Eufisky - The lost book

1-1+1-1+...＝1/2？

来源：马明辉QQ空间
 
在一个寒风凛冽的晚上，我一手拿着烤红薯，一手玩着手机。
 
当时我正在看熊哥的微信公众号: Xionger的数学小屋，他在上面发了北大18年研究生考试题(侵删)。
 
当然了，我早已对这种考试题失去了兴趣，但唯独最后一题，却很是在意。我与向老师讨论了这个题目，向老师嘛，以前与之讨论过一些数学问题，感觉他特厉害，随手就写了份解答给我。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$ on $[0,+\infty)$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
     Since $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$, it is not hard to see $f'(x)>0$, then
\[ \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\sum\limits_{n=0}^{\infty} \left( \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)} \right)=-\sum\limits_{n=0}^{\infty} g_s(\xi_n), \]
where $g_s(x)=(\frac{1}{f^s(x)})'=-\frac{sf'(x)}{f^{s+1}(x)}$ and $\xi_n \in (2n,2n+1)$. Since
\[ g_s'(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0, \]
we can get
\[ -\sum\limits_{n=0}^{\infty} g_s(\xi_n)\le -\sum\limits_{n=0}^{\infty} g_s(2n)\le -g_s(0)-\frac{1}{2}\sum\limits_{n=1}^{\infty} \int_{2n-2}^{2n} g_s(x)dx \]
\[ =-g_s(0)-\frac{1}{2}\int_{0}^{+\infty} g_s(x)dx=-g_s(0)+\frac{1}{2f^s(0)}\to \frac{1}{2} \]
as $s\to 0_+$. Similarly, we have
\[ -\sum\limits_{n=0}^{\infty} g_s(\xi_n)\ge -\sum\limits_{n=0}^{\infty} g_s(2n+1)\ge -\frac{1}{2}\sum\limits_{n=0}^{\infty} \int_{2n+1}^{2n+3} g_s(x)dx \]
\[ =-\frac{1}{2}\int_{1}^{+\infty} g_s(x)dx=\frac{1}{2f^s(1)}\to \frac{1}{2} \]
as $s\to 0_+$.
\end{proof}
 
想起前段时间刚好做过这样一道题，来源于Tenenbaum解析数论课后题230，证明基于Euler-Maclaurin公式(不了解的可自行wiki)。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=\frac{1}{2}. $
 
\begin{proof}
     Let $f_s(x)=e^{-sx^2}$, by Euler-Maclaurin formula, we have
\[ \sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx. \]
Since $f_s'(x)=-2sxe^{-sx^2}$, $f_s''(x)=2se^{-sx^2}(2sx^2-1)$, we can get
\[ \int_{0}^{N} f_s(x)dx\to \int_{0}^{+\infty} f_s(x)dx=\int_{0}^{+\infty} e^{-sx^2}dx=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}} \]
\[ f_s(0)+f_s(N)=1+e^{-sN^2}\to 1 \]
\[ f_s'(0)+f_s'(N)=-2sNe^{-sN^2}\to 0 \]
\[ \int_{0}^{N} B_2(x)f_s''(x)dx\to \int_{0}^{\infty} B_2(x)f_s''(x)dx=\int_{0}^{\infty} B_2(x)2se^{-sx^2}(2sx^2-1)dx \]
\[ =2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx \]
as $N\to \infty$. Therefore, we have
\[ \sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}-2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx. \]
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(\sqrt{s})$ as $s\to 0_+$, then $\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=2\sum\limits_{n=0}^{\infty} e^{-4sn^2}-\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{1}{2}+O(\sqrt{s})$.\\
Another method: Let $\vartheta(t)=\sum\limits_{n\in\mathbb{Z}}e^{-\pi n^2t}$, $t>0$ be the Jocabi theta function. Since
\[ \sum\limits_{n=1}^{\infty} e^{-\pi n^2t}\le \sum\limits_{n=1}^{\infty} e^{-\pi nt}=\frac{e^{-\pi t}}{1-e^{-\pi t}}=O(e^{-\pi t}),  \]
we can get
\[ \vartheta(t)=1+2\sum\limits_{n=1}^{\infty} e^{-\pi n^2t}=1+O(e^{-\pi t}) \]
and
\[ \vartheta(\frac{1}{t})=\sqrt{t}\theta(t)=\sqrt{t}+O(\sqrt{t}e^{-\pi t}). \]
as $t\to \infty$. Then we have
\[ \sum\limits_{n=0}^{\infty} e^{-sn^2}= \frac{1}{2}+\frac{1}{2}\vartheta(\frac{s}{\pi})=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(s^{-\frac{1}{2}}e^{-\frac{1}{s}}) \]
as $s\to 0_+$.
\end{proof}
 
可能利用Jacobi Theta function也能解决，我没试过。现在问题来了，把指数上的n²换成n³结论还对不对？我问了向老师，他也没什么简单的做法。
 
Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}. \]
 
这种结论一看感觉就是对的呀，但是为什么做不出来啊？这就尴尬了。
 
大概是大二那年十一假期，闲来无事写过一份文档，是讨论Grandi级数的某些意义下的收敛性的，有兴趣可以去看一下http://duodaa.com/blog/index.php/archives/351/。然后呢，我就想起了以前做过的一些题目。这个是借用Γ函数的性质做的，也可以在北大那道题中取f(x)＝x。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^s}=-\frac{1}{2}. $
 
\begin{proof}
     Since
\[ \Gamma(s)=\int_0^{\infty}x^{s-1}e^{-x}dx=\int_0^{\infty}(at)^{s-1}e^{-at}dat=a^s\int_0^{\infty}t^{s-1}e^{-at}dt, \]
we have
\[ a^{-s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}e^{-at}dt. \]
Then
\[ \sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n^s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\sum\limits_{n=1}^{\infty}(-1)^ne^{-nt}dt=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\frac{-e^{-t}}{1+e^{-t}}dt \]
\[ =-\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt=-\frac{\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt}{\int_0^{\infty}t^{s-1}e^{-t}dt}\to -\frac{1}{2} \]
as $s\to 0_+$.
\end{proof}
 
上下极限讨论的细节可见http://www.duodaa.com/?/question/6738。小试身手之后，再看一道比较难的题，来源于biler数学分析问题集3.58。
 
Let $S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}$, prove that $\lim\limits_{s\to 0+}S(s)=\frac{1}{2}$.
 
\begin{proof}
     It is generally known that $\sin z=z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2})$, or
\[ \sin \pi z=\pi z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2}). \]
Take logarithmic derivative, we have
\[ \pi\cot \pi z=\frac{1}{z}+\sum\limits_{n=1}^{\infty}\frac{-2z}{n^2-z^2}=\frac{1}{z}+\sum\limits_{n=1}^{\infty}(\frac{1}{z+n}+\frac{1}{z-n})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{z+n}+\frac{1}{z-n}). \]
Since
\[ \frac{1}{\sin \pi z}-\cot \pi z=\frac{1-\cos \pi z}{\sin \pi z}=\tan \frac{\pi z}{2} \]
and
\[ \pi \tan \frac{\pi z}{2}=\pi\cot (\pi \frac{1-z}{2})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{\frac{1-z}{2}+n}+\frac{1}{\frac{1-z}{2}-n})  \]
\[ =-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n-1)}\right)=-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n+1)}\right) \]
we have
\[ \frac{\pi}{\sin \pi z}=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(-1)^n(\frac{1}{z+n}+\frac{1}{z-n})=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2-n^2}, \]
then
\[ \frac{\pi}{\sinh \pi z}=\frac{i\pi}{\sin i\pi z}=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2+n^2}. \]
It is easy to show
\[ \int_{0}^{\infty}\frac{\pi dt}{\sinh(\pi y \cosh t)}=\int_{0}^{\infty} y \cosh t\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\cosh^2t+n^2}dt\]
\[ =\int_{0}^{\infty} \sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\sinh^2t+(y^2+n^2)}d(y\sinh t) \]
\[ =\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\arctan\frac{y\sinh t}{\sqrt{y^2+n^2}}|_{0}^{\infty} =\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\cdot\frac{\pi}{2}. \]
Let $s=\frac{1}{y^2}$, then we have $y\to+\infty$ as $s\to 0_+$, and
\[ S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{y}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}} \]
\[ =\frac{1}{2}+\frac{1}{\pi}\int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}. \]
\\
Since
\[ \frac{\pi y}{\sinh(\pi y \cosh t)}\to 0\]
as $y\to \infty$, and
\[ \frac{\pi y}{\sinh(\pi y \cosh t)}\le\frac{1}{\cosh t}\in L^1[0,\infty), \]
we have
\[ \int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}\to 0 \]
as $y\to\infty$ by dominated convergence theorem.
 
