S.-T. Yau College Student Mathematics Contests 2022

Problem 1. For $n\ge 1$$n\ge 1$n >= 1n \geq 1$n\ge 1$, we consider the integral

Prove that $\underset{n\to \mathrm{\infty }}{lim}{I}_n$$\underset{n\to \mathrm{\infty }}{lim} I_n$lim_(n rarr oo)I_(n)\lim _{n \rightarrow \infty} I_{n}$\underset{n\to \mathrm{\infty }}{lim}{I}_n$ exists.

Solution: For all positive integers $m$$m$mm$m$ and $n$$n$nn$n$, for all $x,y>0$$x,y>0$x,y > 0x, y>0$x,y>0$, we check that

Thus, $J_n=n{I}_n$$J_n=n{I}_n$J_(n)=nI_(n)J_{n}=n I_{n}$J_n=n{I}_n$ satisfies

It is well-known that $\underset{n\to \mathrm{\infty }}{lim}\frac{J_n}{n}$$\underset{n\to \mathrm{\infty }}{lim} \frac{J_n}{n}$lim_(n rarr oo)(J_(n))/(n)\lim _{n \rightarrow \infty} \frac{J_{n}}{n}$\underset{n\to \mathrm{\infty }}{lim}\frac{J_n}{n}$ exists.

Problem 2. Let $U\subset \mathbf{C}$$U\subset \mathbf{C}$U subCU \subset \mathbf{C}$U\subset \mathbf{C}$ be a non-empty open set and $f:U\to U$$f:U\to U$f:U rarr Uf: U \rightarrow U$f:U\to U$ be a non-constant holomorphic function. Prove that, if $f\circ f=f$$f\circ f=f$f@f=ff \circ f=f$f\circ f=f$, then $f(z)\equiv z$$f(z)\equiv z$f(z)-=zf(z) \equiv z$f(z)\equiv z$ for all $z\in U$$z\in U$z in Uz \in U$z\in U$.

Solution: Since $f$$f$ff$f$ is not a constant map, $V:=f(U)\subset U$$V:=f(U)\subset U$V:=f(U)sub UV:=f(U) \subset U$V:=f(U)\subset U$ is an nonempty open set (the open mapping property). Thus, for $z\in V$$z\in V$z in Vz \in V$z\in V$, we have $z=f(z)$$z=f(z)$z=f(z)z=f(z)$z=f(z)$. This implies that $f(z)\equiv z$$f(z)\equiv z$f(z)-=zf(z) \equiv z$f(z)\equiv z$.

Problem 3. Let $X\subset \mathbf{R}$$X\subset \mathbf{R}$X subRX \subset \mathbf{R}$X\subset \mathbf{R}$ be a set with positive (Lebesgue) measure. Show that we can find an arithmetic progression of 2022 terms in $X$$X$XX$X$, i.e., there exists $x_1,\dots ,x_{2022}\in X$$x_1,\dots ,x_{2022}\in X$x_(1),dots,x_(2022)in Xx_{1}, \ldots, x_{2022} \in X$x_1,\dots ,x_{2022}\in X$ so that the $x_{i+1}-x_i$$x_{i+1}-x_i$x_(i+1)-x_(i)x_{i+1}-x_{i}$x_{i+1}-x_i$ 's are all equal and positive, $i=1,\dots ,2021$$i=1,\dots ,2021$i=1,dots,2021i=1, \ldots, 2021$i=1,\dots ,2021$.

Solution: We use $m$$m$mm$m$ to denote the Lebesgue measure. Let $x\in X$$x\in X$x in Xx \in X$x\in X$ be a Lebesgue point; therefore, there exists an interval $I$$I$II$I$ so that $x\in I$$x\in I$x in Ix \in I$x\in I$ and $\frac{m(InX)}{m(I)}\ge 1-\epsilon$$\frac{m(InX)}{m(I)}\ge 1-\epsilon$(m(InX))/(m(I)) >= 1-epsi\frac{m(I n X)}{m(I)} \geq 1-\varepsilon$\frac{m(InX)}{m(I)}\ge 1-\epsilon$ and $ϵ$$ϵ$epsilon\epsilon$ϵ$ will be determined at the end of the proof. By translating and rescaling, we may assume that $I=[0,1]$$I=[0,1]$I=[0,1]I=[0,1]$I=[0,1]$. We divide $I$$I$II$I$ into 2022 intervals:

