## Problems of the Miklós Schweitzer Memorial Competition

1、AOPS论坛

2、匈牙利语版本

3、英文版本

4、其它试题

Suppose that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true :
$$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0.$$
Is it true that the only solution to this is the constant function ?

For any $y \in {\bf R}^+$, there thus exists $x$ arbitrarily close to $1$ for which $f(x) \neq f(y)$, hence $f((x+y)/2) = f(\sqrt{xy})$.  By continuity, this implies that $f((1+y)/2) = f(\sqrt{y})$ for all $y \in {\bf R}^+$.  Making the substitution $z := (1+y)/2$, we conclude that $f(z) = f(g(z))$ for all $z \in {\bf R}^+$, where $g(z) := \sqrt{2z-1}$.  The function $g$ has the fixed point $z=1$ as an attractor, so on iteration and by using the continuity of $f$ we conclude that $f(z)=f(1)$ for all $z \in {\bf R}^+$, so $f$ is indeed constant.

(1950年)令$a>0,d>0$,设$$f(x)=\frac{1}{a}+\frac{x}{a(a+d)}+\cdots+\frac{x^n}{a(a+d)\cdots(a+nd)}+\cdots$$给出$f(x)$的封闭解.

\begin{align*}\sum_{n=0}^{\infty}{\frac{x^n}{\prod\limits_{k=0}^n{\left( a+kd \right)}}}&=\frac{\Gamma \left( \frac{a}{d} \right)}{d}\sum_{n=0}^{\infty}{\frac{\left( x/d \right) ^n}{\Gamma \left( \frac{a}{d}+n+1 \right)}}\\&=\frac{\Gamma \left( \frac{a}{d} \right)}{d}\gamma \left( \frac{a}{d},\frac{x}{d} \right) \left( \frac{d}{x} \right) ^{a/d}\frac{e^{x/d}}{\Gamma \left( \frac{a}{d} \right)}=\left( \frac{d}{x} \right) ^{a/d}\frac{e^{x/d}}{d}\gamma \left( \frac{a}{d},\frac{x}{d} \right) ,\end{align*}

$g(x) = x^a f(x^d)$ satifies $g'(x) = x^{a-1} + x^{d-1} g(x)$. Solve the associated differential equation and conclude.

$$\int_0^1{\frac{\arctan x}{x\sqrt{1-x^2}}dx}.$$

Let $n$ be a positive integer. Prove that, for $0<x<\frac\pi{n+1}$,
$$\sin{x}-\frac{\sin{2x}}{2}+\cdots+(-1)^{n+1}\frac{\sin{nx}}{n}-\frac{x}{2}$$
is positive if $n$ is odd and negative if $n$ is even.

Since

\begin{align*}f_n(x) &= \sin{x} - \frac {\sin{2x}}{2} + \cdots + ( - 1)^{n + 1}\frac {\sin{nx}}{n} - \frac {x}{2},\\f_n'(x) &= - \mbox{Re}\left(\sum_{n = 1}^{n}z^n\right) - \frac12.\end{align*}

After some simplifications we get
$$f_n'(x) = \frac {( - 1)^{n + 1}}{2}((1 - \cos(x))\frac {\sin((n + 1)x)}{\sin(x)} + \cos((n + 1)x))$$
and $$f_n''(x) = \frac {( - 1)^{n}}{2}\frac {(n + 1)\sin(nx) + n\sin((n + 1)x)}{1 + \cos(x)}.$$
The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \frac {\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\frac {\pi}{n + 1}) > 0$.

We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even.
Let's restrict to the case $n$ is even.
We prove $f_{2n}(\frac {\pi}{2n + 1}) < 0$.

\begin{align*}f_{2n}\left( \frac{\pi}{2n+1} \right) &=\sum_{k=1}^{2n}{\left( -1 \right)}^{k+1}\frac{\sin \left( \frac{k\pi}{2n+1} \right)}{k}-\frac{\pi}{2\left( 2n+1 \right)}\\&=\frac{\pi}{2n+1}\left( \sum_{k=1}^n{\frac{\sin \left( \frac{\left( 2k-1 \right) \pi}{2n+1} \right)}{\frac{\left( 2k-1 \right) \pi}{2n+1}}}-\sum_{k=1}^n{\frac{\sin \left( \frac{2k\pi}{2n+1} \right)}{\frac{2k\pi}{2n+1}}} \right) -\frac{\pi}{2\left( 2n+1 \right)}.\end{align*}

The function $x \mapsto \frac {\sin(x)}{x}$ is descending on $[0,\pi]$, thus
both sums lay between $a$ and $a + \frac {2\pi}{2n + 1}$, where $a = \int_0^{\pi}\frac {\sin(x)}{x}\,dx$.