\end{proof}
 
下面这个问题又作何解呢？那汤松的实变函数论里面，证明傅里叶级数一些性质的时候，用到了这种级数。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} (-1)^n(\frac{\sin ns}{ns})^2=-\frac{1}{2}$.\\ \\ \\ \\
 
 
好像没什么思路？算一下x²的傅里叶级数就行了。前面的证明几个例子基本上都依赖于一些恒等式，也就是说，每做一个这种类型的题目，就要找到一个恒等式。可万一找不到怎么办？比如多项式那个。
 
给一个题做一个题，效率太低。这就促使我们找到对这类问题一个统一的解法。
 
Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$\text{(iii)}\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0 \text{ as } s\to 0_+,$\\
we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}. \]
 
\begin{proof}
     Since
\[ a_s(0)=\int_{0}^{\infty}a_s'(x)dx=\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx, \]
we have
\[ \sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\sum\limits_{n=0}^{\infty} \left( a_s(2n)-a_s(2n+1) \right)=\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx \]
\[ =\frac{1}{2}a_s(0)+\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx-\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx \]
\[ =\frac{1}{2}a_s(0)+\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx. \]
Since $a_s(0)\to a_0(0)=1$ and
\[ |\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx|\le \sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} |a_s'(x)-a_s'(x+1)| dx\]
\[ \le \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0  \]
as $s\to 0_+$, we can get the conclusion.
 
\end{proof}
 
条件(iii)看起来有点奇奇怪怪的，那我们就把它换成一个弱一点但看起来很明了的条件。
 
Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$(\text{iii}')a_s(x) \text{ is a convex function },$\\
we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}. \]
 
\begin{proof}
     Since $a_s(x)$ is a convex function, $a_s'(x)$ is increasing, we have
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{\infty} a_s'(x+1)-a_s'(x)dx=a_s(0)-a_s(1)\to 1-1=0 \]
as $s\to 0_+$, it implys (iii).
\end{proof}
 
再看看北大的那道题，就成了上述结论的简单推论。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
     If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0 \]
means $a_s(x)$ is convex.
\end{proof}
 
当然了，我们还可以稍微加强一点北大那题的结论，可自行检验这个要更强一点。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
     If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}\ge 0 \]
means $a_s(x)$ is convex.
\end{proof}
 
现在我们就有能力解决下面的问题了。
 
Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
We have solved the case $m=1$ before, and we only consider $m\ge 2$. Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}. \]
By lemma, we have
\[ (\frac{f'^2}{f''})'\ge (1+\frac{1}{m-1}-\varepsilon)f'>0 \]
and
\[ \frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C, \]
then $\frac{f''}{f'^2}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$. If $s$ is small enough, then there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, and
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx\]
\[ =-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)). \]
Now, we only need to prove that $a_s(x_s)-a_s(x_s+1)\to 0$ as $s\to 0_+$. Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
\[ |a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s). \]
Since
\[ f'(\xi_s)=f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}(\xi_s-x_s)^{m-1}+\frac{f^{m+1}(\eta_s)}{m!}(\xi_s-x_s)^m \]
\[ \le f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}<<f'(x_s) \]
and
\[ s=\frac{f''(x_s)}{f'^2(x_s)}<<\frac{1}{f(x_s)}, \]
we have $sf'(\xi_s)<<\frac{f'(x_s)}{f(x_s)}\to 0$ as $s\to 0_+$.
\end{proof}
 
这里才用了Vinogradov记号，$f＜＜g$来表示$|f|≤C|g|$对某个正常数$C$成立。如果把上述首一多项式改成指数函数，结论还成立吗？
 
Suppose $f(x)>0$, $f'''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
If there is $x_0\ge 0$ s.t. $f''(x_0)\le 0$, then $f''(x)\le 0$ for $x\ge x_0$, we have solved this case. So we can assume $f''(x)>0$, and $f'(x)>0$. If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}. \]
Since $f''(x)$ is decreasing, and $f'(x)$ is strictly increasing, then $\frac{f''(x)}{f'^2(x)}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$.\\
If $a_s''(x)>0$ for all $x\ge 0$, then $a_s(x)$ is convex.\\
If there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, then
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx\]
\[ =-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)). \]
Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
\[ |a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s). \]
Since $f'(\xi_s)-f'(x_s)=f''(\eta_s)(\xi_s-x_s)\le f''(0)$, where $\eta_s \in(x_s, \xi_s)$, and
\[ f'(x_s)=\sqrt{\frac{f''(x_s)}{s}}\le \sqrt{\frac{f''(0)}{s}}, \]
we have
\[ |a_s(x_s+1)-a_s(x_s)|\le s(f''(0)+\sqrt{\frac{f''(0)}{s}})=sf''(0)+\sqrt{sf''(0)}\to 0\]
as $s\to 0_+$.
\end{proof}
 
也许你觉得这个应该也是成立的，但可惜的是，Hardy有个结论却说极限一定不存在。
 
Prove that
$$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!^s}=\frac{1}{2}. $$
 
 
Lemma: Suppose $f\in C^{m+1}[0,\infty)$ and $f, f',\cdots, f^{(m)}>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. Prove that
$$\limsup\limits_{x\to +\infty} \frac{f(x)f''(x)}{f'^2(x)}\le 1-\frac{1}{m}.$$
 
\begin{proof}
We can deal with this problem by induction. The case $m=1$ is trivial, so we can assume $m\ge 2$. Since
\[ f(x)=f(0)+f'(0)x+\frac{f''(\xi)}{2}x^2\ge f'(0)x, \]
we can get $f(x)\to\infty$ as $x\to \infty$. Fixed $\varepsilon>0$, we have
\[ \frac{f'f'''}{f''^2}\le 1-\frac{1}{m-1}+\varepsilon \]
for $x\ge x_0$ by induction. Since
\[ (\frac{f'^2}{f''})'=\frac{2f'f''^2-f'^2f'''}{f''^2}=(2-\frac{f'f'''}{f''^2})f'\ge (1+\frac{1}{m-1}-\varepsilon)f', \]
we have
\[ \frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C, \]
that means
\[ \frac{ff''}{f'^2}\le \frac{f}{(1+\frac{1}{m-1}-\varepsilon)f-C}\to \frac{1}{1+\frac{1}{m-1}-\varepsilon} \]
as $x\to\infty$. Let $\varepsilon\to 0$, we can obtain
$$\limsup\limits_{x\to +\infty} \frac{ff''}{f'^2}\le 1-\frac{1}{m}.$$
\end{proof}
 
这里才用了Vinogradov记号，$f＜＜g$来表示$|f|≤C|g|$对某个正常数$C$成立。如果把上述首一多项式改成指数函数，结论还成立吗？
 
Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}. \]
By lemma, we can get $a_s''(x)>0$ for $x\ge x_0$, that means $a_s(x)$ is convex.
\end{proof}
 
现在我们就有能力解决有关首一多项式的那道题了。
 
Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}. \]
 
\begin{proof}
     Let $a_s(x)=e^{-sp(x)}$, then (i), (ii) are trivial. We only need to show that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=2N}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2} \]
for some positive integer $N$, or equivalent
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n+2N)}=\frac{1}{2}, \]
so we can consider $p(x+2N)$ instead of $p(x)$. Without loss of generality, we can assume the coefficients of $p(x)$ are positive.
 