Let $X_k=(I_k\cap X)-\frac{k-1}{2022}$$X_k=\left(I_k\cap X\right)-\frac{k-1}{2022}$X_(k)=(I_(k)nn X)-(k-1)/(2022)X_{k}=\left(I_{k} \cap X\right)-\frac{k-1}{2022}$X_k=(I_k\cap X)-\frac{k-1}{2022}$ be the translation of $I_k\cap X$$I_k\cap X$I_(k)nn XI_{k} \cap X$I_k\cap X$ and $X_k\subset I_1,k=1,\dots ,2022$$X_k\subset I_1,k=1,\dots ,2022$X_(k)subI_(1),k=1,dots,2022X_{k} \subset I_{1}, k=1, \ldots, 2022$X_k\subset I_1,k=1,\dots ,2022$. We know that

Thus,

We may take $ϵ=2023$$ϵ=2023$epsilon=2023\epsilon=2023$ϵ=2023$, thus, $\bigcap _{1\le k\le 2022}{X}_k\ne \varnothing$$\bigcap _{1\le k\le 2022} X_k\ne \varnothing$nnn_(1 <= k <= 2022)X_(k)!=O/\bigcap_{1 \leq k \leq 2022} X_{k} \neq \varnothing$\bigcap _{1\le k\le 2022}{X}_k\ne \varnothing$. Let $x_1\in \bigcap _{1\le k\le 2022}{X}_k$$x_1\in \bigcap _{1\le k\le 2022} X_k$x_(1)innnn_(1 <= k <= 2022)X_(k)x_{1} \in \bigcap_{1 \leq k \leq 2022} X_{k}$x_1\in \bigcap _{1\le k\le 2022}{X}_k$. Then $x_k=x_1+\frac{k-1}{2022}$$x_k=x_1+\frac{k-1}{2022}$x_(k)=x_(1)+(k-1)/(2022)x_{k}=x_{1}+\frac{k-1}{2022}$x_k=x_1+\frac{k-1}{2022}$ is the arithmetic progression.

Problem 4. Let $C([0,1])$$C([0,1])$C([0,1])C([0,1])$C([0,1])$ be the space of all continuous $\mathbf{C}$$\mathbf{C}$C\mathbf{C}$\mathbf{C}$-valued functions equipped with $L^{\mathrm{\infty }}$$L^{\mathrm{\infty }}$L^(oo)L^{\infty}$L^{\mathrm{\infty }}$-norm. Let $\mathbf{P}\subset$$\mathbf{P}\subset$Psub\mathbf{P} \subset$\mathbf{P}\subset$ $C([0,1])$$C([0,1])$C([0,1])C([0,1])$C([0,1])$ be a closed linear subspace. Assume that the elements of $\mathbf{P}$$\mathbf{P}$P\mathbf{P}$\mathbf{P}$ are polynomials. Prove that $\dim \mathbf{P}<\mathrm{\infty }$$\dim \mathbf{P}<\mathrm{\infty }$dim P < oo\operatorname{dim} \mathbf{P}<\infty$\dim \mathbf{P}<\mathrm{\infty }$.

Solution: Let $I=\{(x,y)\in [0,1{]}^2\mid x\ne y\}$$I=\left\{(x,y)\in [0,1{]}^2\mid x\ne y\right\}$I={(x,y)in[0,1]^(2)∣x!=y}I=\left\{(x, y) \in[0,1]^{2} \mid x \neq y\right\}$I=\{(x,y)\in [0,1{]}^2\mid x\ne y\}$. For each $(x,y)\in I$$(x,y)\in I$(x,y)in I(x, y) \in I$(x,y)\in I$, we define a mapping

Therefore, we have

Since $\mathbf{P}$$\mathbf{P}$P\mathbf{P}$\mathbf{P}$ is closed, we can apply the Banach-Steinhaus Theorem: there exists $C>0$$C>0$C > 0C>0$C>0$, so that

We consider the unit ball of $\mathbf{P}$$\mathbf{P}$P\mathbf{P}$\mathbf{P}$ :

Hence, for all $u\in B$$u\in B$u in Bu \in B$u\in B$, we have

Thus, the family $B$$B$BB$B$ is equicontinuous. By the Arzelà-Ascoli Theorem, it is compact. Thus, $\mathbf{P}$$\mathbf{P}$P\mathbf{P}$\mathbf{P}$ is finite-dimensional.

