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 class="MathJax_Preview">5aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mn">5border="0" src="about:blank">
out of the following 6id="MathJax-Element-2-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mn">6border="0" src="about:blank">
problems.

 

 

1. Let A=[a_{ij}] id="MathJax-Element-3-Frame" aria-readOnly="true"
class="MathJax MathJax_Processed" role="textbox">id="MathJax-Span-7" class="math">>>class="mrow">class="mi">Aid="MathJax-Span-10" class="mo">=id="MathJax-Span-11" class="mo">[>>>id="MathJax-Span-13" class="mi">aclass="MathJax_strut" border="0" src="about:blank">id="MathJax-Span-14" class="texatom">class="mrow">id="MathJax-Span-16" class="mi">iid="MathJax-Span-17" class="mi">jclass="MathJax_strut" border="0" src="about:blank"> id="MathJax-Span-18"
class="mo">]border="0" src="about:blank">
be a real symmetric n\times nid="MathJax-Element-4-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">nid="MathJax-Span-22" class="mo">×id="MathJax-Span-23" class="mi">nclass="MathJax_strut" border="0" src="about:blank">
matrix. Define f:\bbR^n\to\bbRid="MathJax-Element-5-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">f id="MathJax-Span-27"
class="mo">:class="msubsup">>>class="texatom">class="texatom">id="MathJax-Span-33" class="mi">Rclass="MathJax_strut" border="0" src="about:blank"> id="MathJax-Span-34"
class="mi">nborder="0" src="about:blank">id="MathJax-Span-35" class="mo">→class="texatom">class="texatom">id="MathJax-Span-40" class="mi">R class="MathJax_strut" border="0" src="about:blank">>
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)id="MathJax-Element-7-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">f id="MathJax-Span-95"
class="mo">∈class="msubsup">>>id="MathJax-Span-97" class="mi">Lclass="MathJax_strut" border="0" src="about:blank"> id="MathJax-Span-98"
class="mn">1border="0" src="about:blank">id="MathJax-Span-99" class="mo">(>>>class="texatom">class="texatom">id="MathJax-Span-105" class="mi">Rclass="MathJax_strut" border="0" src="about:blank"> id="MathJax-Span-106"
class="mi">nborder="0" src="about:blank">id="MathJax-Span-107" class="mo">)class="MathJax_strut" border="0" src="about:blank">
if and only if the matrix Aid="MathJax-Element-8-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">Aborder="0" src="about:blank">
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 Aid="MathJax-Element-10-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">Aborder="0" src="about:blank">
is positive definite.

Solution: Since class="MathJax_Preview">Aaria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">Aborder="0" src="about:blank">
is real, symmetric, we may find an orthogonal matrix P id="MathJax-Element-12-Frame" aria-readOnly="true"
class="MathJax MathJax_Processed" role="textbox">id="MathJax-Span-187" class="math">>>class="mrow">class="mi">P class="MathJax_strut" border="0" src="about:blank">>
such that \bex P^tAP=\diag(\lm_1,\cdots,\lm_n),
\eex

with \lm_iid="MathJax-Element-14-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">>>class="texatom">id="MathJax-Span-231" class="mi">λclass="MathJax_strut" border="0" src="about:blank"> id="MathJax-Span-232"
class="mi">iborder="0" src="about:blank">class="MathJax_strut" border="0" src="about:blank">
being the eigenvalues of Aid="MathJax-Element-15-Frame" aria-readOnly="true" class="MathJax MathJax_Processed"
role="textbox">class="math">>>class="mrow">class="mi">Aborder="0" src="about:blank">
. 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 class="MathJax_Preview">\lm_i>0id="MathJax-Element-17-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, which is equivalent to that Aid="MathJax-Element-18-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is positive definite. If Aid="MathJax-Element-19-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
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 id="MathJax-Element-23-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
be a simply connected region in the complex plane and V\neq
\bbCaria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
. Let a,bid="MathJax-Element-25-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
be two distinct points in Vid="MathJax-Element-26-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
. Let \phi_1,\phi_2id="MathJax-Element-27-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
be two one-to-one holomorphic maps of class="MathJax_Preview">Varia-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
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) id="MathJax-Element-32-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
. Consequently, \phi\equiv \phi_2^{-1}\circ
\phi_1\in \Aut(\lap)aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
has two fixed points a,bid="MathJax-Element-34-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
. Let \varphi_a(\zeta)=\cfrac{-\zeta+a}{1-\bar
a\zeta}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
, then \varphi_a\circ \phi\circ
\varphi_aaria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
have two fixed points 0id="MathJax-Element-37-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
and \cfrac{-b+a}{1-\bar ab}id="MathJax-Element-38-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
. 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]id="MathJax-Element-40-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
consider a subset \bex E=\sed{x;\ \mbox{ in the
decimal expansion of }x\mbox{ there is no }4}, \eex

show that Eid="MathJax-Element-42-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is measurable and calculate its measure.