Since $a_s'(x)=-sp'(x)e^{-sp(x)}<<sx^{m-1}e^{-sx^m}$, we have $a_s(x)-a_s(x+1)=-a_s'(\xi)<<sx^{m-1}e^{-sx^m}$, where $\xi\in (x,x+1)$. Then
\[ \int_{0}^{\infty} |sp'(x)e^{-sp(x)}-sp'(x)e^{-sp(x+1)}|dx<<s^2\int_{0}^{\infty} x^{2(m-1)}e^{-sx^m}dx\]
\[ =s^{\frac{1}{m}}\int_{0}^{\infty} x^{2(m-1)}e^{-x^m}dx \to 0 \]
as $s\to 0_+$. Since $p'(x)-p'(x+1)<<x^{m-2}$, we have
\[ \int_{0}^{\infty} |sp'(x)e^{-sp(x+1)}-sp'(x+1)e^{-sp(x+1)}|dx<<s\int_{0}^{\infty} x^{m-2}e^{-sx^m}dx\]
\[ =s^{\frac{1}{m}}\int_{0}^{\infty} x^{m-2}e^{-x^m}dx \to 0 \]
as $s\to 0_+$. That means (iii).
\end{proof}
 
 
Suppose $f\in C^1[0,\infty)$, $f'\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
\begin{proof}
     Let $a_s(x)=f(sx)$, then (i), (ii) are trivial. Since $a_s'(x)=sf'(sx)$, we have
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=s\int_{0}^{\infty} |f'(sx)-f'(s(x+1))|dx\]
\[ =\int_{0}^{\infty} |f'(x)-f'(s+x)|dx=||f'-f_s'||_{L^1} \to 0 \]
as $s\to 0_+$.
\end{proof}
 
这个结论简洁又好用。随便取一个函数就可以出一道题，拿出来坑人真是不亦乐乎。
 
Let $f(x)=\frac{1}{\sqrt{1+x^2}}, (\frac{\sin x}{x})^2, 2(\frac{1}{x}-\frac{1}{e^x-1})$ and so on, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]\\
 
下面说明在某种意义下，可导的条件是不能必不可少的。这里的条件要求导数可积，如果说函数是单调递减的，那么导数自然是可积的，所以对于此类函数而言，连续可微似乎是多于的条件？然而并不是这样。
 
Suppose $F(x)$ is the Cantor-Lebesgue function, let $f(x)=F(1-x)$, then it is obvious that
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=1\]
for $s=\frac{1}{3^n}$, and
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2} \]
for $s=\frac{2}{3^n}$, where $n\ge 1$.\\
 
然后就猜测会不会对于Cantor-Lebesgue函数而言，对任意s，无穷级数的值都不小于1/2？感觉很有可能是对的，但是费了半天劲没证出来，拿给小罗他也没证出来。最后发现8/45的地方，级数的值是小于1/2的。行吧，竟然有反例。其实我当时的第一想法没想到Cantor-Lebesgue函数，而是强行构造了另一个函数。
 
Suppose $g(x), 0\le x\le 1$ is the unique continuous solution of the following equation
 
$ g(x)=\frac{1}{4}f(2x), 0\le x\le\frac{1}{2},$\\
 
$g(x)=\frac{1}{4}+\frac{3}{4}f(2x-1), \frac{1}{2}\le x\le 1.$\\
or equivalent,
 
\[ g: \sum\limits_{n=1}^{\infty} \frac{a_n}{2^n}\mapsto  \sum\limits_{n=1}^{\infty} \frac{a_n 3^{S_{n-1}}}{4^n} \]
where $a_n=0, 1$, and $S_n=\sum\limits_{i=1}^{n} a_i$, $S_0=0$. Let $f(x)=g(1-x)$, then it is obvious that
 \[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{3}{4} \]
for $s=\frac{1}{2^n}$, where $n\ge 1$.\\
 
对于这个函数而言，会不会极限刚好等于3/4呢？肯定不可能，猜也能猜到极限存在肯定就是1/2。
 
Suppose $f\in C[0,\infty)$ is an decreasing function with $f(0)=1$ and $f(\infty)=0$. If the limit
\[  \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\lambda \]
exists, then $\lambda=\frac{1}{2}$.
 
其实呢，我们有下述更一般些的结论。
 
Suppose $f\in C[0,\infty)$ is an increasing function with $f(0)=1$ and $f(\infty)=0$, then
\[  \liminf\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)\le\frac{1}{2}\le \limsup\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns).\]
 
\begin{proof}
     Let $f_1(x)=\int_{0}^{1}f(xt)dt=\frac{1}{x}\int_{0}^{x} f(t)dt$, then we can get the conclusion.
\end{proof}
 
再积分一次，还可以把连续弱化为在0处连续，结论也是一样的。槊神说我的做法跟他不谋而合。
 
如果函数导数不可积但是二阶导可积呢？比如下面这个例子？
 
Let $f(x)=\frac{\sin\sqrt{x}}{\sqrt{x}}$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
幸好，我们也有类似的结论，其证明也都是很类似的。
 
Suppose $f\in C^2[0,\infty)$, $f''\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
\begin{proof}
     Since $f''\in L^1[0,\infty)$, $f'(\infty)$ exists, then we have $f'(\infty)=0$ by $f(\infty)=0$. Let $f_s(x)=f(sx)$, by Euler-Maclaurin formula, we have
\[ \sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx. \]
Since
$f_s(0)=f(0)=1$, $f_s'(x)=sf'(sx)$, $f_s''(x)=s^2f''(sx)$,
\[ \int_{0}^{N} f_s(x)dx=\frac{1}{s}\int_{0}^{sN}f(x)dx, \]
\[\int_{0}^{N} B_2(x)f_s''(x)dx=s^2\int_{0}^{N} B_2(x)f''(sx)dx=s\int_{0}^{sN} B_2(\frac{x}{s})f''(x)dx \]
we have
\[ \sum\limits_{n=0}^{N} f_{2s}(n)=\frac{1}{2s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{6} \]
\[ -s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx\]
and
\[ \sum\limits_{n=0}^{2N} f_{s}(n)=\frac{1}{s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{12} \]
\[ -\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx\]
Since $f''\in L^1[0,\infty)$, we can get $f'$ is bounded.
Thus,
\[ \sum\limits_{n=0}^{2N} (-1)^nf(ns)=2\sum\limits_{n=0}^{N} f(2ns)-\sum\limits_{n=0}^{2N} f(ns)=2\sum\limits_{n=0}^{N} f_{2s}(n)-\sum\limits_{n=0}^{2N} f_s(n) \]
\[ =\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{4}-2s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx \]
\[ +\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx\]
Let $N\to \infty$, we have
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+s\frac{f'(0)}{4}-2s\int_{0}^{\infty} B_2(\frac{x}{2s})f''(x)dx+\frac{s}{2}\int_{0}^{\infty} B_2(\frac{x}{s})f''(x)dx \]
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+O(s)$.
\end{proof}
 
有兴趣还可以得到高阶导可积对应的情形。最后再考虑一个例子，它的任意阶导数不可积，所以我们只能截断一下。
 
Let $f(x)=\frac{\sin x}{x}$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
\begin{proof}
     Let $f_N(x)=f(x)\chi_{[0,N\pi]}$, it is not difficult to show that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf_N(ns)=\frac{1}{2}, \]
then we have
\[ \limsup\limits_{s\to 0+}|\sum\limits_{n=0}^{\infty} (-1)^nf(ns)-\frac{1}{2}|=\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} (-1)^nf(ns)|  \]
\[ =\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}| \]
By Abel formula, take $a_n=(-1)^n\sin ns=\sin n(\pi+s)$, $b_n=\frac{1}{ns}$, we have
\[ |\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}|\le \frac{A}{N\pi}. \]
Let $N\to \infty$, we can get the conclusion.
\end{proof}
 
证明过程中用到了Abel分部求和公式。
 
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-se^n}=\frac{1}{2}? \]
 
也许你觉得这个应该也是成立的，但可惜的是，Hardy有个结论却说极限一定不存在。
 
 
For any $a>1$, the limit
\[ \lim\limits_{x\to 1-}\sum\limits_{n=0}^{\infty} (-1)^nx^{a^n} \]
does not exist.
 