Problem 5. Let $\mathrm{\Omega }\subset {\mathbf{R}}^3$$\mathrm{\Omega }\subset {\mathbf{R}}^3$Omega subR^(3)\Omega \subset \mathbf{R}^{3}$\mathrm{\Omega }\subset {\mathbf{R}}^3$ be a bounded domain with smooth boundary. Assume that $u\in C\left(\overline{{\mathbf{R}}^3-\mathrm{\Omega }}\right)$$u\in C\left(\overline{{\mathbf{R}}^3-\mathrm{\Omega }}\right)$u in C( bar(R^(3)-Omega))u \in C\left(\overline{\mathbf{R}^{3}-\Omega}\right)$u\in C\left(\overline{{\mathbf{R}}^3-\mathrm{\Omega }}\right)$ is a harmonic function on ${\mathbf{R}}^3-\mathrm{\Omega }$${\mathbf{R}}^3-\mathrm{\Omega }$R^(3)-Omega\mathbf{R}^{3}-\Omega${\mathbf{R}}^3-\mathrm{\Omega }$ so that ${u|}_{\mathrm{\Omega }}=1$${\left.u\right|}_{\mathrm{\Omega }}=1$u|_(Omega)=1\left.u\right|_{\Omega}=1${u|}_{\mathrm{\Omega }}=1$ and $\underset{|x|\to \mathrm{\infty }}{lim}|u(x)|=0$$\underset{|x|\to \mathrm{\infty }}{lim} |u(x)|=0$lim_(|x|rarr oo)|u(x)|=0\lim _{|x| \rightarrow \infty}|u(x)|=0$\underset{|x|\to \mathrm{\infty }}{lim}|u(x)|=0$.

Prove that for such $u,\underset{|x|\to \mathrm{\infty }}{lim}|x|u(x)$$u,\underset{|x|\to \mathrm{\infty }}{lim} |x|u(x)$u,lim_(|x|rarr oo)|x|u(x)u, \lim _{|x| \rightarrow \infty}|x| u(x)$u,\underset{|x|\to \mathrm{\infty }}{lim}|x|u(x)$ exists.

Solution: Let $\phi (x)\in C^{\mathrm{\infty }}\left({\mathbf{R}}^3\right)$$\phi (x)\in C^{\mathrm{\infty }}\left({\mathbf{R}}^3\right)$varphi(x)inC^(oo)(R^(3))\varphi(x) \in C^{\infty}\left(\mathbf{R}^{3}\right)$\phi (x)\in C^{\mathrm{\infty }}\left({\mathbf{R}}^3\right)$ so that $\phi \equiv 0$$\phi \equiv 0$varphi-=0\varphi \equiv 0$\phi \equiv 0$ on an open neighborhood of $\mathrm{\Omega }$$\mathrm{\Omega }$Omega\Omega$\mathrm{\Omega }$ and $\phi \equiv 1$$\phi \equiv 1$varphi-=1\varphi \equiv 1$\phi \equiv 1$ for $|x|\ge R$$|x|\ge R$|x| >= R|x| \geq R$|x|\ge R$ where $R>0$$R>0$R > 0R>0$R>0$ is a sufficiently large number. Therefore, we can regard $\phi \cdot u$$\phi \cdot u$varphi*u\varphi \cdot u$\phi \cdot u$ as a smooth function defined on ${\mathbf{R}}^3$${\mathbf{R}}^3$R^(3)\mathbf{R}^{3}${\mathbf{R}}^3$. Hence,

where $\rho \equiv 0$$\rho \equiv 0$rho-=0\rho \equiv 0$\rho \equiv 0$ for $|x|\ge R$$|x|\ge R$|x| >= R|x| \geq R$|x|\ge R$. Therefore, for sufficiently large $|x|$$|x|$|x||x|$|x|$, we have

Therefore,

Since $|y|\le R,\frac{|x|}{|y-x|}$$|y|\le R,\frac{|x|}{|y-x|}$|y| <= R,(|x|)/(|y-x|)|y| \leq R, \frac{|x|}{|y-x|}$|y|\le R,\frac{|x|}{|y-x|}$ converges uniformly to 1 as $|x|\to \mathrm{\infty }$$|x|\to \mathrm{\infty }$|x|rarr oo|x| \rightarrow \infty$|x|\to \mathrm{\infty }$, the conclusion follows.

