数学分析精讲回忆版

作者：黄景娟

一 判断题

1-1

一阶导等于 $0$, $x_0$ 存在严格局部极小值当且仅当二阶导大于 $0$

错误 ❌

1-2

${\mathbb{R}}^n$ 任意范数等价

1-3

$x\in (0,1)$, $x^n$ 不一致收敛到 $0$

1-4

等度连续则一致连续

1-5

$a>0$, ${\log }_{a}x$ 不是凸函数

1-6

紧空间存在可数稠密子集

1-7

$f_n$ 连续 + 逐点收敛不能推 $f$ 连续

1-8

偏导数存在不能推可微

1-9

可逆线性算子是 $L({\mathbb{R}}^n)$ 的开集

1-10

还有一个不记得了

二 积分

2-1

${\int }_0^{\pi }{\sin }^{3}xdx$

2-2

${\int }_1^{2}x\ln xdx$

三

$f$ 连续可微, $f(1)=0$, ${\int }_0^1{f}^4(x)dx=1$, 证明 ${\int }_0^1{f}^{\prime }(x{)}^{2}dx{\int }_0^1{x}^2{f}^6(x)dx>\frac{1}{16}$

解:

用 Cauchy 不等式即

$$({\int }_a^{b}f(x)g(x)dx{)}^2\le {\int }_a^b{f}^2(x)dx\cdot {\int }_a^b{g}^2(x)dx$$

放缩, 接着用分部积分证明不等式

四

给出 $f(x,y)$ 和 $v$ 和 $x_0$, 求 $f^{\prime }(x_0)(v)$ 要用矩阵, 转为梯度算然后 8 分没了

五

一致收敛和一致 Cauchy 等价证明, 改为了 $sup|f_n-f_m|$, 本质一样

7.8 Theorem

The sequence of functions $\{f_n\}$, defined on $E$, converges uniformly on $E$ if and only if for every $\epsilon >0$ there exists an integer $N$ such that $m\ge N$, $n\ge N,x\in E$ implies

$$|f_n(x)-f_m(x)|\le \epsilon$$

mark as euqation (13)

Proof

Suppose $\{f_n\}$ converges uniformly on $E$, and let $f$ be the limit function. Then there is an integer $N$ such that $n\ge N,x\in E$ implies

$$|f_n(x)-f(x)|\le \frac{\epsilon }{2},$$

so that

$$|f_n(x)-f_m(x)|\le |f_n(x)-f(x)|+|f(x)-f_m(x)|\le \epsilon$$

if $n\ge N,m\ge N,x\in E$. Conversely, suppose the Cauchy condition holds.

Then, the sequence $\{f_n(x)\}$ converges, for every $x$, to a limit which we may call $f(x)$. Thus the sequence $\{f_n\}$ converges on $E$, to $f$. We have to prove that the convergence is uniform

Let $\epsilon >0$ be given, and choose $N$ such that (13) holds. Fix $n$, and let $m\to \mathrm{\infty }$ in (13) . Since $f_m(x)\to f(x)$ as $m\to \mathrm{\infty }$, this gives

$$|f_n(x)-f(x)|\le \epsilon$$

mark as euqation (14)

for every $n\ge N$ and every $x\in E$, which completes the proof

六

证明, 题干条件改为f连续可微, 集合凸有界闭, 删去范数有界

9.19 Theorem

Suppose $f$ maps a convex open set $E\subset R^n$ into $R^m$ , $f$ is differentiable in $E$ , and there is a real number $M$ such that

$$\parallel {\mathbf{f}}^{\prime }(\mathbf{x})\parallel \le M$$

for every $\mathbf{x}\in E$ .Then

$$|\mathbf{f}(\mathbf{b})-\mathbf{f}(\mathbf{a})|\le M|\mathbf{b}-\mathbf{a}|\text{ for all }\mathbf{a}\in E,\mathbf{b}\in E\text{. }$$

Proof

Fix $\mathbf{a}\in E,\mathbf{b}\in E$ .Define

$$\gamma (t)=(1-t)\mathbf{a}+t\mathbf{b}$$

for all $t\in R^1$ such that $\gamma (t)\in E$ .Since $E$ is convex, $\gamma (t)\in E$ if $0\le t\le 1$

Put

$$\mathbf{g}(t)=\mathbf{f}(\gamma (t))$$

Then

$${\mathbf{g}}^{\prime }(t)={\mathbf{f}}^{\prime }(\gamma (t)){\gamma }^{\prime }(t)={\mathbf{f}}^{\prime }(\gamma (t))(\mathbf{b}-\mathbf{a}),$$

so that

$$|{\mathbf{g}}^{\prime }(t)|\le \parallel {\mathbf{f}}^{\prime }(\gamma (t))\parallel |\mathbf{b}-\mathbf{a}|\le M|\mathbf{b}-\mathbf{a}|$$

for all $t\in [0,1]$ .By Theorem 5.19,

$$|\mathbf{g}(1)-\mathbf{g}(0)|\le M|\mathbf{b}-\mathbf{a}|$$

But $\mathbf{g}(0)=\mathbf{f}(\mathbf{a})$ and $\mathbf{g}(1)=\mathbf{f}(\mathbf{b})$ .This completes the proof

七

不动点定理证明 题干改为 $d(Tx,Ty)<d(x,y)$

反证法, 用课件方法证明, 然后11分没了, 可自行查百度

9.23 Theorem

If $X$ is a complete metric space, and if $\phi$ is a contraction of $X$ into $X$ , then there exists one and only one $x\in X$ such that $\phi (x)=x$

Proof

Pick $x_0\in X$ arbitrarily, and define $\{x_n\}$ recursively, by setting

$$x_{n+1}=\phi (x_n)(n=0,1,2,\dots )$$

Choose $c<1$ so that (43) holds.For $n\ge 1$ we then have

(43) 推测为上文某一条等式, 这里没引用

$$d(x_{n+1},x_n)=d(\phi (x_n),\phi (x_{n-1}))\le cd(x_n,x_{n-1})$$

Hence induction gives

$$d(x_{n+1},x_n)\le c^{n}d(x_1,x_0)(n=0,1,2,\dots )$$

If $n<m$ , it follows that

$$\begin{array}{rl}d(x_n,x_m)& \le \sum _{i=n+1}^{m}d(x_i,x_{i-1})\\ & \le (c^n+c^{n+1}+\cdots +c^{m-1})d(x_1,x_0)\\ & \le [(1-c{)}^{-1}d(x_1,x_0)]c^n\end{array}$$

Thus $\{x_n\}$ is a Cauchy sequence.Since $X$ is complete, $lim{x}_n=x$ for some $x\in X$

Since $\phi$ is a contraction, $\phi$ is continuous(in fact, uniformly continuous) on $X$ .Hence

$$\phi (x)=\underset{n\to \mathrm{\infty }}{lim}\phi (x_n)=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=x$$