也就是说，我们不能让指数上的部分增长的太快。其实我也忘了Hardy这个结论咋证的了，就不贴证明了。。。
 
当然了，也可以考虑x的傅里叶级数。所以做一些题目的时候千万别被套路了！就像Dirichlet有个定理是说(a,q)＝1，那么等差数列a+nq中有无穷个素数一样，经常被拿来出题，比如4n+1，6n+5这种，要是不想想根本原因，不知道要被坑多少次。
 
这类例子很多，希望能起到抛砖引玉的作用，以后看问题能看到更深刻的地方，别老是被人家给套路了。如果想不到，更甚说不愿去想更深层的东西，庸俗的人生大多都相仿。
 
(2018年中科大考研题) Suppose $a_n>0$ and
\[ \left| \sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2} \right|\le |\tan x| \]
for $x\in (-1,1)$, prove that $a_n=o(n^2)$ as $n\to\infty$.
\begin{proof}
Since $\sum\limits_{n=1}^{\infty} \cfrac{\sin (a_nx)}{n^2}$ is uniform convergence, we have
\[ \int_{0}^{t}\sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2}dx=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}\int_{0}^{t} \sin (a_nx)dx=\sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n} \]
and
\[ \int_{0}^{t} \tan x dx=-\ln\cos t, \]
then we can get
\[ \sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n}\le -\ln\cos t  \]
for $t\in (0,1)$, so that
\[ 2\sum\limits_{n=1}^{N} \frac{1-\cos(a_nt)}{n^2a_nt^2}\le -\frac{2\ln\cos t}{t^2}  \]
for any $N\in\mathbb{N}$. Let $t\to 0_+$, then
\[ \sum\limits_{n=1}^{N} \frac{a_n}{n^2}\le 1. \]
Let $N\to\infty$, it is obvious that
\[ \sum\limits_{n=1}^{\infty} \frac{a_n}{n^2}\le 1. \]
\end{proof}

数学分析 逆逆 Comments(0) 2018年7月22日 18:40

第一届熊赛分析与方程部分试题

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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}
 
%\begin{solution}
 
%\end{solution}
 
\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}
 
%\begin{enumerate}[itemsep=0pt,parsep=0pt,label=(\arabic*)]
%\item
%\end{enumerate}
 
%\begin{newproof}
 
%\end{newproof}
 
 
 
 
\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.\\
(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
\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
$$h'(x)=f(x+1)-f(x)\to 0$$
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 (Dominated convergence theorem), we have
$$g(x)=\int_{0}^{1} f(x)-f(x+t)\to 0 \text{ as } n\to\infty.$$
\end{proof}
 
 
\end{enumerate}
 
\end{document}

数学分析 熊赛 逆逆 Comments(0) 2018年7月17日 18:14

逆神的数学分析题答案总算补全了

9月5号逆神在数学竞赛交流群里给了一份试题，建模结束后自己才真正仔细思考起来，经过各位大神的指教，终于能够把所有试题的答案给补全，难免存在错误，联系2609480070@qq.cm进行纠正.

数学分析练习题

来源：逆蝶

整理：1729

2014年9月5日

 

 

下面的习题均来自大学生数学竞赛群群友逆蝶提供的数学分析练习题，难免存在错误，请联系（2609480070）进行纠正！

1.已知$a_1=a_2=1,a_{n+2}=2a_{n+1}+3a_n,n=1,2,\ldots$，求幂级数$\sum\limits_{n=1}^\infty{a_n x^n}$的收敛半径，收敛域以及和函数.

解.\begin{align*}&{a_{n + 2}} + {a_{n + 1}} = 3\left( {{a_{n + 1}} + {a_n}} \right) \\\Rightarrow &{a_{n + 1}} + {a_n} = \left( {{a_2} + {a_1}} \right) \times {3^{n - 1}} = 2 \times {3^{n - 1}}\\&{a_{n + 1}} - \frac{1}{2} \times {3^n} = \left( { - 1} \right)\left( {{a_n} - \frac{1}{2} \times {3^{n - 1}}} \right) \\\Rightarrow &{a_n} - \frac{1}{2} \times {3^{n - 1}} = \frac{1}{2}{\left( { - 1} \right)^{n - 1}}.\end{align*}

即

\begin{align*}\sum\limits_{n = 1}^\infty {{a_n}{x^n}} &= \frac{1}{2}\sum\limits_{n = 1}^\infty {\left( {{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}} \right){x^n}} \\&= \frac{1}{6}\sum\limits_{n = 1}^\infty {{{\left( {3x} \right)}^n}} - \frac{1}{2}\sum\limits_{n = 1}^\infty {{{\left( { - x} \right)}^n}} \\&= \frac{1}{2}\frac{x}{{1 - 3x}} + \frac{1}{2}\frac{x}{{1 + x}} = \frac{{x\left( {1 - x} \right)}}{{\left( {1 + x} \right)\left( {1 - 3x} \right)}}.\end{align*}

由\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n} + {{\left( { - 1} \right)}^n}}}{{{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}}} = 3\]

知幂级数收敛半径为$\frac13$，收敛域为$\left(-\frac13,\frac13\right)$.

 

2.计算级数\[\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \]的和.

解.由Stolz公式

\[\mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 2}} = 0.\]

我们有

\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \\&= \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left[ {\left( {n + 1} \right)\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right) - n\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{{n + 1}}} \right) + \frac{n}{{n + 1}}} \right]} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} - \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^2}}}} \\&= 1 - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} + \frac{{{\pi ^2}}}{6} - 1 = \frac{{{\pi ^2}}}{6}.\end{align*}

3.设$\alpha$是实数，计算\[\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} .\]

解.令$x=\tan t$，我们知

\begin{align*}\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} &= \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} \\&= \frac{1}{2}\left( {\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} + \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} } \right)\\&= \frac{\pi }{4}.\end{align*}

4.设$f(x)$在$[0,\pi]$连续，求证：不能同时有\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} < \frac{\pi }{4},\int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} < \frac{\pi }{4}.\]又问何时上面的两个不等式成为等式？

证.注意到

\[a^2+b^2\geq \frac{(a-b)^2}{2}.\]

我们有

\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} + \int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} \ge \int_0^\pi {\frac{{{{\left( {\sin x - \cos x} \right)}^2}}}{2}dx} = \frac{\pi }{2}.\]

由抽屉原理知题给两式不能同时成立.由取等条件知当且仅当

\[f\left( x \right) - \sin x = \cos x - f\left( x \right) \Rightarrow f\left( x \right) = \frac{{\sin x + \cos x}}{2}\]

时取等成立.

5.设$f(x)$在$[0,\infty]$上有$n+1$阶连续导函数，且$f(0)\geq 0,f'(0)\geq0,\ldots,f^{(n)}(0)\geq0.$又对任意$x>0$，有$f(x)\leq f^{(n+1)}(x)$.求证：$f(x)\geq0$.

证.1. （gg）若$f(x)$在$x=0$的某个领域$(0,\xi)$内，满足$f(x)>0$.不妨设存在某个$x>0$，有$f(x)<0$，则此时由连续性，存在某个$x_1>0$，使得$f(x_1)=0$.当$x\in(0,x_1)$时，有$f(x)>0$，则$f^{n+1}(x)\geq f(x)>0$.易推得$f(x)$在$(0,x_1)$上为增函数，$f(x_1)>0$，故此时假设不成立；

2. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，下证矛盾.构造$g\left( x \right) = \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} /{e^x}$，则$g(x)$为增函数，所以$g\left( x \right) \ge g\left( 0 \right) \ge 0 \Rightarrow \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} \ge 0$.由于$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，则必存在某个$\xi_1$，使得当$x\in(0,\xi_1)$时，有$f(x)<0$.对$k=1,2,\ldots,n$，均存在$\xi_k>0$，使得当$x\in(0,\xi_k)$时，使得$f^{(k)}(x)<0$.取$\eta=\min\{\xi_1,\xi_2,\ldots,\xi_n\}$.当$x\in(0,\eta)$时，有$\sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)}<0$，矛盾；

3. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)=0$，且$f(x)$不恒为0，易知此时可推得$(0,\xi)$内，有$f^{(k)}(x)=0,k=1,2,3,\ldots,n$.可转化为在$x=\xi$为初始点的情况，这时我们可采用类似1.,2.的讨论；

4. 若$f(x)=0$，则命题得证.