Solution: Just as the
construction of the Cantor set, E id="MathJax-Element-43-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
is measurable and complete. Moreover, \bex
m\,E=1-\sum_{i=1}^\infty\frac{9^{i-1}}{10^i}=0. \eex

 

 

 

4. Let 1 id="MathJax-Element-45-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
L^p([0,1],\rd m)id="MathJax-Element-46-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
be the completion of C[0,1]id="MathJax-Element-47-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
with the norm: \bex \sen{f}_p=\sex{\int_0^1
|f(x)|^p\rd m}^\frac{1}{p}, \eex

where \rd mid="MathJax-Element-49-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is the Lebesgue measure. Show that \bex
\lim_{\lm\to\infty}\lm^p\cdot m\sex{x;\ |f(x)|>\lm}=0. \eex

Solution: Let class="MathJax_Preview">\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 class="MathJax_Preview">\ve>0aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
, there exists a \delta>0id="MathJax-Element-55-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
such that for any measurable A\subset
[0,1]aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
with m\,A<\deltaid="MathJax-Element-57-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, we have \bex \int_A |f(x)|^p\rd m<\ve.
\eex

For this \delta>0id="MathJax-Element-59-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, we have by \eqref{4:eq}id="MathJax-Element-60-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
that \bex \exists\ \vLa>0,\st \lm\geq \vLa\ra
m\, E_\lm<\delta. \eex

Cosequently, when \lm\geq \vLaid="MathJax-Element-62-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, \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} id="MathJax-Element-64-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
, \nu=1,2,\cdots,nid="MathJax-Element-65-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
or \nu=1,2,\cdotsid="MathJax-Element-66-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is an orthonomal basis in an inner product space class="MathJax_Preview">Haria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
. Let Eid="MathJax-Element-68-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
be the closed linear subspace spanned by class="MathJax_Preview">\mathfrak{F}id="MathJax-Element-69-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
. For any x\in Hid="MathJax-Element-70-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
show that the following are equivalent:

(1) x\in
Earia-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
;

(2) \dps{\sen{x}^2=\sum_\nu
|\sef{x,e_\nu}|^2}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
;

(3) \dps{x=\sum_\nu
(x,e_\nu)e_\nu}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
. let H=L^2[0,2\pi]id="MathJax-Element-74-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
with the inner product \bex
\sef{f,g}=\frac{1}{\pi}\in_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 id="MathJax-Element-77-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
spaned by \mathfrak{F}id="MathJax-Element-78-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is Hid="MathJax-Element-79-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
.

Solution: This is the standard result in
Functional Analysis.

 

 

 

6. Let \scrH=L^2[0,1] id="MathJax-Element-80-Frame" aria-readOnly="true"
class="MathJax MathJax_Processing" role="textbox">
relative to the Lebesgue measure and define \bex
(Kf)(s)=\int_0^s f(t)\rd t \eex

for each fid="MathJax-Element-82-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
in \scrHid="MathJax-Element-83-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
. Show that Kid="MathJax-Element-84-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
is a compact operator without eigenvalues.

Solution:

(1) Suppose that class="MathJax_Preview">\sed{f_n}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
is bounded in L^2[0,1]id="MathJax-Element-86-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
(with bound Mid="MathJax-Element-87-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
). 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 class="MathJax_Preview">\lmaria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
is an eigenvalue of Kid="MathJax-Element-93-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
with the corresponding eigen-function 0\neq f\in
L^2[0,1]aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
. 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 0id="MathJax-Element-96-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
(otherwise, \dps{\int_0^s f(t)\rd t=0,\ \forall\
s;\ f\equiv 0}aria-readOnly="true" class="MathJax MathJax_Processing" role="textbox">
), and \bex \sedd{\ba{ll}
f‘(s)&=\cfrac{1}{\lm}f(s),\\ f(0)&=0. \ea} \eex

This implies that f\equiv 0id="MathJax-Element-99-Frame" aria-readOnly="true" class="MathJax MathJax_Processing"
role="textbox">
, which is a contradiction. 

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,布布扣,bubuko.com