问题2012061001
if $f(x)$ is continuous at $x = a$ and $f(a) > 0$, show that the domain of $f$ contains an open interval about $a$ where $f(x) > 0$.
来源
Introduction to Calculus and Analysis
分析
这个说明了微分的局部性质, 从一点的性质，能够得到一个邻域的性质.
解答
$f(x)$在点$a$连续, 于是$\forall \epsilon > 0$, 存在$\delta > 0$, 使得$|x – a| < \delta$时, $|f(x) – f(a)| < \epsilon$. 特别的对于$\epsilon = \frac{f(a)}{2} > 0$, 存在这样的$\delta_0$, 使得
$$\begin{aligned}|f(x) – f(a)| < \frac{f(a)}{2} \\\frac{f(a)}{2} < f(x) < \frac{3f(a)}{2} \\f(x) > \frac{f(a)}{2} > 0\end{aligned}$$
问题2012061002
In the definition of continuity show that the centered intervals
$$|f(x) – f(x_0)| < \epsilon \quad \text{and} \quad |x – x_0| < \delta$$
may be replaced by an arbitrary open interval containing $f(x_0)$ and a sufficiently small open interval containing $x_0$, as indicated on p.33.
来源
Introduction to Calculus and Analysis
分析
这道题目中,P.33是指英文版的书本中的脚注中提到的. 也就是需要证明两者是等价的.
解答
必要性
对于任意包含$f(x_0)$的开区间$B$, 存在一个$\epsilon$, 使得$B \supset \{ f(x) | |f(x) – f(x_0)| < \epsilon \}$, 对于这个$\epsilon$, 存在$\delta > 0$, 使得$|x – x_0| < \delta$时, $|f(x) – f(a)| < \epsilon$, 取$A = (a – \delta, a+\delta)$, 则
$$f(A) \subset \{ f(x) | |f(x) – f(x_0)| < \epsilon \} \subset B$$
结论成立.充分性.
也就是说, 对于任意的包含$f(x_0)$的开区间$B$, 总是存在包含$x_0$的开区间$A=(\alpha, \beta)$, 使得$\forall x \in A$, $f(x) \in B$, 或者说$f(A) \subset B$. $\forall \epsilon > 0$, 取$B = \{f(x) | |f(x) – f(x_0)| < \epsilon\}$, 则存在开区间$A = (\alpha, \beta)$, 使得$f(A) \subset B$, 令$\delta = \min\{x_0 – \alpha, \beta – x_0\}$, 则$|x – x_0| < \delta$时, $x \in A$, 于是$f(x) \in B$, $f(x)$连续.
问题2012061003
Let $f(x)$ be continuous for $0 \le x \le 1$,. Suppose further that $f(x)$ assumes rational values only and that $f(x) = \frac{1}{2}$ when $x = \frac{1}{2}$. Prove that $f(x) = \frac{1}{2}$ everywhere.
来源
Introduction to Calculus and Analysis
分析
这里需要用到连续函数的介质性质和有理数的稠密性, 注意, 有介质性质的不一定是连续的, 函数的导函数拥有介质性.
解答
反证法, 假设存在$x_0$, $f(x_0) \neq \frac{1}{2}$, 不妨设$x_0 < \frac{1}{2}$, 令
$$\begin{aligned}m &= \min\{f(x_0), \frac{1}{2}\} \\ M &= \max\{f(x_0), \frac{1}{2}\} \end{aligned}$$
那么在$(m, M)$上存在无理数$c$, 由于连续性, 存在$x’ \in (x_0, \frac{1}{2})$, 使得$f(x’) = c$, 这与题设$f(x)$的值都是有理数矛盾.
问题2012061004
(a) Let $f(x)$ be defined for all values of $x$ in the following manner:
$$f(x) = \left\{ \begin{aligned} &0, \quad x \quad \text{irrational} \\ &1, \quad x \quad \text{rational} \end{aligned} \right.$$
Prove that $f(x)$ is everywhere discontinuous.(b) On the other hand, consider
$$g(x) = \left\{ \begin{aligned} &0, \quad x \quad \text{irrational} \\ & \frac{1}{q}, \quad x = \frac{p}{q} \quad \text{rational in lowest terms}. \end{aligned} \right.$$
(The rational number $p/q$ is said to be in lowest terms if the integers $p$ and $q$ have no common factor larger than $1$, and $q > 0$. Thus $f(16 / 29) = 1/29$.) Prove that $g(x)$ is continuous for all irrational values and discontinuous for all rational values.
来源
Introduction to Calculus and Analysis
分析
间断点的分类如下(这里设$x_0$是间断点):
(1) 若$f(x_0+0)$与$f(x_0-0)$都存在, 则称$x_0$为$f(x)$的第一类间断点, 此时, 若$$f(x_0+0) = f(x_0-0) \neq f(x_0),$$ 则称此间断点为可去间断点; 否则称其为跳跃间断点.
(2) 若$f(x_0+0)$与$f(x_0-0)$至少有一个不存在时, 则称$x_0$为$f(x)$的第二类间断点.
解答
(a) 对于任意$x_0$, 选择$\epsilon = \frac{1}{2}$, 对于任意的$\delta > 0$, 满足$|x – x_0| < \delta$的有理数和无理数都是无穷多个, 于是存在$x_1$为有理数, $x_2$为无理数在$x_0$的$\delta$邻域中, 此时$|f(x_1) – f(x_2)| = 1 > \epsilon$. 因此$f(x)$在$x_0$不收敛, 自然不可能连续. 它的左右极限都不存在, 属于第二类间断点.
(b)首先对于有理数$x_0 = \frac{p}{q}$, $q > 0$来说, 取$\epsilon = \frac{1}{2q} > 0$, 对于任意的$\delta >0 $, 有无数个无理数满足$|x – x_0| < \delta$, 即存在$x_1$为无理数在$x_0$的$\delta$邻域中, 此时
$$|f(x_1) – f(x_0)| = \frac{1}{q} > \epsilon.$$
也就是说$\lim\limits_{x \rightarrow x_0}{f(x)} \neq f(x_0)$.
对于无理数$x_0$来说, 对于任意的$\epsilon > 0$, 存在自然数$N > 0$使得, $\epsilon > \frac{1}{N}$, 在区间$(x_0 – 1, x_0 + 1)$上, 只有有限个数$\frac{p}{q}$, 满足$0 < q \le N$, 不妨设这些数为$a_1$, $\cdots$, $a_m$, 令$$\delta = \min\{1, |a_1 – x_0|, \cdots, |a_m – x_0| \},$$ 于是当$|x – x_0| < \delta$时, 所有的有理数$\frac{p}{q}$, $q > N$,
$$|f(x) – f(x_0)| < |f(x)| < \frac{1}{q} < \frac{1}{N} < \epsilon.$$
也就是说$\lim\limits_{x \rightarrow x_0}{f(x)} = f(x_0)$.
这里面有理数点作为间断点, 同样是第二类间断点.
今天的题目涉及连续性的概念.