综上，命题成立.

------------------------------------------

（逆蝶）对于$x\in(0,1]$，由Taylor公式，存在$x_1\in(0,1)$使\[f\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} + \frac{{{f^{\left( {n + 1} \right)}}\left( {{x_1}} \right)}}{{\left( {n + 1} \right)!}}{x^{n + 1}}.\]根据条件得\[f\left( x \right) \ge f\left( {{x_1}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}.\]同样将$f(x_1)$展开，可得$x_2\in(0,x_1)$使得\[f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}}.\]继续这个过程，可得$(0,x)$中严格递减序列$\{x_k\}$使得\[f\left( {{x_k}} \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]于是\[f\left( x \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}} \cdots \frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]因为$x$及$x_k$都在$[0,1]$中，上式右端当$k\to+\infty$时趋于0，于是对于$x\in[0,1]$有$f(x)\geq0$.由此\[f'\left( x \right) = f'\left( 0 \right) + f''\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)!}}{x^{n - 1}} + \frac{{{f^{\left( {n + 1} \right)}}\left( \xi \right)}}{{n!}}{x^n} \ge \frac{{f\left( \xi \right)}}{{n!}}{x^n} \ge 0,\]其中$\xi\in(0,x)$.归纳可证$f^{(k)}\geq0,x\in[0,1],k=1,2,\ldots,n+1$.对函数$g(x)=f(x+1)$重复以上过程可知$f(x)\geq0,x\in[1,2]$.用归纳法可证对任意自然数$m$，$f(x)$在$[m,m+1]$上非负.于是结论得证.

6. 设$f(x)$是$[0,+\infty)$上连续函数，满足$0<f(x)<1$，而且无穷积分在$\int_0^{+\infty}{f(x)\, dx}$和$\int_0^{+\infty}{xf(x)\, dx}$都收敛.求证：\[\int_0^{ + \infty } {xf\left( x \right)\, dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)\, dx} } \right)^2}.\]

证.令\[g\left( y \right) = \int_0^y {xf\left( x \right)dx} - \frac{1}{2}{\left( {\int_0^y {f\left( x \right)dx} } \right)^2},y>0.\]

我们得到\[g'\left( y \right) = yf\left( y \right) - f\left( y \right)\int_0^y {f\left( x \right)dx} = f\left( y \right)\left( {y - \int_0^y {f\left( x \right)dx} } \right).\]

又$0<f(x)<1$，我们得到\[0 < \int_0^y {f\left( x \right)dx} < \int_0^y {dx} = y.\]因此我们有$g'(y)>0,g(y)>g(0)=0.$再令$y\to+\infty$即可.

7. 设$0<\alpha\leq1,\beta>0,\alpha+\beta>1$，$f(x)$是$[1,+\infty)$的正函数，且$\int_1^{+\infty}{f(x)\, dx}$收敛.求证：$\int_1^{+\infty}{\frac{{(f(x))}^\alpha}{x^\beta}\, dx}$收敛.

证.当$\alpha=1$时，由\[0 < \int_1^{ + \infty } {\frac{{f\left( x \right)}}{{{x^\beta }}}dx} \le \int_1^{ + \infty } {f\left( x \right)dx} .\]可知积分收敛.

当$0<\alpha<1$时.

（Holder积分不等式）若函数$f(x)$与$g(x)$在区间$[a,b]$上连续非负，且$p>1,\frac1p+\frac1q=1$，则有不等式\[\int_a^b {f\left( x \right)g\left( x \right)dx} \le {\left( {\int_a^b {{{\left[ {f\left( x \right)} \right]}^p}dx} } \right)^{\frac{1}{p}}}{\left( {\int_a^b {{{\left[ {g\left( x \right)} \right]}^q}dx} } \right)^{\frac{1}{q}}}.\]

取$p=\frac1\alpha,q=\frac1{1-\alpha},f(x)={[f(x)]}^\alpha,g(x)=\left(\frac1x\right)^\beta$，对于任给的正数$A$，我们有

\[0 < \int_1^A {\frac{{{{\left( {f\left( x \right)} \right)}^\alpha }}}{{{x^\beta }}}dx} \le {\left( {\int_1^A {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^A {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }} \le {\left( {\int_1^{ + \infty } {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }}.\]

再注意到$\int_1^{ + \infty } {f\left( x \right)dx} ,\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} \left( {\frac{\beta }{{1 - \alpha }} > 1} \right)$均收敛即可得证.

8. 设$\{a_n\}$是正的递增数列.求证：级数$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛的充分必要条件是$\{a_n\}$有界.

证.1. ($\Rightarrow$)注意到\[\ln(1+x)<x,x>0.\]我们有\[A = \sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} > \sum\limits_{n = 1}^\infty {\ln \frac{{{a_{n + 1}}}}{{{a_n}}}} = \mathop {\lim }\limits_{n \to \infty } \ln \frac{{{a_n}}}{{{a_1}}}.\]故\[\mathop {\lim }\limits_{n \to \infty } {a_n} = B \le {a_1}{e^A}.\]

2. ($\Leftarrow$)又\[\sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} = \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}} < \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n} - {a_1}}}{{{a_1}}} = \frac{1}{{{a_1}}}\mathop {\lim }\limits_{n \to \infty } {a_n} - 1.\]故$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛.

9. 设$\alpha>0,\{a_n\}$是递增正数列.求证：级数$\sum\limits_{n=1}^\infty \frac{a_{n+1}-a_n}{a_{n+1}a_n^\alpha}$收敛.

证.1. 当$0<\alpha<1$时，有\[\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} \le \sum\limits_{n = 1}^\infty {\frac{1}{\alpha }\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{\alpha a_1^\alpha }} - \frac{1}{\alpha }\mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\]

事实上，由Lagrange中值定理，我们有\[\frac{{a_{n + 1}^\alpha - a_n^\alpha }}{{{a_{n + 1}} - {a_n}}} = \alpha {\xi ^{\alpha - 1}} > \alpha a_{n + 1}^{\alpha - 1},\]其中$\xi\in(a_n,a_{n+1})$.

因此，根据$a_n$单调递增的性质，对于其有界和无界两种情况，不等式右端的级数都是收敛的，由此得证.

2. 当$\alpha\geq1$时，又有

\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} &= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_n^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \le \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_{n + 1}^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{a_1^\alpha }} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\end{align*}

同理，不等式右端的级数亦是收敛的，证毕.

10. 设$0<\alpha<1$，证明数列\[{a_n} = \frac{1}{{1 + {n^\alpha }}} + \frac{1}{{2 + {n^\alpha }}} + \cdots + \frac{1}{{n + {n^\alpha }}},n = 1,2, \cdots \]发散.

证.注意到\[x>\ln(x+1).\]我们有\[\frac{1}{{k + {n^\alpha }}} > \ln \frac{{k + 1 + {n^\alpha }}}{{k + {n^\alpha }}}.\]

因此\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k + {n^\alpha }}}} > \mathop {\lim }\limits_{n \to \infty } \ln \frac{{n + 1 + {n^\alpha }}}{{1 + {n^\alpha }}} \to \infty .\]

所以此数列发散.

