
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 $f^{(n)}(0)=i^nint_{0}^{infty} t^ne^{-t}varphi(t)dt=0, n=0,1,2,cdots$, 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
$$h(t)equiv g_1(t), tin [0,T],$$
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 $T oinfty$, 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 $xin A$. Then the complement set of $sigma(x)$ is connected.
egin{proof}
Let us decompose $sigma(x)^c$ into its connected components, obtaining an unbounded component $Omega_{infty}$ together with a sequence of holes $Omega_1, Omega_2, cdots,$
$$sigma(x)^c=Omega_{infty}cupOmega_1cupOmega_2cupcdots.$$
Let $Omega=Omega_1cupOmega_2cupcdots$. If $sigma(x)^c$ is not connected, the $Omega
e emptyset$. Suppose $lambdainOmega$, then for arbitrary polynomial $p(z)$, since $p(z)$ is analytic, we have
$$|p(lambda)|le max_{zin sigma(x)} |p(z)|=maxlimits_{omegain Sp(A)} |omega(p(x))|le ||p(x)||$$
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 $omegain Sp(A)$. Then $lambda=omega(x)insigma(x)$, 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 $Y=cap{A: AinOmega}$ is connected.
egin{proof}
If $Y$ is not connected, then are open set $B$ and $C$, with $Bcap C=emptyset$, $Bcap Y
e emptyset$ and $Ccap Y
e emptyset$. Consider the set $Y_1=cap{A-(Bcup C): AinOmega}$, then $Y_1=Y-(Bcup C)=emptyset$. Since $A$ is connected, if $A-(Bcup C)=emptyset$, or $Asubset Bcup C$, then $Asubset B$, or $Asubset C$, it is impossible. Thus $A-(Bcup C)
eemptyset$. Since $A-(Bcup C)$ is compact, and finite intersection is not empty, then $Y_1
eemptyset$. It is a contradiction.
end{proof}
8. Suppose the measurable set $Asubset mathbb{R}$ with $0<m(A)<infty$. Let $f(x,r)=m(Acap[x-r,x+r])/2r$, then there exists $xinmathbb{R}$ s.t.
$$0<liminflimits_{r o 0_+}f(x,r)le limsuplimits_{r o 0_+}f(x,r)<1.$$
egin{proof}
Since $0<m(A)<infty$, there are interval $I_1, I_2$ with $|I_1|=|l_2|=2r_0$ s.t. $m(Acap I_1)>frac{1}{2}|I_1|$, $m(Acap I_2)<frac{1}{2}|I_2|$.
Since $f(x,r)$ is continuous about $x$, there exists $x_0$ with $f(x_0,r_0)=frac{1}{2}$. Since we have
$$m(Acap [x_0-r_0,x_0+r_0])=m(Acap [x_0-r_0, x_0])+m(Acap [x_0, x_0+r_0])=r_0,$$
there exists $x_1in [x_0-frac{r_0}{2},x_0+frac{r_0}{2}]$ with $f(x_1,frac{r_0}{2})=frac{1}{2}$. Or equivalently, there exists $x_1inmathbb{R}$ with
$$|x_1-x_0|lefrac{r_0}{2}, f(x_1,frac{r_0}{2})=frac{1}{2}.$$
Similarly, there exists $x_ninmathbb{R}$ with
$$|x_n-x_{n-1}|lefrac{r_0}{2^n}, f(x_n,frac{r_0}{2^n})=frac{1}{2}.$$
Let $x=limlimits_{n oinfty} x_n$, then $|x-x_n|=|sumlimits_{k=n+1}^{infty} (x_k-x_{k-1})|le sumlimits_{k=n+1}^{infty} |x_k-x_{k-1}|lefrac{r_0}{2^n}$. For any $r<r_0$, there exists unique $Nge 1$ 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