Thus $$f_{2n}\left(\frac {\pi}{2n + 1}\right) < \frac {\pi}{2n + 1}\cdot\frac {2\pi}{2n + 1} - \frac {\pi}{2(2n + 1)} < 0.$$

## 2015年丘赛分析组个人赛试题

1. Let $f_n\in L^2(R)$ be a sequence of measurable functions over the line, $f_n\rightarrow f$ almost everywhere. Let $||f_n||_{L^2}\rightarrow||f||_{L^2}$, prove that $||f_n-f||_{L^2}\rightarrow 0$. Proof.(Weingarten) $L^2(\mathbf R)$ is a Hilbert space, and $\|f_n\|_{L^2}\to\|f\|_{L^2}$ as $n\to\infty$ , so we only need to prove $f_n\in f$ weakly in $L^2(\mathbf R)$, that is, for each $g\in L^2(\mathbf R)$, there holds

$\lim_{n\to\infty}\int_{\mathbf R}f_ng=\int_{\mathbf R} fg.$

For each $\epsilon>0$, there exists $R>0$, such that $\int_{|x|>R}|g|^2<\epsilon^2,$

and by the absolute continuity of integration of $g$, there exists a positive $\delta$, such that: for any Lebesgue measurable subset $E$ of $\mathbf R$ with $m(E)<\delta$, there holds $\int_E|g|^2<\epsilon^2.$

By the Egoroff's thoerem, there exists a subset $E_\delta$ of $(-R,R)$ with $m((-R,R)\setminus E_\delta)<\delta$, such that the convergence $\lim\limits_{n\to\infty}f_n=f$ is uniform on the $E_\delta$, so there exists $N\in\mathbf Z_+$, such that $(\forall n>N,x\in E_\delta),(|f_n-f|<\epsilon/\sqrt{2R}).$

Assume $M=\|g\|_{L^2}+\|f\|_{L^2}+\sup\limits_n\|f_n\|_{L^2}$, hence $\forall n>N$, we have the following estimations

\begin{align*}\int_{\mathbf R}|f_n-f|\cdot|g|&=\left(\int_{E_\delta}+\int_{(-R,R)\setminus E_\delta}+\int_{|x|>R}\right)|f_n-f|\cdot|g|\\&\leq\sqrt{\int_{E_\delta}|f_n-f|^2\cdot\int_{E_\delta}|g|^2}+\sqrt{\int_{(-R,R)\setminus E_\delta}|f_n-f|^2\cdot\int_{(-R,R)\setminus E_\delta}|g|^2}\\&+\sqrt{\int_{|x|>R}|f_n-f|^2\cdot\int_{|x|>R}|g|^2}\\&\leq M\epsilon+2\sqrt{2}M\epsilon.\end{align*}

the proof is finished.

2. Let $f$ be a continuous function on $[a,b]$, define $M_n=\int ^b_a f(x)x^n{\rm d}x$. Suppose that $M_n=0$ for all $n$, show that $f(x)=0$ for all $x$.
3. Determine all entire functions $f$ that satisfying the inequality $$|f(z)|\leq|z|^2|{\rm Im}(z)|^2$$ for $z$ sufficlently large.
4. Describe all holomorphic functions over the unit disk $D=\{z||z|\leq 1\}$ which maps the boundary of the disk into the boundary of the disk.
5. Let $T:H_1\rightarrow H_2, Q:H_2\rightarrow H_1$ be bounded linear operators of Hilbert spaces $H_1,\ H_2$. Let $QT={\rm Id}-S_1,TQ={\rm Id}-S_2$ where $S_1$ and $S_2$ are compact operators. Prove ${\rm Ker}T=\{v\in H_1,Tv=0\},{\rm Coker}T=H_2/\overline{{\rm Im}T}$, where ${\rm Im}T=\{Tv\in H_2,v\in H_1\}$ are finite dimensional and ${\rm Im}(T)$ is closed in $H_2$.
6. Let $H_1$ be the Sobolev space on the unit interval $[0,1]$, i.e. the Hilbert space consisting of functions $f\in L^2([0,1])$ such that $$||f||_1^2=\sum^\infty_{n=-\infty}(1+n^2)|\hat f(n)|^2<\infty;$$ where $$\hat f(n)=\frac 1 {2\pi}\int_0^1f(x)e^{-2\pi inx}{\rm d}x$$ are Fourier coefficients of $f$. Show that there exists constant $C>0$ such that $$||f||_{L^\infty}\leq C||f||_1$$ for all $f\in H_1$, where $||\cdot||_{L^\infty}$ stands for the usual supremum norm. (Hint: Use Fourier series.)