11. 计算\[\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{1^2} + \sqrt n + {n^2}}} + \frac{n}{{{2^2} + 2\sqrt n + {n^2}}} + \cdots + \frac{n}{{{n^2} + n\sqrt n + {n^2}}}} \right).\]

解.注意到

\begin{align*}&\mathop {\lim }\limits_{n \to \infty } \left[ {\sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} - \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} } \right] = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{\left( {{k^2} + k\sqrt n + {n^2}} \right)\left( {{k^2} + {n^2}} \right)}}} \\&\le \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{{n^2} \cdot {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n\sqrt n }}{{{n^2} \cdot {n^2}}} \cdot \frac{{n\left( {n + 1} \right)}}{2} = 0.\end{align*}

因此

\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{{{\left( {\frac{k}{n}} \right)}^2} + 1}}} = \int_0^1 {\frac{1}{{{x^2} + 1}}dx} = \frac{\pi }{4}.\]

12. 设$f(x)$是$[0,2\pi]$上可导的凸函数，$f'(x)$有界.试证\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nx\, dx} \ge 0.\]

证.由分部积分，我们有

\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nxdx} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)d\frac{{\sin nx}}{n}} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} .\]

又由第二积分中值定理，我们有

\begin{align*}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} &= f'\left( {0 + } \right)\int_0^\xi {\sin nxdx} + f'\left( {2\pi - } \right)\int_\xi ^{2\pi } {\sin nxdx} \\&= \frac{{1 - \cos n\xi }}{n}\left[ {f'\left( {0 + } \right) - f'\left( {2\pi - } \right)} \right] \le 0.\end{align*}

故\[{a_n} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} \ge 0.\]

13. 设$\{a_n\}$是正数列使得$\sum\limits_{n=1}^\infty{\frac{1}{a_n}}$收敛.求证\[\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} \le 2\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} ,\]而且上式右端的系数2是最佳的.

证.由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}}  \ge {\left( {1 + 2 +  \cdots  + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

即\[\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}} \le \frac{4}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}}}  &\le 4\sum\limits_{n = 1}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} }  = 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}} } \\&\le 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} }  = 2\sum\limits_{m = 1}^\infty  {\frac{1}{{{a_m}}}} .\end{align*}

这里用到了\[\frac{1}{{n{{\left( {n + 1} \right)}^2}}} \le \frac{1}{2}\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}} = \frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{n{{\left( {n + 1} \right)}^2}}}} \right].\]

 

注意到$a_n=n^\alpha,\alpha>1$时有

\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]

14. 设$f(x)$是$[0,+\infty)$上正的连续函数，且$\int_0^{+\infty}{\frac{1}{f(x)}\, dx}$收敛.记$F(x)=\int_0^x{f(t)\, dt}$.求证\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} < 2\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} ,\]且上式右端的系数2是最佳的.

证.（陈洪葛）由$Cauchy-Schwarz$不等式，得到\[\left( {\int_0^x {f\left( t \right)dt} } \right) \cdot \left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right) \ge {\left( {\int_0^x {tdt} } \right)^2} = \frac{1}{4}{x^4}.\]

所以

\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} \le \int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} .\]

注意到\[\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2f\left( x \right)}} = 0.\]

以及\[\frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \int_0^\xi {\frac{1}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{1}{{f\left( t \right)}}dt} .\]

故\begin{align*}&\int_0^A {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} = \int_0^A {\left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right)d\left( { - \frac{2}{{{x^2}}}} \right)} \\&= \int_0^A {\frac{2}{{f\left( x \right)}}dx} - \frac{2}{{{A^2}}}\int_0^A {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^A {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^A {\frac{{{t^2}}}{{f\left( t \right)}}dt}.\end{align*}

令$A\to+\infty$，得到

\[\int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} \le \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^{ + \infty } {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx}.\]

另外，当我们取$f(x)=x^a+1(a>1)$时，有$\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} = \frac{\pi }{{a\sin \frac{\pi }{a}}}$收敛.此时有

\[\mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} }}{{\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{{x^{a + 1}}/\left( {a + 1} \right) + x}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{1}{{{x^a}/\left( {a + 1} \right) + 1}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} {\left( {a + 1} \right)^{\frac{1}{a}}} = 2.\]

15. 设$f(x)$在$\mathbb{R}$上有二阶导函数，$f(x),f'(x),f''(x)$都大于零，假设存在正数$a,b$使得$f''(x)\leq af(x)+bf'(x)$对一切$x\in\mathbb{R}$成立.

1. 求证：$\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right) = 0$;

2. 求证：存在常数$c$使得$f'(x)\leq cf(x)$;

3. 求使上面不等式成立的最小常数$c$.

证.1. 显然由单调有界定理知$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right),\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)$均存在，不妨设$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right)=c_1,\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)=c_2$.因此我们有$f(x)\geq c_1\geq0,f'(x)\geq c_2\geq0$.又\[f\left( 0 \right) = \int_x^0 {f'\left( x \right)dx} + f\left( x \right) \ge - {c_2}x + f\left( x \right) \Rightarrow f\left( x \right) \le {c_2}x + f\left( 0 \right).\]我们知$c_2=0$，否则令$x\to-\infty$，矛盾.

2. 令

\begin{align*}&g\left( x \right) = \left( {\frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right) - f'\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}},\\&g'\left( x \right) = \left( {af\left( x \right) + bf'\left( x \right) - f''\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}} \ge 0.\end{align*}

又$\lim\limits_{n\to-\infty} g(x)=0$.故$g(x)\geq g(-\infty)=0$.即\[f'\left( x \right) \le \frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right).\]故存在常数$c={\frac{{b + \sqrt {4a + {b^2}} }}{2}}$使得$f'(x)\leq cf(x)$成立.

 

3. 我觉得应该把“求使上面不等式成立的最小常数$c$”改成“求使上面不等式成立的最大常数$c$”.事实上，我们取：

\[\left\{ \begin{array}{l}h\left( x \right) = {e^{cx}}\\h'\left( x \right) = c{e^{cx}}\\h''\left( x \right) = {c^2}{e^{cx}}\end{array} \right.\left( {c > 0} \right).\]

只需保证\[c^2\leq bc+a\Rightarrow 0<c\leq {\frac{{b + \sqrt {4a + {b^2}} }}{2}}.\]

故常数$c$的最大值为${\frac{{b + \sqrt {4a + {b^2}} }}{2}}$.

16. 设$f(x)$是$\mathbb{R}$上有下界或者有上界的连续函数且存在正数$a$使得\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt} \]为常数.求证：$f(x)$必为常数.

证.（Slade）由题意，$f(x)$必存在上下界，否则，在等式\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt}=C\]两端同时取$x\to \infty$，等式左端无界，而右端为常数，矛盾.

再之，注意到$\int_{x-1}^x f(t)\,dt$是可微的，故$f$也是可微的.又\[f'\left( x \right) = a\left( {f\left( {x - 1} \right) - f\left( x \right)} \right) \Rightarrow \left| {f'\left( x \right)} \right| \le a\left( {\left| {f\left( {x - 1} \right)} \right| + \left| {f\left( x \right)} \right|} \right).\]由于$f$有界，故$f'$亦有界.记$N=[a]+1$，则$0<\frac aN<1$.令$x=\frac x N$，我们得到\[f'\left( {\frac{x}{N}} \right) = a\left( {f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right)} \right).\]

固定$x$，由Lagrange中值定理，存在${a_1} \in \left( {\frac{x}{N} - 1,\frac{x}{N}} \right)$使得\[f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right) = - \frac{1}{N}f'\left( {{a_1}} \right),\]即$f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right)$.类似地，我们有数列$\{a_n\}$使得\[f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right) = {\left( { - \frac{a}{N}} \right)^2}f'\left( {{a_2}} \right) = \cdots = {\left( { - \frac{a}{N}} \right)^n}f'\left( {{a_n}} \right).\]令$n\to\infty$，我们有$f'\left( {\frac{x}{N}} \right) = 0 $即$f'\left( x \right) = 0$.故$f(x)$必为常数.