$$f(x,r)=frac{m(Acap[x-r,x+r])}{2r}gefrac{m(Acap[x_{N+1}-frac{r_0}{2^{N+1}}, x_{N+1}+frac{r_0}{2^{N+1}}])}{2frac{r_0}{2^{N-1}}}$$
$$=frac{1}{4}f(x_{N+1},frac{r_0}{2^{N+1}})=frac{1}{8}.$$
On the other hand, we have
$$m(Acap[x-r,x+r])le m(Acap[x_{N+1}-frac{r_0}{2^{N+1}}, x_{N+1}+frac{r_0}{2^{N+1}}])$$
$$+m([x-r,x+r]-[x_{N+1}-frac{r_0}{2^{N+1}}, x_{N+1}+frac{r_0}{2^{N+1}}])$$
$$=frac{r_0}{2^{N+1}}+2r-frac{2r_0}{2^{N+1}}=2r-frac{r_0}{2^{N+1}}le 2r-frac{1}{4}r=frac{7}{4}r,$$
then
$$f(x,r)=frac{m(Acap[x-r,x+r])}{2r}lefrac{7}{8}.$$
That is to say
$$frac{1}{8}leliminflimits_{r o 0_+}f(x,r)le limsuplimits_{r o 0_+}f(x,r)lefrac{7}{8}.$$
end{proof}
9. Suppose ${f_n}_{n=1}^{infty}$ is a bounded sequence in $L^p$ with $1le p<infty$. If $f_n o f$ a.e., then $fin L^p$ and
$$limlimits_{n oinfty}int |f_n|^p-|f_n-f|^p=int |f|^p.$$
egin{proof}
We denote $M=mathop{sup}_{nge 1}int |f_n|^p<infty$.Since $f_n o f$ a.e., we have $|f_n|^p o|f|^p$ a.e. and by Fatou Lemma
$$int |f|^plemathop{underline{lim}}limits_{n oinfty}int |f_n|^ple M<infty,$$
that is to say $fin L^p$. For $forall a,bge 0$, we have
$$|a^p-b^p|=pxi^{p-1}|a-b|le pmax{a,b}^{p-1}|a-b|$$
by Lagrange Mean Value Theorem, where $xi$ is a real number between $a$ and $b$. Then we obtain
$$||f_n|^p-|f_n-f|^p|le pmax{|f_n|,|f_n-f|}^{p-1}|f|.$$
Fixed $varepsilon>0$. Suppose $A$ is a measurable set with $m(A)<infty$, and the follow inequality holds
$$int_{A^c} |f|^ple varepsilon.$$
There is a $delta>0$ such that
$$int_B |f|^p<varepsilon ext{ whenever } m(B)<delta$$
by absolute continuity. Since $f_n o f$ a.e. on $A$ and $m(A)<infty$, we can find a measurable subset $asubset A$ such $m(Asetminus a)<delta$ and $f_n o f$ uniformly on $a$ by Egorov Theorem. Then we have
$$int_a |f_n|^p-|f_n-f|^p o int_a |f|^p,$$
as $n oinfty$, and
$$int_{a^c} |f|^p=int_{Asetminus a} |f|^p+int_{A^c} |f|^p< 2varepsilon.$$
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
$int_{a^c} ||f_n|^p-|f_n-f|^p|le pint_{a^c}max{|f_n|,|f_n-f|}^{p-1}|f|
le p(3M)^{frac{1}{p'}}(int_{a^c} |f|^p)^{frac{1}{p}}\
le p(3M)^{frac{1}{p'}}(2varepsilon)^{frac{1}{p}},$ by H"{o}lder inequality. Thus we obtain
$$mathop{overline{lim}}limits_{n oinfty}|int |f_n|^p-|f_n-f|^p-int |f|^p|le (1+p(3M)^{frac{1}{p'}})(2varepsilon)^{frac{1}{p}}.$$
end{proof}
10. Suppose $D_n(t)$ are the Dirichlet kernels, and $F_N(t)$ is the $N$-th Fej$acute{ ext{e}}$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 $int_{mathbb{T}} L_N(t)dtle 4pi$. If $fin L^1(mathbb{T})$ and the $N$-th Ces$grave{ ext{a}}$ro mean of Fourier series is
$$sigma_N(f)(x)=frac{S_0(f)(x)+cdots+S_{N-1}(f)(x)}{N},$$
then $sigma_N(f)(x) o f(x)$ for every $x$ in the Lebesgue set of $f$.
