[家里蹲大学数学杂志]第276期丘成桐大学生数学竞赛2013年分析与方程团体赛试题参考解答
S.-T.YauCollege Student Mathematics Contests 2013
Analysis and Differential Equations Team
Please solve $5$ out of the following $6$ problems.
1. Supppose $\lap=\sed{z\in\bbC;|z|<1}$ is the open unit disk in the complex plane. Show that for any holomorphic function $f:\lap\to \lap$, $$\bee\label{1.eq} \frac{|f'(z)|}{1-|f(z)|^2}\leq\frac{1}{1-|z|^2}, \eee$$for all $z$ in $\lap$. If equality holds in \eqref{1.eq} for some $z_0\in \lap$, show that $f\in \Aut(\lap)$ and that $$\bex \frac{|f'(z)|}{1-|f(z)|^2}=\frac{1}{1-|z|^2}, \eex$$ for all $z\in\lap$.
Solution: This is just the Schwarz-Pick Lemma. For any $z\in \lap$, consider the mapping $$\beex \bea \varphi:\lap\to \lap,&\quad\xi\mapsto\cfrac{-\xi+z}{1-\bar z\xi};\\ f:\lap\to \lap,&\quad\eta\mapsto f(\eta);\\ \psi:\lap\to\lap,&\quad\zeta\mapsto \cfrac{-\zeta+f(z)}{1-\overline{f(z)}\zeta}. \eea \eeex$$ Then $\psi\circ f\circ \varphi:\lap\to\lap$ with $0$ being fixed. By Schwarz Lemma, $$\beex \bea |(\psi\circ f\circ \varphi)'(0)|&\leq 1,\\ |\psi'(f(z))|\cdot |f'(z)|\cdot |\varphi'(0)|&\leq 1,\\ \cfrac{1}{1-|f(z)|^2}\cdot |f'(z)|\cdot (1-|z|^2)&\leq 1,\\ \cfrac{|f'(z)|}{1-|f(z)|^2}&\leq\cfrac{1}{1-|z|^2}. \eea \eeex$$ And the equality holds for some $z_0\in \lap$, iff ($\varphi^{-1}=\varphi,\psi^{-1}=\psi$) $$\beex \bea \psi\circ f\circ \varphi&\in \Aut(\lap),\\ f=\psi\circ(\psi\circ f\circ \varphi)\circ \varphi&\in \Aut(\lap). \eea \eeex$$
2. Let $f$ be a function of bounded variation on $[a,b]$, $f_1$ its generalized derivative as a measure, i.e. $$\bex f(x)-f(a)=\int_a^x f_1(y)\rd y \eex$$ for every $x\in [a,b]$ and $f_1(x)$ is an integrable function on $[a,b]$. Let $f'$ be its weak derivative as a generalized function, i.e. $$\bex \int_a^b f(x)g'(x)\rd x=-\int_a^bf'(x)g(x)\rd x, \eex$$ for every smooth $g(x)$ on $[a,b]$, $g(a)=g(b)=0$. Show that
(1) If $f$ is absolutely continuous, then $f'=f_1$.
(2) If the weak derivative $f'$ of $f$ is an integrable function on $[a,b]$, then $f(x)$ is equal to an absolutely continuous function outside a set of measure zero.
Solution:
(1) If $f$ is absolutely continuous, then $$\beex \bea \int_a^b f(x)g'(x)\rd x&=\int_a^b \sez{f(a)+\int_a^x f_1(y)\rd y}g'(x)\rd x\\ &=-\int_a^b \frac{\rd }{\rd x} \sez{f(a)+\int_a^x f_1(y)\rd y}g(x)\rd x\\ &=-\int_a^b f_1(x)g(x)\rd x. \eea \eeex$$ Thus, $$\bex \int_a^b [f'(x)-f_1(x)]g(x)\rd x=0. \eex$$ Approximation arguments then yields that $f'=f_1$.
(2) Let $$\bex g(x)=f(a)+\int_a^xf'(t)\rd t. \eex$$ Then $g$ is absolutely continuous. Since $f'=g'$ $\ae$ (weak derivative), and $f(a)=g(a)$, we have $f=g$ $\ae$ as desired.
3. Show that the convex hull of the roots of any polynomial contains all its critical points as well as all the zeros of higher derivatives of the polynomial. Here the convex hull of a given bounded set in the plane is the smallest convex set containing the given set in the plane.
Solution: Let $$\bex p(z)=b\prod_{i=1}^n (z-a_i) \eex$$ be a polynomial of degree $n$ ($b\neq 0$). For $z$ satisfying $p'(z)=0$,
(1) if $p(z)=0$, then $z=a_i$ for some $i=1,2,\cdots,n$;
(2) if $p(z)\neq 0$, then $$\beex \bea 0&=\cfrac{p'(z)}{p(z)}=\sum_{i=1}^n\cfrac{1}{z-a_i} =\sum_{i=1}^n\cfrac{\bar z-\bar a_i}{|z-a_i|^2},\\ z\sum_{i=1}^n \cfrac{1}{|z-a_i|^2} &=\sum_{i=1}^n \cfrac{1}{|z-a_i|^2}a_i,\\ z&=\sum_{i=1}^n \cfrac{\cfrac{1}{|z-a_i|^2}}{\sum_{j=1}^n \cfrac{1}{|z-a_j|^2}}a_i. \eea \eeex$$ Inductively, we have also that the zeros of higher derivatives of $p(z)$ are all in the convex hull of $\sed{a_i}_{i=1}^n$.