17. 设$f:[0,+\infty)\to [0,+\infty)$且对任意$x\geq0$有$f\circ f(x)=af(x)+bx$，其中$a<0,b>0$.求$f(x).$

证.先证明几个引理：

引理1.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若方程中的常数$b\neq0$，则$f:\mathbb{R}\to\mathbb{R}$即单射又是满射，即是一个一一映射.

引理2.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若$b\neq0$，且$\lambda$的多项式$\lambda^2-a\lambda-b$的两个根$r$与$s$不相等，则对任意$x\in\mathbb{R}$及任意$n\in\mathbb{Z}$均有

\[{f^n}\left( x \right) = \frac{{{s^n}\left( {f\left( x \right) - rx} \right) + {r^n}\left( {sx - f\left( x \right)} \right)}}{{s - r}}.\]

事实上，当$n$为非正整数时下一式成立：\[{f^n}\left( x \right) = \frac{{{s^n}\left( {{f^{ - 1}}\left( x \right) - \frac{x}{r}} \right) + {r^n}\left( {\frac{x}{s} - {f^{ - 1}}\left( x \right)} \right)}}{{\frac{1}{s} - \frac{1}{r}}}.\]以$f(x)$代替$x$，以$n-1$代替$n$，可推知$n$为非正整数时亦成立.

引理3.设$g_n\in C^0(\mathbb{R},\mathbb{R})$，且${(-1)}^n g_n(x)$对$x$单调递增（$n=0,1,2,\ldots$）.若极限$g(x)=\lim\limits_{n\to\infty}g_n(x)$对任$x\in \mathbb{R}$均存在，则极限函数$g$是个常值函数.

事实上，因$g(x)=\lim\limits_{n\to\infty}g_{2n}(x)$，而$g_{2n}(x)$对$x$递增.又因$g(x)=\lim\limits_{n\to\infty}g_{2n+1}(x)$，而$\lim\limits_{n\to\infty}g_{2n+1}(x)$对$x$递减，故$g(x)$对$x$递减.于是$g$只能是个常值函数.

回到原题，我们有定理：设$r$及$s$是$\lambda$的二次多项式$\lambda^2-a\lambda-b=0$的两个根.若$r<0<s$，且$r\neq -s$，则$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解的充分必要条件是$f(x)=rx$（对任$x\in\mathbb{R}$），或者$f(x)=sx$（对任$x\in\mathbb{R}$），或者，当$s=1$时还可以是$f(x)=rx+c$（对任$x\in\mathbb{R}$，$c$可以是任一个给定的实数）.

定理的充分性部分是显然的.下面证明定理的必要性部分，设$f$是方程$f(f(x))=af(x)+bx$的一个解，据引理1知$f$严格单调.若$f$严格递增，可推出

\[sx - f\left( x \right) = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| > s,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| < s.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/r^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可知极限函数$sx-f(x)$恒取常值，即存在某$c\in\mathbb{R}$使得$f(x)=sx+c$对任意$x\in\mathbb{R}$成立，把$f(x)$的这一表达式代入得到$s^2x+sc+c=(r+s)(sx+c)-rsx$，推出$c=0$.因此，当$f$递增时$f(x)=sx$（对任意$x\in\mathbb{R}$）.

若$f$严格递减，可推出\[f\left( x \right) - rx = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s > \left| r \right|,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s < \left| r \right|.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/s^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可同样推出$f(x)=rx+c$（对任意$x\in\mathbb{R}$）.将此表达式代入得到$r^2x+rc+c=(r+s)(rx+c)-rsx$，可化简为$(s-1)c=0$.因此，当$s\neq1$时$c=0$，当$s=1$时$c$可为任意给定的实数.

另外：注意到$f:[0,+\infty)\to [0,+\infty)$这一条件，我们知$f(x)=sx(s>0)$.可参阅：

[1]关于迭代函数方程$f^2(x)=af(x)+bx$的通解，麦结华，数学研究与评论第17卷第1期83-90页，1997年2月.

[2]J.Matkowski and Zhang Weinian, Method of characteristics for functional equations in polynomial

form, Acta Math.Sinica, New Series.

18. 设$f(x)$在$(-\infty,+\infty)$上连续，且对任意$x$有$f(2x-f(x))=x$.求证：$f(x)\equiv x+c$，其中$c$为常数.

证.（逆蝶）令$g(x)=2x-f(x)$，则$f[g(x)]=x$，显然$g$是单射且为增函数.若$g$有上界，则$\lim\limits_{x\to+\infty}{g(x)}=a$存在.又$f$连续，$\lim\limits_{x\to+\infty}{f[g(x)]}=f(a)$，矛盾.从而$g$无上界也无下界.由于$g$连续，$\forall x\in R,\exists y$使得$x=g(y)$.从而$f(x)=f[g(y)]=y$，于是$g[f(x)]=g(y)=x$.由$g$递增知$f$递增.

若$\exists x_1,x_2$使得$f(x_1)-x_1\neq f(x_2)-x_2$，则由于$f(x)-x$的连续性知其可以取遍$f(x_1)-x_1$及$f(x_2)-x_2$之间的任何数.

设$d,d'$在它们之间，且$d/d'>1$是正无理数，设$d=f(x)-x,d'=f(x')-x'$.令$x_0=x,x_{n+1}=2x_n-f(x_n)$，则$f(x_{n+1})=x_n$.归纳可得$x_n=x+nd$，同理$x'_n=x'+nd'$.

令$x_m>x'_n,x_{m+1}<x'_n+1$，即$x+md>x'+nd,x+d+md<x'+d'+nd'$.即$(x'-x)/d<(md')/d-n<(x'-x)/d+(d/d'-1)$.由$d/d'$为无理数知上式关于$m,n$有解，这与$f$递增矛盾.

下面是一个推广：对$\forall m\neq0$，若一个连续函数$f:\mathbb{R}\to\mathbb{R}$满足函数方程\[f\left(2x-\frac{f(x)}{m}\right)=mx,\]则有$f(x)=m(x-c)$.

证.（陈洪葛）我们设$g(x)=2x-\frac{f(x)}{m}$，显然$g(x)$是连续函数且有\[g(g(x))=2g(x)-x,\forall x\in\mathbb{R}.\]

若$g(x_1)=g(x_2)$，则有$g(g(x_1))=g(g(x_2))$，我们得到\[x_1=x_2.\]故$g(x)$是一个单射，而我们知道，若$g(x)$是一个连续的单射，则$g(x)$严格单调（关于这点可以用反证法证明）.因此，$g(x)$有2种情况，严格递增或者严格递减.下面证明$g(x)$只能严格递增.

（反证）设$g(x)$严格递减，则对于$x_1<x_2$，我们有$g(x_1)>g(x_2)$，接着又有$g(g(x_1))<g(g(x_2))$，二者等价于\[2g(x_1)-x_1<2g(x_2)-x_2\Leftrightarrow 2[g(x_1)-g(x_2)]<x_1-x_2.\]上面不可能成立，因为左边大于0而右边小于0，故$g(x)$只能严格递增.

改写$g(g(x))=2g(x)-x$为\[g(g(x))-g(x)=g(x)-x.\]递推后得到\[g^n(x)=ng(x)-(n-1)x,(n\geq1),\]这里$g^{(n)}(x)$表示$n$次复合.那么有\[g^n(x)-g^n(0)=n[g(x)-x-g(0)]+x\Leftrightarrow\frac{g^n(x)-g^n(0)}{n}=g(x)-x-g(0)+\frac{x}{n}.\]而$g(x)$严格递增，$g^n(x)$也严格递增，故对上式令$n\to\infty$，由$g(x)$的单调性，我们得到

\begin{align*}g(x)&\leq x+g(0),&&x<0\\g(x)&\geq x+g(0),&&x>0.\end{align*}

这样，我们得到$g(x)$的值域也是$\mathbb{R}$，故$g(x)$是一个一一映射，且$g^{-1}$存在.现在，用$x=g^{-1}(g^{-1}(y))$代入原来的方程，则有\[g^{-1}(g^{-1}(y))=2g^{-1}(y)-y.\]$g^{-1}(y)$同样满足这个方程，则用相同的手段，我们得到

\begin{align*}g^{-1}(y)&\leq y+g(0),&&y<0\\g^{-1}(y)&\geq y+g(0),&&y>0.\end{align*}

现在，用$x=g^{-1}(y)$代入\[g(g(x))-g(x)=g(x)-x\]得到\[g(y)-y=y-g^{-1}(y).\]令$y=0$得到$g^{-1}(0)=-g(0)$.