egin{proof}
Since $D_N(t)=sumlimits_{n=-N}^{N}e^{int}=frac{sin(N+frac{1}{2})t}{sinfrac{t}{2}}$, we have
$$F_N(t)=frac{1}{N}frac{sin^2frac{Nt}{2}}{sin^2frac{t}{2}}=frac{1}{N}frac{1-cos Nt}{1-cos t}.$$
Since $|D_N(t)|le 2N+1$, we can get $F_N(t)le N$. For $0<x<frac{pi}{2}$, we have $sin xgefrac{2}{pi}x$, then
$$F_N(t)=frac{1}{N}frac{sin^2frac{Nt}{2}}{sin^2frac{t}{2}}lefrac{1}{N}frac{1}{sin^2frac{t}{2}}lefrac{pi^2}{Nt^2}.$$
That mens $F_N(t)le L_N(t)$. And
$$int_{mathbb{T}} L_N(t)dt=2int_0^{pi}L_N(t)dt=2int_0^{frac{pi}{N}}Ndt+2int_{frac{pi}{N}}^{pi}frac{pi^2}{Nt^2}dt=4pi-frac{2pi}{N} le 4pi.$$
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 $sigma_N(f)(x)=(f*F_N)(x) o f(x)$ 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
$$(e-1)a_0+(e+1)a_1=0.$$
egin{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
$$a_n=p_{n-2}a_1+q_{n-2}a_0, nge 2.$$
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 $frac{p_n}{q_n} o [b_1,b_2,cdots]$ as $n oinfty$. 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 $n oinfty$, 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
$$|a_n|=|a_1|cdot|p_{n-2}-q_{n-2}[b_1,b_2,cdots]|lefrac{|a_1|}{q_{n-1}} o 0.$$
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) $x_n^2-2ln S_n$ is increasing and $x_n^2-2ln S_{n-1}$ is decreasing for $nge 2$.\
(b) $x_n^2-2ln n-lnln n$ is convengence.\
(c) $limlimits_{n oinfty} cfrac{ln n}{ln ln n}left(cfrac{x_n}{sqrt{2ln n}}-1
ight)=frac{1}{4}$.
egin{proof}
For $nge 2$, we have
$$(x_{n+1}^2-2ln S_n)-(x_n^2-2ln S_{n-1})=(x_n+frac{1}{S_n})^2-x_n^2+2lnfrac{S_{n-1}}{S_n}$$
$$=frac{2x_n}{S_n}+frac{1}{S_n^2}+2ln(1-frac{x_n}{S_n})<frac{2x_n}{S_n}+frac{1}{S_n^2}-2(frac{x_n}{S_n}+frac{x_n^2}{2S_n^2})=frac{1-x_n^2}{2S_n^2}le 0.$$
Similarly,
$$(x_{n+1}^2-2ln S_{n+1})-(x_n^2-2ln S_n)=(x_n+frac{1}{S_n})^2-x_n^2-2lnfrac{S_{n+1}}{S_n}$$
$$=frac{2x_n}{S_n}+frac{1}{S_n^2}-2ln(1+frac{x_{n+1}}{S_n})>frac{2x_n}{S_n}+frac{1}{S_n^2}-2(frac{x_{n+1}}{S_n}-frac{x_{n+1}^2}{2S_n^2}+frac{x_{n+1}^3}{3S_n^3})$$
$$=frac{x_{n+1}^2-1}{S_n^2}-frac{2}{3}frac{x_{n+1}^3}{S_n^3}=frac{x_{n+1}^2}{S_n^2}(1-frac{1}{x_{n+1}^2}-frac{2x_{n+1}}{3S_n}).$$
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 $frac{2x_{n+1}}{3S_n}lefrac{2}{3}$, 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 $x_n^2-2ln S_n$ is increasing, we can get $x_n^2-2ln S_nge x_2^2-2ln S_2=4-2ln 3>1$. Since $x_nge 1$, we have $S_nge n$, then
$$x_n^2ge 1+2ln S_nge 1+2ln n.$$
Since $x_n^2-2ln S_{n-1}$ is decreasing, we can get $x_n^2-2ln S_nle x_{n+1}^2-2ln S_nle x_2^2-2ln S_2=4$, then
$$x_n^2le 4+2ln S_n.$$
Since $S_nge n$, we can get $x_nle 1+1+frac{1}{2}+cdots+frac{1}{n-1}le 2+ln n$, then $S_nle nx_nle n(2+ln n)$, and
$$x_n^2le 4+2ln n+2ln(2+ln n).$$
