[家里蹲大学数学杂志]第277期丘成桐大学生数学竞赛2012年分析与方程团体赛试题参考解答

S.-T.YauCollege Student Mathematics Contests 2012

Analysis and Differential Equations Team

Please solve $5$ out of the following $6$ problems.

1. Let $A=[a_{ij}]$ be a real symmetric $n\times n$ matrix. Define $f:\bbR^n\to\bbR$ by $$\bex f(x_1,\cdots,x_n)=\exp\sex{-\frac{1}{2}\sum_{i,j=1}^n a_{ij}x_ix_j}. \eex$$ Prove that $f\in L^1(\bbR^n)$ if and only if the matrix $A$ is positive definite. Compute $$\bex \int_{\bbR^n} \exp\sex{-\frac{1}{2}\sum_{i,j=1}^n a_{ij}x_ix_j+\sum_{i=1}^n b_ix_i}\rd x \eex$$ when $A$ is positive definite.

Solution: Since $A$ is real, symmetric, we may find an orthogonal matrix $P$ such that $$\bex P^tAP=\diag(\lm_1,\cdots,\lm_n), \eex$$ with $\lm_i$ being the eigenvalues of $A$. Then $$\beex \bea \int_{\bbR^n} f(x)\rd x &=\int_{\bbR^n}\exp\sex{-\frac{1}{2}x^tAx}\rd x\\ &=\int_{\bbR^n}\exp\sex{-\frac{1}{2}\sum_{i=1}^n \lm_iy_i^2}\rd y\quad(x=Py)\\ &=\int_{\bbR^n}\prod_{i=1}^n \exp\sex{-\frac{1}{2}\lm_iy_i^2}\rd y\\ &=\prod_{i=1}^n \int_{\bbR} \exp\sex{-\frac{1}{2}\lm_iy_i^2}\rd y_i \eea \eeex$$ is finite if and only if all $\lm_i>0$, which is equivalent to that $A$ is positive definite. If $A$ is positive definite, then $$\beex \bea I&\equiv \int_{\bbR^n} \exp\sex{-\frac{1}{2}\sum_{i,j=1}^n a_{ij}x_ix_j+\sum_{i=1}^n b_ix_i}\rd x\\ &=\int_{\bbR^n} \exp\sex{ -\frac{1}{2}\sum_{i=1}^n \lm_i^2y_i^2 +\sum_{i=1}^n c_iy_i }\rd y\quad\sex{x=Py,\ b^tP=c^t}\\ &=\prod_{i=1}^n \int_{\bbR} \exp\sex{-\frac{1}{2}\lm_iy_i^2+c_iy_i}\rd y_i\\ &=\prod_{i=1}^n \int_{\bbR} \exp\sez{-\frac{1}{2}\lm_i \sex{y_i-\frac{c_i}{\lm_i}}^2+\frac{c_i^2}{2\lm_i}}\rd y_i\\ &=\exp\sex{\frac{1}{2}\sum_{i=1}^n\frac{c_i^2}{\lm_i}} \cdot \prod_{i=1}^n \exp\sex{-\frac{1}{2}\lm_iy_i^2}\rd y_i. \eea \eeex$$ Due to the fact that $$\beex \bea P^tAP=\diag(\lm_1,\cdots,\lm_n) &\ra A=P\,\diag(\lm_1,\cdots,\lm_n)P^{-1}\\ &\ra A^{-1}=P\,\diag\sex{\frac{1}{\lm_1},\cdots,\frac{1}{\lm_n}}P^{-1}\\ &\ra \tilde a_{jk}=\sum_i p_{ji}\frac{1}{\lm_i}p_{ki},\\ c^t=b^tP&\ra c=P^tb\\ &\ra c_i=\sum_j p_{ji}b_j\\ &\ra c_i^2=\sum_{j,k}p_{ji}b_jp_{ki}b_k\\ &\ra \sum_i\frac{c_i^2}{\lm_i} =\sum_{jk}b_kb_j\sum_ip_{ji}\frac{1}{\lm_i}p_{ki}\\ &\quad\ =\sum_{jk}b_j\tilde a_{jk}b_k =b^tA^{-1}b, \eea \eeex$$ we have $$\bex I=\exp\sex{\frac{1}{2}b^tA^{-1}b}\cdot \prod_{i=1}^n \frac{\sqrt{2\pi}}{\sqrt{\lm_i}} =\frac{(2\pi)^\frac{n}{n}}{|A|^\frac{1}{2}} e^{\frac{1}{2}b^tA^{-1}b}. \eex$$

2. Let $V$ be a simply connected region in the complex plane and $V\neq \bbC$. Let $a,b$ be two distinct points in $V$. Let $\phi_1,\phi_2$ be two one-to-one holomorphic maps of $V$ onto itself. If $$\bex \phi_1(a)=\phi_2(a),\mbox{ and } \phi_1(b)=\phi_2(b), \eex$$ show that $$\bex \phi_1(z)=\phi_2(z),\quad \forall\ z\in V. \eex$$

Solution: By the Riemann mapping theorem, we may assume without loss of generality that $$\bex V=\lap=\sed{z\in\bbC;\ |z|<1}. \eex$$ And hence $\phi_1,\phi_2\in \Aut(\lap)$. Consequently, $\phi\equiv \phi_2^{-1}\circ \phi_1\in \Aut(\lap)$ has two fixed points $a,b$. Let $\varphi_a(\zeta)=\cfrac{-\zeta+a}{1-\bar a\zeta}$, then $\varphi_a\circ \phi\circ \varphi_a$ have two fixed points $0$ and $\cfrac{-b+a}{1-\bar ab}$. By the Schwarz Lemma, we have $$\beex \bea \varphi_a\circ \phi\circ \varphi_a=\id\ra \phi=\id \ra \phi_1=\phi_2. \eea \eeex$$

3. In the unit interval $[0,1]$ consider a subset $$\bex E=\sed{x;\ \mbox{ in the decimal expansion of }x\mbox{ there is no }4}, \eex$$ show that $E$ is measurable and calculate its measure.