假设$g(0)\geq0$，则对$x>0$有$g(x)\geq x+g(0)>0$，则对$y=g(x)>0$有$x>g(x)+g^{-1}(0)=g(x)-g(0)$.故得到

\[g(x)=x+g(0),x>0\]同理可得\[g(x)=x+g(0),x<0.\]这样我们得到$f(x)=m(x-g(0))$对$x\in\mathbb{R}$成立.

19. 设$0<a<2$.求证：不存在$(-\infty,+\infty)$上连续的函数$f(x)$，使得对任意$x$有$f(ax-f(x))=x$.

证.我们设$g(x)=ax-f(x)$，显然$g(x)$是连续函数且有\[g(g(x))=ag(x)-x,\forall x\in\mathbb{R}.\]假设此方程有一连续函数解$g:\mathbb{R}\to\mathbb{R}$且多项式方程$r^2-ar+1=0$有一对复数特征根\[{r_1} = a - ib = S\exp \left( { - i\theta } \right),{r_2} = a + ib = S\exp \left( {i\theta } \right)，\]其中$a,b\in \mathbb{R},b>0,S>0$及$\theta\in(0,\pi)$.易知$f$是单调的且$f^2$是严格递增的.并且对于$x\neq0$有$f(x)\neq x$.因此当$f$是严格递增时，数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$对于任意固定的$x\neq0$同样是严格递增的.因此，我们有

\begin{align*}{f^n}\left( x \right) &= \frac{{r_2^n}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right) + \frac{{r_1^n}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)\\&= \frac{1}{b}{S^n}\sin \theta \cdot f\left( x \right) - \frac{1}{b}{S^{n + 1}}\sin \left( {n - 1} \right)\theta \cdot x.\end{align*}

则

\[{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = r_2^nU\left( x \right) + r_1^nV\left( x \right),\]

其中$U\left( x \right) = \frac{{{r_2} - 1}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right),V\left( x \right) = \frac{{{r_1} - 1}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)$.显然$\overline U\left( x \right) = V\left( x \right)$，故对于固定的$x\neq0$我们可令

\[U\left( x \right) = T\exp \left( {it} \right)\text{和}V\left( x \right) = T\exp \left( { - it} \right),\]

其中$T\geq0$且$t\in[0,2\pi]$.因此\[\begin{array}{l}{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = {S^n}T\left[ {\exp \left( {i\left( {n\theta + t} \right)} \right) + \exp \left( { - i\left( {n\theta + t} \right)} \right)} \right]\\= 2{S^n}T\cos \left( {n\theta + t} \right)\end{array}.\]

因为当$T>0$时$S>0$，与数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$的性质矛盾；当$T=0$时我们有$U(x)=V(x)=0$，因此对所有$x\neq0$我们有$f(x)=r_1x=r_2x$，由此我们得到$r_1=r_2$这一矛盾的结论.证毕.

20. 求证：不存在可微函数$f:(0,+\infty)\to(0,+\infty)$满足方程\[f'(x)=f\circ f(x),x\in(0,+\infty).\]

证.（刘畅）由$f(x)>0$知$f(f(x))>0$，从而有$f'(x)>0$，显然$\lim\limits_{x\to+\infty}f(x)=+\infty$（否则得到$\lim\limits_{x\to+\infty}f'(x)=0$，矛盾），因此$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f(f(x))=+\infty$.当$x$充分大时，我们有$f'(x)>1$，故存在足够大的$n\in\mathbb{N_+}$使得$f(n)>n+1$，我们有\[f\left( {n + 1} \right) - f\left( n \right) = \int_n^{n + 1} {f'\left( x \right)dx} = \int_n^{n + 1} {f\left( {f\left( x \right)} \right)dx} > f\left( {n + 1} \right),\]矛盾.

21. 设正数列$\{a_n\}$满足$\varliminf\limits_{n\to+\infty}{a_n}=1,\varlimsup\limits_{n\to+\infty}{a_n}<+\infty,\lim\limits_{n\to+\infty}\sqrt[n]{a_1a_2\ldots a_n}=1.$求证：

\[\lim\limits_{n\to+\infty}\frac{a_1+a_2+\ldots+a_n}{n}=1.\]

证.（别人解答）记$x_n=\ln a_n$，由题设条件，

\begin{align*}&\varliminf\limits_{n\to+\infty}x_n=0,\varlimsup\limits_{n\to+\infty}x_n\leq A<+\infty(A>0),\\&\lim\limits_{n\to\infty}\frac1n\sum_{k=1}^n x_k=0.\end{align*}

假设所有的$x_n\geq0$，则当$x\leq\ln2$时，成立不等式$e^x\leq1+2x$.对于固定的$n$，记${S_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} \le \ln 2} \right.} \right\},{T_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} > \ln 2} \right.} \right\}$，则

\[\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{x_k}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{x_k}} \ge \frac{{\left| {{T_n}} \right|}}{n}\ln 2 \ge 0.\]

由此即知\[\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{T_n}} \right|}}{n} = 0.\]

从而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{e^{{x_k}}}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{e^{{x_k}}}} \le 1 +\frac{{\left| {{T_n}} \right|}}{n}\left( {C+ {e^A}} \right) + \frac{2}{n}\sum\limits_{k = 1}^n {{x_k}} .\]而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \ge {e^{\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} }}.\]由迫敛性即得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]在一般情况下，作序列\[{z_n} = \left\{ \begin{array}{l}- {x_n},{x_n} < 0\\0,{x_n} \ge 0\end{array} \right.,n = 1,2, \cdots \]由题设可知\[\mathop {\lim }\limits_{n \to \infty } {z_n} = 0.\]故$y_n=x_n+z_n\geq0$，得到$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{y_k}} = 0.$从而$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}} = 1$，再由$\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \le \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}}$，可得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]

22. 设$f(x)$在$\mathbb{R}$上有二阶连续导数且满足方程\[f^3+{(f')}^3=1.\]求证：$f=1$.

证.（刘畅）显然$f=1$满足题意.当$f\neq1$时，我们有

\[f' = \sqrt[3]{{1 - {f^3}}},f'' = \frac{{ - {f^2}f'}}{{\sqrt[3]{{{{\left( {1 - {f^3}} \right)}^2}}}}} = \frac{{ - {f^2}}}{{\sqrt[3]{{1 - {f^3}}}}}.\]

若$\exists x_0\in\mathbb{R}$使$f(x_0)>1$，则$f'(x_0)<0$，故对$x\leq x_0$都有$f'(x)<0,f''(x)>0$,从而$f(-\infty)=+\infty$.若$\exists x_1>x_0$使得$f(x_1)<1$，则对$x\leq x_1$都有$f'(x)>0,f''(x)<0$，故$f(-\infty)=-\infty$，矛盾.

故对$x>x_0$，均有$f(x)\geq1$.

若$\exists a>x_0$使得$f(a)=1$（其中$a$为使得$f(x)=1$的最小实数），则我们有$x>a$均有$f(x)=1$，这时有$f''(x)$在$x=a$处不连续，矛盾.

若不$\exists b>x_0$使得$f(b)=1$，则对于$x\in\mathbb{R}$均有$f(x)>1$，设$f(+\infty)=1$（否则有$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) < 0$，与$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0$矛盾），而\[\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0,\mathop {\lim }\limits_{x \to + \infty } f''\left( x \right) = + \infty .\]亦矛盾.证毕.

 

 

 

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数学分析 大学生数学竞赛 逆逆 Comments(1) 2014年9月30日 03:44