Thus, it is obvious that
$$frac{x_n}{sqrt{ln n}} osqrt{2},$$
and we have
$$frac{S_n}{nsqrt{ln n}} o sqrt{2}$$
by Stolz formula. Since $x_n^2-2ln S_nle 4$, we can get $x_n^2-2ln S_n$ is convengence, then
$$x_n^2-2ln n-lnln n=x_n^2-2ln S_n+2lnfrac{S_n}{nsqrt{ln n}}$$
is convengence. That implys (b).\
Since $x_n^2=2ln n+lnln n+a+o(1)$, we can get
$$frac{x_n^2}{2ln n}-1=frac{ln ln n}{2ln n}+frac{a+o(1)}{2ln n},$$
then
$$cfrac{ln n}{ln ln n}left(cfrac{x_n}{sqrt{2ln n}}-1
ight)=(cfrac{x_n}{sqrt{2ln n}}+1)^{-1}cfrac{ln n}{ln ln n}left(frac{x_n^2}{2ln n}-1
ight)$$
$$ o frac{1}{2}cdotfrac{1}{2}=frac{1}{4}.$$
Moreover, we have
$$limlimits_{n oinfty} ln ln nleft(cfrac{ln n}{ln ln n}left(cfrac{x_n}{sqrt{2ln n}}-1
ight)-frac{1}{4}
ight)=frac{a}{4}.$$
end{proof}
13. Suppose $fin C[0,infty)$ and for $forall age0$, we have
$$limlimits_{x oinfty} f(x+a)-f(x)=0.$$
Then there exist $gin C[0,infty)$ and $hin C^1[0,infty)$ with $f=g+h$, such that
$$limlimits_{x oinfty} g(x)=0, ext{ } limlimits_{x oinfty} h'(x)=0.$$
egin{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
$$x_n, y_n oinfty, |x_n-y_n| o 0, |f(x_n)-f(y_n)|gevarepsilon_0$$
as $n oinfty$. 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
$$varphi(x), phi(x) o 0$$
as $n oinfty$ due to $limlimits_{x oinfty} f(x+a)-f(x)=0$. 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 $varphi_n, phi_n o 0$ uniformly on $A_{varepsilon}$ by Egorov Theorem. Take a integer $N$ such that $forall nge N$ and $forall xin A_{varepsilon}$, we have $|x_n-y_n|<1-2varepsilon$ and
$$|varphi_n(x)|lecfrac{varepsilon_0}{3},|phi_n(x)|lecfrac{varepsilon_0}{3}.$$
Since $m((x_n+A_{varepsilon})cap(y_n+A_{varepsilon}))=m(x_n+A_{varepsilon})+m(y_n+A_{varepsilon})-m((x_n+A_{varepsilon})cup(y_n+A_{varepsilon}))ge 2(1-A_{varepsilon})-(1+|x_n-y_n|)=1-2varepsilon-|x_n-y_n|>0$, there is a point $xin(x_n+A_{varepsilon})cap(y_n+A_{varepsilon})$. We have $x-x_n,x-y_nin A_{varepsilon}$, and then
$$|varphi(x-x_n)|=|f(x)-f(x_n)|lecfrac{varepsilon_0}{3}, |phi(x-y_n)|=|f(x)-f(y_n)|lecfrac{varepsilon_0}{3},$$
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
$$h'(x)=f(x+1)-f(x) o 0$$
as $x oinfty$. Since $f$ is uniformly continous, there is positive $M$ such $forall x,yge 0$ with $|x-y|le 1$, we have $|f(x)-f(y)|le M$. Since $f(x)-f(x+t) o 0$ as $x oinfty$ and $|f(x)-f(x+t)|le M$ for $forall tin [0,1]$, by DCT, we have
$$g(x)=int_{0}^{1} f(x)-f(x+t) o 0 ext{ as } n oinfty.$$
end{proof}
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egin{thebibliography}{99}
%ibitem{AF12}%
%Antunes, P., Freitas, P.: Optimal spectral rectangles and lattice ellipses. emph{Proc. Royal Soc. London Ser. A.} extbf{469} (2012), 20120492.
end{thebibliography}
end{document}