4. Let $D\subset \bbR^3$ be open domain. Show that every smooth field ${\bf F}=(P,Q,R)$ over $D$ can be written as ${\bf F}={\bf F}_1+{\bf F}_2$ such that $$\bex \rot {\bf F}_1={\bf 0},\quad \Div {\bf F}_2=0, \eex$$ where $$\bex \rot {\bf F}=\sex{\frac{\p R}{\p y}-\frac{\p Q}{\p z},\frac{\p P}{\p z}-\frac{\p R}{\p x},\frac{\p Q}{\p x}-\frac{\p P}{\p y}},\quad \Div {\bf F}=\frac{\p P}{\p x}+\frac{\p Q}{\p y}+\frac{\p R}{\p z}. \eex$$
Solution: The following Poisson equation $$\bex \sedd{\ba{ll} \lap u=\Div {\bf F},&\mbox{in }D\\ u=0,&\mbox{on }\p D \ea} \eex$$ has a unique solution $u$. Setting $$\bex {\bf F_1}=\n u,\quad {\bf F_2}={\bf F}-{\bf F_1}. \eex$$ Then $\rot {\bf F_1}={\bf0}$, $\Div {\bf F_2}=\Div {\bf F}-\Div{\bf F_1}=\Div{\bf F}-\lap u=0$.
5. Let $\scrH$ be a Hilbert space and $A$ a compact self-adjoint linear operator over $\scrH$. Show that there exists an ortho-normal basis of $\scrH$ consisting of eigenvectors $\varphi_n$ of $A$ with non-zero eigenvalues $\lm_n$ such that every vector $\xi\in \scrH$ can be written as: $$\bex \xi=\sum_k c_k\varphi_k+\xi', \eex$$ where $\xi'\in \ker A$, i.e., $A\xi'=0$. We also have $$\bex A\xi=\sum_k \lm_kc_k\varphi_k. \eex$$ If there are infinitely many eigenvalues then $\dps{\lim_{n\to\infty}\lm_n=0}$.
Solution: This is the classical result in Functional Analysis.
6. A function $f:\bbR\to \bbR$ is called convex if $$\bex f(\lm x+(1-\lm)x')\leq \lm f(x)+(1-\lm)f(x') \eex$$ for $0\leq \lm\leq 1$ and each $x,x'\in\bbR$, and is called strictly convex if $$\bex f(\lm x+(1-\lm)x')<\lm f(x)+(1-\lm)f(x') \eex$$ for $0<\lm<1$. We assume $|f(x)|<\infty$ whenever $|x|<\infty$.
(1) Show that a convex function $f$ is continuous and the function $$\bex g(y)=\max_{x\in\bbR} [xy-f(x)] \eex$$ is a well-defined convex function over $\bbR$.
(2) Show that a convex function $f$ is differentiable except at most countably many points.
(3) $f$ is differentiable everywhere if both $f$ and $g$ are strictly convex.
Solution:
(1) Without loss of generality, we need only to prove that $f$ is continuous at $0$. Noticing that $$\beex \bea &\quad f(x)=f(x\cdot 1+(1-x)\cdot 0)\leq x\cdot f(1)+(1-x)\cdot f(0)\\ &\ra f(x)-f(0)\leq x\cdot [f(1)-f(0)],\\ &\quad f(0)=f\sex{\cfrac{x}{x+1}\cdot (-1)+\cfrac{1}{x+1}\cdot x} \leq \cfrac{x}{x+1}\cdot f(-1)+\cfrac{1}{x+1}\cdot f(x) \\ &\ra f(x)-f(0)\geq x\cdot [f(0)-f(-1)], \eea \eeex$$ we have $$\bex x\cdot [f(0)-f(-1)]\leq f(x)-f(0)\leq x\cdot [f(1)-f(0)], \eex$$ which implies the continuity of $f$.
(2) It is easy to see that $f$ is differentiable from left and from right, and the set where $f$ is not differentiable $$\bex Z=\sed{(f_-'(a),f_+'(a));\ f_-'(a)<f_+'(a)} \eex$$ is countable (Compare with the set $\bbQ$).
(3) Fix $y\in\bbR$. If there exists an $x_0$ such that $f'(x_0)=y$, then $$\beex \bea &\quad f(x)-f(x_0)\geq y(x-x_0)\\ &\ra xy-f(x)\leq x_0y-f(x_0). \eea \eeex$$ Otherwise, we would have an $x_0$ such that $$\bex f_-'(x_0)\leq y\leq f_+'(x_0)\quad (f_-'(x_0)<f_+'(x_0)). \eex$$ In this case, $$\beex \bea x<x_0&\ra \cfrac{f(x)-f(x_0)}{x-x_0}\leq f_-'(x_0)\leq y\ra xy-f(x)\leq x_0y-f(x_0),\\ x>x_0&\ra \cfrac{f(x)-f(x_0)}{x-x_0}\geq f_-'(x_0)\leq y\ra xy-f(x)\leq x_0y-f(x_0). \eea \eeex$$ Thus $g$ is well defined and the maximum is achieved at $x_0$.
(4) For $0\leq \tt\leq 1$, $y<y'$, $$\beex \bea &\quad x((1-\tt)y+\tt y')-f(x)\\ &=(1-\tt)[xy-f(x)]+\tt[xy'-f(x)]\\ &\leq (1-\tt)g(y)+\tt g(y'). \eea \eeex$$ This verifies the convexity of $g$.
(5) We prove the third statement by contradiction. Suppose that $f$ is strictly convex and and is not differentiable at some $x_0$. Then for $f_-'(x_0)<y<f_+'(x_0)$, $$\bex g(y)=x_0y-f(x_0), \eex$$ which is linear, contradicting to the strict convexity of $g$.