108學年度｜微積分模組班｜Class 01｜Homework 3.

Homework 3.

• Date: 108/10/03 (Week 04)
• Place: 新生教學館 301

Problem 1.

• Let $f(x)$ be a strictly increasing continuous function on $[a, b]$. That is, for any $x_1 < x_2 \in [a, b]$ we have $f(x_1) < f(x_2)$.

1. Show that $f(b)$ is the maximum and $f(a)$ is the minimum.
2. Show that for any $y_0 \in [f(a), f(b)]$, there exists $x_0$ such that $f(x_0) = y_0$.
3. Show that the inverse function $f^{−1}$ is defined everywhere on $[f(a), f(b)]$.
4. Show that $f^{-1}$ is continuous on $[f(a), f(b)]$.

Solution.

1. Since $x < b$ for all $x \in [a,b)$, we have $f(x) < f(b)$ for all $x \in [a,b)$, so $f(b)$ is the maximum, similarly, $f(a)$ is the minimum.
2. Since $y_0 \in [f(a),f(b)]$ and $f(x)$ is continuous, by Intermediate value theorem, there exist $x_0 \in [a,b]$ such that $f(x_0) = y_0$.
3. The previous problem gives that $f \colon [a,b] \to [f(a),f(b)]$ is surjective, and it is injective by the property strictly increasing (Why?). Hence, the function $f \colon [a,b] \to [f(a),f(b)]$ is bijective and there exist an inverse function $f^{-1} \colon [f(a),f(b)] \to [a,b]$.
4. Pick $y_0 \in (f(a),f(b))$, suppose $f(x_0)=y_0$ with $a < x_0 < b$, pick $$0< \epsilon < \min\{ x_0-a, b-x_0 \}$$ (The restriction of $\epsilon$ is only to ensure that such $\epsilon$ doesn't give a interval larger than $(a,b)$, it actually doesn't matter if we have a larger $\epsilon >0$) For this $\epsilon$, we have: $$a<x_0-\epsilon<x_0<x_0+\epsilon<b$$, and since $f(x)$ is strictly increasing, we have: $$f(a)<f(x_0-\epsilon)<f(x_0)<f(x_0+\epsilon)<f(b)$$ Let $\delta= \min \{f(x_0)-f(x_0-\epsilon), f(x_0+\epsilon)-f(x_0)\}$. If $|y-y_0|=|y-f(x_0)|<\delta$, we have : \begin{align*} f(x_0-\epsilon)<y<f(x_0+\epsilon) \overset{(*)}{\Longleftrightarrow} & g(f(x_0-\epsilon))<g(y)<g(f(x_0+\epsilon)) \\ \Longleftrightarrow & x_0-\epsilon<g(y)<x_0+\epsilon \end{align*}, $(*)$ holds since $g(y)$ is also strictly increasing. (Why?)

Problem 2.

• Suppose that there are sequences $\{a_n\}_{n = 1}^{\infty}$, $\{b_n\}_{n = 1}^\infty$ with the following properties:
1. $\{a_n\}_{n = 1}^{\infty}$ is increasing.
2. $\{b_n\}_{n = 1}^{\infty}$ is decreasing.
3. $a_n \leqslant b_n \leqslant a_n + \dfrac{1}{2^n}$.
Show that both sequences converge with the same limit

Solution.

• Note that $a_n \leqslant b_1$ for all $n \in \mathbb{N}$, hence $\{a_n\}_{n = 1}^{\infty}$ is a bounded increasing sequence, where by Monotone convergence theorem, $\displaystyle \lim_{n \to \infty}a_n$ exist. Similarly, we have $b_n \geqslant a_1$ for all $n \in \mathbb{N}$, again by Monotone convergence theorem, $\displaystyle \lim_{n \to \infty}b_n$ exist. Moreover, by the inequality $a_n \leqslant b_n \leqslant a_n + \dfrac{1}{2^n}$ and squeeze theorem, the two limit coincides.

Problem 3.

• Given $f_n(x) = x^n$ on $[0, 1]$ for any $n \geqslant 1$, we can define a function $f(x)$ on $[0, 1]$ realizing ”limit of sequence of functions $\{f_n\}_{n = 1}^\infty$". It is defined as $\displaystyle f(a) := \lim_{n \to \infty} a^n$ for any $a \in [0, 1]$. Show that $f(x)$ is not continuous everywhere on $[0, 1]$. (This is an example that ”a limit of continuous functions is not necessarily continuous”).

Solution.

• For $a \in [0,1)$, we have $\displaystyle \lim_{n \to \infty}a^n = 0$. (Verify yourself!) As for $a = 1$, we have $\displaystyle \lim_{n \to \infty}a^n = 1$, so the "limit of sequence of functions $\{f_n\}_{n = 1}^\infty$" is: $$f(x) = \begin{cases} 0 &, x \in [0,1)\\ 1 &, x=1\end{cases}$$ Which is obviously not continuous at $x = 1$.

Pointwise convergence of function:$f(x)$  $\displaystyle = \lim_{n \to \infty} $$f_n(x)$