Problem 6. Let $f(x,y)\in C^1\left({\mathbf{R}}^2\right)$$f(x,y)\in C^1\left({\mathbf{R}}^2\right)$f(x,y)inC^(1)(R^(2))f(x, y) \in C^{1}\left(\mathbf{R}^{2}\right)$f(x,y)\in C^1\left({\mathbf{R}}^2\right)$. We assume that there exists $C>0$$C>0$C > 0C>0$C>0$ so that for all $(x,y)\in {\mathbf{R}}^2,|\frac{\mathrm{\partial }f}{\mathrm{\partial }y}(x,y)|\le C$$(x,y)\in {\mathbf{R}}^2,\left|\frac{\mathrm{\partial }f}{\mathrm{\partial }y}(x,y)\right|\le C$(x,y)inR^(2),|(del f)/(del y)(x,y)| <= C(x, y) \in \mathbf{R}^{2},\left|\frac{\partial f}{\partial y}(x, y)\right| \leq C$(x,y)\in {\mathbf{R}}^2,|\frac{\mathrm{\partial }f}{\mathrm{\partial }y}(x,y)|\le C$. Prove that the following ODE has a globally defined solution for all $y(0)=y_0\in \mathbf{R}$$y(0)=y_0\in \mathbf{R}$y(0)=y_(0)inRy(0)=y_{0} \in \mathbf{R}$y(0)=y_0\in \mathbf{R}$ :

In addition, we assume that $f$$f$ff$f$ is 1-periodic in $x$$x$xx$x$, i.e., for all $(x,y)\in {\mathbf{R}}^2$$(x,y)\in {\mathbf{R}}^2$(x,y)inR^(2)(x, y) \in \mathbf{R}^{2}$(x,y)\in {\mathbf{R}}^2$, we have $f(x+1,y)=f(x,y)$$f(x+1,y)=f(x,y)$f(x+1,y)=f(x,y)f(x+1, y)=f(x, y)$f(x+1,y)=f(x,y)$. Prove that if (1) admits a globally defined bounded solution, then (1) admits a periodic solution.

Solution: The global existence is easy: fix an interval $[0,a)$$[0,a)$[0,a)[0, a)$[0,a)$, we have

where $M=\underset{x\in [0,a]}{sup}|f(x,y(0))|$$M=\underset{x\in [0,a]}{sup} |f(x,y(0))|$M=s u p_(x in[0,a])|f(x,y(0))|M=\sup _{x \in[0, a]}|f(x, y(0))|$M=\underset{x\in [0,a]}{sup}|f(x,y(0))|$. By Gronwall's inequality, $y$$y$yy$y$ is bounded all the way up to $[0,a]$$[0,a]$[0,a][0, a]$[0,a]$. We can then extend $f$$f$ff$f$ across $a$$a$aa$a$. This shows the solution can be defined globally.

Assume that $\phi$$\phi$varphi\varphi$\phi$ is a bounded solution. We may assume that $\phi (1)\ne \phi (0)$$\phi (1)\ne \phi (0)$varphi(1)!=varphi(0)\varphi(1) \neq \varphi(0)$\phi (1)\ne \phi (0)$. Otherwise, $\phi$$\phi$varphi\varphi$\phi$ is a periodic solution. Without loss of generality, we may assume that $\phi (1)>\phi (0)$$\phi (1)>\phi (0)$varphi(1) > varphi(0)\varphi(1)>\varphi(0)$\phi (1)>\phi (0)$. By comparing two solutions $\phi (x)$$\phi (x)$varphi(x)\varphi(x)$\phi (x)$ and $\phi (x+1)$$\phi (x+1)$varphi(x+1)\varphi(x+1)$\phi (x+1)$ of (1), we see that $\phi (0)<\phi (1)<\cdots <\phi (n)<\cdots$$\phi (0)<\phi (1)<\cdots <\phi (n)<\cdots$varphi(0) < varphi(1) < cdots < varphi(n) < cdots\varphi(0)<\varphi(1)<\cdots<\varphi(n)<\cdots$\phi (0)<\phi (1)<\cdots <\phi (n)<\cdots$. Thus, by the boundedness of $\phi$$\phi$varphi\varphi$\phi$, we may assume that

Therefore, the solution to (1) with $y_{\ast }$$y_{\ast }$y_(**)y_{*}$y_{\ast }$ as the initial data is a 1-periodic solution.