Solution: Just as the construction of the Cantor set, $E$ is measurable and complete. Moreover, $$\bex m\,E=1-\sum_{i=1}^\infty\frac{9^{i-1}}{10^i}=0. \eex$$

4. Let $1<p<\infty$ $L^p([0,1],\rd m)$ be the completion of $C[0,1]$ with the norm: $$\bex \sen{f}_p=\sex{\int_0^1 |f(x)|^p\rd m}^\frac{1}{p}, \eex$$ where $\rd m$ is the Lebesgue measure. Show that $$\bex \lim_{\lm\to\infty}\lm^p\cdot m\sex{x;\ |f(x)|>\lm}=0. \eex$$

Solution: Let $$\bex E_\lm=\sed{x;\ |f(x)|>\lm}. \eex$$ Then $$\bex \lm^p\cdot m\, E_\lm\leq \int_{E_\lm}|f(x)|^p\rd m \leq \int_0^1 |f(x)|^p\rd m \eex$$ implies that $$\bee\label{4:eq} mE_\lm\to0\quad(\lm\to\infty). \eee$$By the absolute continuity, for any $\ve>0$, there exists a $\delta>0$ such that for any measurable $A\subset [0,1]$ with $m\,A<\delta$, we have $$\bex \int_A |f(x)|^p\rd m<\ve. \eex$$ For this $\delta>0$, we have by \eqref{4:eq} that $$\bex \exists\ \vLa>0,\st \lm\geq \vLa\ra m\, E_\lm<\delta. \eex$$ Cosequently, when $\lm\geq \vLa$, $$\bex \lm\cdot m\, E_\lm\leq \int_{E_\lm}|f(x)|^p\rd m<\ve. \eex$$ This completes the proof of the problem.

5. let $\mathfrak{F}=\sed{e_\nu}$, $\nu=1,2,\cdots,n$ or $\nu=1,2,\cdots$ is an orthonomal basis in an inner product space $H$. Let $E$ be the closed linear subspace spanned by $\mathfrak{F}$. For any $x\in H$ show that the following are equivalent:

(1) $x\in E$;

(2) $\dps{\sen{x}^2=\sum_\nu |\sef{x,e_\nu}|^2}$;

(3) $\dps{x=\sum_\nu (x,e_\nu)e_\nu}$. let $H=L^2[0,2\pi]$ with the inner product $$\bex \sef{f,g}=\frac{1}{\pi}\int_0^{2\pi}f(x)g(x)\rd x, \eex$$ $$\bex \mathfrak{F}=\sed{\frac{1}{2},\cos x,\sin x,\cdots,\cos nx,\sin nx,\cdots} \eex$$ be an orthonormal basis. Show that the closed linear subspace $E$ spaned by $\mathfrak{F}$ is $H$.

Solution: This is the standard result in Functional Analysis.

6. Let $\scrH=L^2[0,1]$ relative to the Lebesgue measure and define $$\bex (Kf)(s)=\int_0^s f(t)\rd t \eex$$ for each $f$ in $\scrH$. Show that $K$ is a compact operator without eigenvalues.

Solution:

(1) Suppose that $\sed{f_n}$ is bounded in $L^2[0,1]$ (with bound $M$). By reflexivity, $$\bex \exists\ \sed{n_k},\st f_{n_k}\rhu f. \eex$$ Thus by the definition of weak compactness, $$\bex (Kf_{n_k})(s) =\int_0^1 \chi_{[0,s]}(t)f_{n_k}(t)\rd t \to \int_0^1 \chi_{[0,s]}(t)f(t)\rd t=(Kf)(s),\quad 0\leq s\leq 1. \eex$$ Consequently, by Lebesgue's dominated convergence theorem and the fact that $$\beex \bea |Kf_{n_k}(s)-Kf(s)|^2 &=\sev{\int_0^s [f_{n_k}(t)-f(t)]\rd t}^2\\ &\leq s\cdot \int_0^s |f_{n_k}(t)-f(t)|^2\rd t\\ &\leq 2\int_0^1 [|f_{n_k}(t)|^2+|f(t)|^2]\rd t\\ &\leq 4M\\ &\in L^1[0,1], \eea \eeex$$ we have $$\bex \sen{Kf_{n_k}-Kf}_{L^2}\to 0\quad(k\to\infty). \eex$$

(2) Suppose $\lm$ is an eigenvalue of $K$ with the corresponding eigen-function $0\neq f\in L^2[0,1]$. Then $$\bex Kf=\lm f\ra \int_0^s f(t)\rd t=\lm f(s),\quad 0\leq s\leq 1. \eex$$ Consequently, $\lm\neq 0$ (otherwise, $\dps{\int_0^s f(t)\rd t=0,\ \forall\ s;\ f\equiv 0}$), and $$\bex \sedd{\ba{ll} f'(s)&=\cfrac{1}{\lm}f(s),\\ f(0)&=0. \ea} \eex$$ This implies that $f\equiv 0$, which is a contradiction.