竞赛 - Eufisky - The lost book

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 ?

陶哲轩解答：Yes. If $f$ were not constant, then (since ${\bf R}^+$ is connected) it could not be locally constant, thus there exists $x_0 \in {\bf R}^+$ such that $f$ is not constant in any neighbourhood of $x_0$. By rescaling (replacing $f(x)$ with $f(x_0 x)$) we may assume without loss of generality that $x_0=1$.

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.

来源：这里

(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.

也就是求\[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{\prod\limits_{k = 0}^n {\left( {a + kd} \right)} }}} .\]

解.首先有$$\prod_{k=0}^n{\frac{1}{a+kd}}=\frac{\Gamma \left( \frac{a}{d} \right)}{d^{n+1}\Gamma \left( \frac{a}{d}+n+1 \right)},$$

又因为$$\gamma \left( s,x \right) =\sum_{k=0}^{\infty}{\frac{x^se^{-x}x^k}{s\left( s+1 \right) ...\left( s+k \right)}}=x^s\,\Gamma \left( s \right) \,e^{-x}\sum_{k=0}^{\infty}{\frac{x^k}{\Gamma \left( s+k+1 \right)}},$$我们有

\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*}

其中$\displaystyle\Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t$为the upper incomplete gamma function,而$\displaystyle\gamma(s,x) = \int_0^x t^{s-1}\,e^{-t}\,{\rm d}t$为the lower incomplete gamma function.参考这里.

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

令$a\in (0,\pi)$,设$n$为正整数.证明$$\int_0^{\pi}{\frac{\cos \left( nx \right) -\cos \left( na \right)}{\cos x-\cos a}dx}=\pi \frac{\sin \left( na \right)}{\sin a}.$$

求$$\int_1^{\frac{\sqrt{5}+1}{2}}{\left( \frac{\arctan x}{\arctan x-x} \right) ^2dx},$$

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

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

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.
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 $x$.
Determine all entire functions $f$ that satisfying the inequality $$|f(z)|\leq|z|^2|{\rm Im}(z)|^2$$ for $z$ sufficlently large.
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.
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 $H_2$.
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.)

Problems of the Miklós Schweitzer Memorial Competition

2015年丘赛分析组个人赛试题