Eufisky - The lost book

## 1-1+1-1+...＝1/2？

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}

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}

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
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}.$

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}

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}

Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} (-1)^n(\frac{\sin ns}{ns})^2=-\frac{1}{2}$.\\ \\ \\ \\

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}

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}

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}

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}

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}

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}

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}

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}

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}

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
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$\\

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 $n\ge 1$.\\

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 $n\ge 1$.\\

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 $\lambda=\frac{1}{2}$.

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}

Let $f(x)=\frac{\sin\sqrt{x}}{\sqrt{x}}$, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$

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}

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}

$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-se^n}=\frac{1}{2}?$

For any $a>1$, the limit
$\lim\limits_{x\to 1-}\sum\limits_{n=0}^{\infty} (-1)^nx^{a^n}$
does not exist.

(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}

## 第一届熊赛分析与方程部分试题

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%\title{中国科学技术大学\\
%2015年硕士学位研究生入学考试试题}
%\author{(线性代数与解析几何)}
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{\erhao \CJKfamily{kd}{第一届Xionger网络数学竞赛} }\\
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{\sanhao \textbf{分析与方程部分试题解答} }\\
\vspace{0.2cm}
2018年6月8日}
%{\sanhao \textbf{解答：Eufisky (Xiongge)}}\\
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\item Suppose $S_n=a_1+\cdots+a_n$, and $T_n=S_1+\cdots+S_n$. If $\lim\limits_{n\to\infty} \frac{1}{n}T_n=s$ and $na_n$ is bounded, then $\lim\limits_{n\to\infty} S_n=s.$
\begin{proof}
We may assume that $\lim\limits_{n\to\infty} \frac{1}{n}T_n=0$, or we can replace $a_1$ by $a_1-s$. For $\forall \varepsilon>0$, there is a large number $N$, $\forall n\ge N$, we have $|T_n|\le \varepsilon n.$ Since
$$T_{n+k}-T_n=S_{n+1}+\cdots+S_{n+k}$$
$$=kS_n+(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}),$$
we have
$$kS_n=T_{n+k}-T_n-(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}).$$
Suppose $|na_n|\le C$, then
$$k|S_n|\le (2n+k)\varepsilon+C\left(\frac{k}{n+1}+\frac{k-1}{n+2}+\cdots+\frac{1}{n+k}\right)\le (2n+k)\varepsilon+C\frac{k^2}{n},$$
or equivalent,
$$|S_n|\le \left(\frac{2n}{k}+1\right)\varepsilon+C\frac{k}{n}.$$
Take $k=[\sqrt{\varepsilon}n]>\frac{1}{2}\sqrt{\varepsilon}n$, then we have
$$|S_n|\le \left(\frac{4}{\sqrt{\varepsilon}}+1\right)\varepsilon+C\sqrt{\varepsilon}=\left(4+\sqrt{\varepsilon}+C\right)\sqrt{\varepsilon}.$$
\end{proof}

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

## 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.

2014年9月5日

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.

\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*}

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.

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

\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$\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} .$

\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？

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

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

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

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：$f(x)\geq0$.

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.

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（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：$\int_0^{ + \infty } {xf\left( x \right)\, dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)\, dx} } \right)^2}.$

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.

（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$\int_a^b {f\left( x \right)g\left( x \right)dx} \le {\left( {\int_a^b {{{\left[ {f\left( x \right)} \right]}^p}dx} } \right)^{\frac{1}{p}}}{\left( {\int_a^b {{{\left[ {g\left( x \right)} \right]}^q}dx} } \right)^{\frac{1}{q}}}.$

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

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.

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.

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*}

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.

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

\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*}

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

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

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

\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*}

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.

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

\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*}

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

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.

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

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

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

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$c$.

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*}

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：

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

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.

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$f(x).$

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

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

[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.

（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.

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

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

$g(x)=x+g(0),x>0$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$f(ax-f(x))=x$.

\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*}

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

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

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

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：

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

\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*}

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

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：$f=1$.

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